TypeScript definite assignment assertions

TypeScript never stops improving, although most changes over the past year have been “non syntactical” – i.e. there have been a huge swathe of improvements to how types are handled, and a large slice of improvements to make the tooling even better. It has a been a while, though, since we got a new character to decorate our code. The wait is over, thanks to the TypeScript Definite Assignment Assertion. Let’s take a look at it with a short example.

Warning Triangle

No definite assignment

The new feature is related to the following improved compile-time check. In the example below, I forgot to assign a value to the wordsPerMinute property. This can happen when you forget to add a default value, or when you forget to initialize it in the constructor, or (as below) when you forget to map a parameter to the property (remember, you don’t need to manually map constructor parameters !).

Whatever the reason, if you compile using the strict flag (I keep telling you to use it), you’ll get the following error, known as a definite assignment error because there is no definite assignment:

app.ts(2,13): error TS2564: Property ‘wordsPerMinute’ has no initializer and is not definitely assigned in the constructor.

Fixing, and definite assignment assertions

The correct fix is probably to assign this.wordsPerMinute = wordsPerMinute in the constructor – but in some cases, you may be doing something funky where the dependency will be resolved in a way the compiler is unable to determine.

When you need to allow a property with no definite assignment, you can use the definite assignment assertion . This is a very grand name for adding a bang (!) to the property name.

This will only work in TypeScript 2.7 and newer.

On the whole, unless you have an iron-clad reason to use it – you’ll probably want to avoid the definite assignment assertion. In most cases, the real value of this feature lies in the part that detects unassigned properties.

Warning Triangle, Public Domain. Wikipedia .

Steve Fenton

Steve Fenton is an Octonaut at Octopus Deploy and six-time Microsoft MVP for developer technologies. He’s a Software Punk and writer.

Categories:

  • Programming

Three Ways to Improve Software Development

Testing NPM publish with a dry run

A drawer full of junk

Code organisation and junk

Definite Assignment Assertions (!)

The Definite Assignment Assertions or also called non-null assertion operator tells the TypeScript compiler that a value typed cannot be null or undefined which is a way to override the compiler's analysis and inform it that a variable will be assigned a value before it is used.

blog-title#highlight"> Using The "Definite Assignment Assertion" To Define Required Input Bindings In Angular 7.1.1

In an Angular application, Directives represent a special type of Class. In a sense, they are a standard JavaScript Class with properties and methods. But, unlike a Class that gets created and consumed explicitly in your own logic, Directive classes are created, consumed, and managed implicitly by the Angular renderer. As such, properties that might otherwise be required as a constructor argument are, instead, provided through template input bindings. Which begs the question: how do you define a "required" property / input binding that is not actually available at the time of instantiation? As explained in the strictPropertyInitialization flag clean-up issue on GitHub , people on the Angular team have started to use TypeScript's "Definite Assignment Assertion" to denote required Class properties as "defined" even when they won't truly be defined until after the Directive's class constructor has been executed. I had never seen the "Definite Assignment Assertion" before, so I wanted to try it out for myself in a Component Directive in Angular 7.1.1.

Run this demo in my JavaScript Demos project on GitHub .

View this code in my JavaScript Demos project on GitHub .

To explore this concept, I created a "Badge" component that accepts a [user] input binding and renders the badge template with name, email, and avatar. The whole reason-to-be for this component is to render the passed-in User object. As such, this component makes no sense without the [user] input. This input binding is therefore required in order for this component to work. Without the user property, this Badge component would be in an "invalid state".

Before learning about the "Definite Assignment Assertion" notation in TypeScript, I might have tried to approach this Component definition by making the "user" property optional (meaning, for example, that it can also be set to null):

In this case, in the BadgeComponent constructor, I have to set the user property to null otherwise TypeScript complains that I didn't fully initialize the class property (thanks to the "strictPropertyInitialization" tsconfig setting). Of course, in the ngOnInit() and ngOnChanges() event-handlers, I am attempting to assert that the value must be defined (as a required input binding). But, when I go to compile this code, I get the following TypeScript error:

ERROR in app/badge.component.ts.BadgeComponent.html(4,10): : Object is possibly 'null'.

As you can see, TypeScript and Angular are complaining that I am attempting to use a potentially null value in my template expressions. Now, all of the logic for this component happens to be in the template; but, if my Class made additional references to the "user" property within its methods, similar errors would be thrown by the TypeScript compiler.

To fix this workflow, Angular internals have started to use the "Definite Assignment Assertion". This is a property name followed by an exclamation point. For example:

public user! : User;

This assertion tells TypeScript that the "user" property (aka, our required input binding) may look like it's not being fully initialized; but, that TypeScript should trust that the application is going to define this value before it is consumed by the given Class. In other words, we are telling TypeScript to treat this as a required property, but to not validate its initialization.

Of course, TypeScript can only make assertions about compile-time truths - it doesn't know anything about the runtime. As such, this won't prevent runtime errors if the "user" property is undefined; but, this assertion will better express the intended use of the Component Class.

Let's take a look at what the BadgeComponent looks like when we add the "Definite Assignment Assertion" notation:

As you can see, we are denoting the "user" property as "defined". Which allows the code to compile even though we've commented-out the constructor initialization. In this case, I'm still using the ngOnInit() and ngOnChanges() life-cycle methods to validate the input binding. But these are here to better report runtime errors and express the intent of the class. After all, if the input binding is missing, something is going to break.

To see this "Definite Assignment Assertion" in action, I've created an App component that attempts to consume the BadgeComponent both with and without a provided [user] input binding:

As you can see, the last of the "bn-badge" elements is omitting the [user] input binding. And, when we run this code in the browser, we get the following output:

The Definite Assignment Assertion in TypeScript makes it easy to denote Angular Directive input bindings as required without having to initialize them.

Clearly, if the [user] input binding is required but omitted from the calling code, the page will still break. After all, the BadgeComponent is in an invalid state. But, the BadgeComponent itself expresses clear intent as to how the input binging is supposed to be used.

When you author an Angular Component, some input bindings are optional and some are required. With optional input bindings, you can provide default or fallback values in order to make the class properties easier to work with. But, when it comes to required input bindings, there is no default or fallback. As such, TypeScript will complain that the required properties aren't properly initialized. Thankfully, the "Definite Assignment Assertion" allows us to tell TypeScript to relax the property validation and trust that the Angular application will define these values before they are used.

Want to use code from this post? Check out the license .

Reader Comments

add definite assignment assertion

Hello, Nice Article. I am actually trying to utilize the definite assertion in my angular project. I have a model class of my object, which is your interface by the way. But Im trying to model the db. So i have a class Rating{rating:number;} And an export class Review which has rating!:Rating; In my ts class, i did review.rating.rating = 0, but it is seeing review.rating as undefined.

Please can you help me

add definite assignment assertion

So, the Definite Assignment Assertion won't actually influence the runtime value of the property - it will only prevent the TypeScript compiler from complaining that the property isn't being initialized. The point of the Definite Assignment Assertion is handle cases in which a property is being set outside of the current context (such as with an input-binding), and we are assuring the compiler that this value will be set by the time it is consumed.

Based on what you are saying, it sounds like you are attempting to reference reviewer.rating before you actually set it.

I believe in love . I believe in compassion . I believe in human rights . I believe that we can afford to give more of these gifts to the world around us because it costs us nothing to be decent and kind and understanding. And, I want you to know that when you land on this site, you are accepted for who you are , no matter how you identify, what truths you live, or whatever kind of goofy shit makes you feel alive! Rock on with your bad self!

Get 42% off the TypeScript bundle

See the bundle then add to cart and your discount is applied.

Write TypeScript like a pro.

Follow the ultimate TypeScript roadmap.

Ultimate Courses

Strict Property Initialization in TypeScript

Todd Motto

by Todd Motto

Feb 28, 2023

2 mins read

Learn TypeScript the right way.

The most complete guide to learning TypeScript ever built. Trusted by 82,951 students .

Todd Motto

with Todd Motto

In later versions of TypeScript there’s the concept of a Strict Class Property Initializer, or Definite Assignment Assertion .

This feature all of a sudden produced type errors on previously working code, so let’s explore why it’s here and why you should be using it.

“Property … has no initializer and is not definitely assigned in the constructor.”

Sound familiar? Welcome to strict initialization and the requirement of definite assignment.

So, what it is? It’s a safer way to declare properties and how you expect the data to be made available in your application.

If you see this error, your property is simply being defined with no value. Or at least, no obvious value.

It’s most likely that you’ve seen this error in an Angular context, for example:

So how do we ‘fix it’? Well, there are a few ways. First, we can declare a ! after the property (using definite assignment):

What it means is: “TypeScript, I know there is no data here yet, but I’m confident it will be there soon, just trust me!”.

This is the preferred option, and I’d encourage you to stick with it.

Alternatively, you can supply a default value, and then it’s not needed (but this involves writing and importing code that doesn’t really do anything, I’m not a fan but it works):

Part of the error we see is “not defined in the constructor”, so by also doing that we can ‘fix’ it (but don’t do this):

Why don’t do this? Well, we’ve removed ngOnInit which is a great practice and you should be using it. Angular knows about ngOnInit , whereas JavaScript knows about the constructor . Angular has no control, therefore we should move to a lifecycle hook.

💥 Read my deep-dive on the difference between NgOnInit and the constructor .

You can (at your own peril) disable this feature in your tsconfig.json file as well, and possibly need to disable other strict TypeScript compiler options:

In a non-Angular context, it’s likely you’ll want to structure your data, properties and initializations better to avoid introducing bugs and other errors.

👋 Hello TypeScript dev! If you enjoyed this post, you’ll love my advanced TypeScript courses that I’ve built over years of experience and hard work.

Using the ! definite assignment in Angular is a safe and acceptable solution to writing components, services, directives and more - as demonstrated with OnInit , we can be sure we’re adopting the correct practice.

Related blogs 🚀

Readonly properties in typescript.

In this post you’ll learn how and when to use the readonly keyword for in TypeScript.

Todd Motto

Mar 9, 2023

TypeScript Classes and Constructors

In this post you’ll learn about TypeScript constructors, how to create classes, and how they differ to their ES5 counterpart of traditional constru...

Mar 7, 2023

Abstract Classes in TypeScript

In this post you’ll learn how to create and use an abstract class in TypeScript.

Mar 6, 2023

Public vs Private Properties and Methods in TypeScript

In this post you’ll learn how to create both public and private members in TypeScript classes, which we can use for both properties and methods.

Mar 5, 2023

Static Properties and Methods in TypeScript

In this post you’ll learn about the static keyword in TypeScript classes so you can confidently use them on properties and methods.

Feb 27, 2023

Readonly Mapped Type in TypeScript

In this post you’ll learn how and when to use the Readonly mapped type TypeScript.

Feb 26, 2023

Free eBooks:

Angular Directives In-Depth eBook Cover

You've got mail! Enjoy.

JavaScript Array Methods eBook Cover

Cookies are used to analyze traffic and optimize experience.

A newer version of this site just became available. Please refresh this page to activate it.

Marius Schulz

Strict Property Initialization in TypeScript

TypeScript 2.7 introduced a new compiler option for strict property initialization checks in classes. If the --strictPropertyInitialization flag is enabled, the type checker verifies that each instance property declared in a class either

  • has a type that includes undefined ,
  • has an explicit initializer , or
  • is definitely assigned to in the constructor.

The --strictPropertyInitialization option is part of the family of compiler options that is enabled automatically when the --strict flag is set. As with all the other strict compiler options, you can set --strict to true and selectively opt out of strict property initialization checks by setting --strictPropertyInitialization to false .

Note that the --strictNullChecks flag must be set (either directly or indirectly via --strict ) in order for --strictPropertyInitialization to have any effect.

Alright, let's see strict property initialization checks in action. Without the --strictPropertyInitialization flag enabled, the following code type-checks just fine, but produces a TypeError at runtime:

The reason for the runtime error is that the username property holds the value undefined because there's no assignment to that property. Therefore, the call to the toLowerCase() method fails.

If we enable --strictPropertyInitialization , the type checker raises an error:

Let's look at four different ways we can properly type our User class to make the type error go away.

# Solution #1: Allowing undefined

One way to make the type error go away is to give the username property a type that includes undefined :

Now, it's perfectly valid for the username property to hold the value undefined . Whenever we want to use the username property as a string, though, we first have to make sure that it actually holds a string and not the value undefined , e.g. using typeof :

Alternatively, we can use optional chaining (the ?. operator) to only call the toLowerCase() method if the username property holds a non-nullish value. We can combine that with nullish coalescing (the ?? operator) to provide the fallback value:

# Solution #2: Explicit Property Initializer

Another way to make the type error go away is to add an explicit initializer to the username property. This way, the property holds a string value right away and is not observably undefined :

# Solution #3: Assignment in the Constructor

Perhaps the most useful solution is to add a username parameter to the constructor, which is then assigned to the username property. This way, whenever an instance of the User class is constructed, the caller has to provide the username as an argument:

We could simplify the User class by removing the explicit assignment to the class field and adding the public modifier to the username constructor parameter:

Note that strict property initialization requires each property to be definitely assigned in all possible code paths in the constructor. The following (contrived) example is therefore not type-correct because in some cases, we leave the username property uninitialized:

# Solution #4: Definite Assignment Assertion

If a class property neither has an explicit initializer nor a type including undefined , the type checker requires that property to be initialized directly within the constructor; otherwise, strict property initialization checks will fail. This is problematic if you want to initialize a property within a helper method or have a dependency injection framework initialize it for you. In these cases, you have to add a definite assignment assertion ( ! ) to that property's declaration:

By adding a definite assignment assertion to the username property, we're telling the type checker that it can expect the username property to be initialized, even if it cannot detect that on its own. It is now our responsibility to make sure the property is definitely assigned to after the constructor returns, so we have to careful; otherwise, the username property can be observably undefined and we're back to the TypeError at runtime.

This article and 44 others are part of the TypeScript Evolution series. Have a look!

add definite assignment assertion

Announcing TypeScript 2.7

add definite assignment assertion

Daniel Rosenwasser

January 31st, 2018 0 0

If you’re not familiar with TypeScript, it’s a language that brings optional static types to JavaScript by building on JavaScript itself. Running TypeScript code through its compiler emits clean readable JavaScript that runs on any browser, and can also make bleeding-edge ECMAScript features you write work on older browsers. That means that you can take advantage of the design time tooling and safety of types (like code completion & navigation), while still leveraging the familiarity, community, and ubiquity of JavaScript.

But if you already know what TypeScript is and want to start using 2.7, go ahead and get it on NuGet or download it over npm:

Visual Studio 2015 users (who have Update 3 ) can install TypeScript 2.7 from here , and Visual Studio 2017 users using version 15.2 or later will be able to get TypeScript by simply installing it from here . Sublime Text Plugin users can install the new version from Package Control .

TypeScript 2.7 will be available for Visual Studio Code very soon, but eager users can get it working pretty easily in the meantime .

We’ve got a lot in 2.7. While you can always take a look at the roadmap , we’ve put together a quick list for a bird’s eye view of this release:

Stricter class property checks

Definite assignment assertions, easier ecmascript module interoperability, unique symbol types and const-named properties, cleaner output in --watch mode, prettier --pretty output, numeric separators, fixed length tuples, in operator narrowing and accurate instanceof, smarter object literal inference.

We’ve also put together a breaking changes section towards the end of this post which existing users should be aware of.

So without further ado, let’s see what this release brings!

TypeScript 2.7 introduces a new strictness flag named --strictPropertyInitialization !

This flag makes sure that each instance property of a class gets set in the constructor body, or by a property initializer. In a sense, it brings some of the definite assignment checks from variables to instance properties in classes. For example:

In the above, baz never gets set, and TypeScript reports an error. If we truly meant for baz to potentially be undefined , we should have declared it with the type boolean | undefined .

There are certain scenarios where properties might be initialized indirectly (perhaps by a helper method or dependency injection library). In those cases, you can convince the type system you know better by using the definite assignment assertions for your properties.

Keep in mind that --strictPropertyInitialization will be turned on along with other --strict mode flags, which can impact your project. You can set the strictPropertyInitialization setting to false in your tsconfig.json ‘s compilerOptions , or --strictPropertyInitialization false on the command line to turn off this checking.

As expressive as we try to make the type system, we understand there are certain times you might know better than TypeScript.

As we mentioned above, definite assignment assertions are a new syntax you can use to convince TypeScript that a property will definitely get assigned. But in addition to working on class properties, TypeScript 2.7 also allows you to use this feature on variable declarations!

If we hadn’t added an exclamation point/bang ( ! ) after x , TypeScript would have reported that x was never initialized. This can be handy in deferred-initialization, or re-initialization scenarios.

Before ECMAScript modules were standardized in ES2015, the JavaScript ecosystem had several different module formats that worked in different ways. Once the standard passed, the community was left with a question of how to best interoperate with existing “legacy” module formats.

TypeScript and Babel took different approaches, and even now, there really isn’t a locked down standard. The short story is that if you’ve used Babel, Webpack, or React Native, and expected different import behaviors than you were used to, we have a new compiler option for you called --esModuleInterop .

Babel and Webpack allow users to import these CommonJS modules as default imports, but also provide each property on the namespace import (unless the module was marked with an __esModule flag).

Because TypeScript’s behavior differs, we added the --allowSyntheticDefaultImports flag in TypeScript 1.8 to allow users to get this behavior for type-checking (but not emit).

In general, TypeScript’s view of CommonJS (and AMD) modules is that namespace imports always correspond to the shape of a CommonJS module object, and that a default import just corresponds to a member on that module named default . Under this assumption, you can create a named import

However, ES namespace imports aren’t callable, so this approach doesn’t always make sense.

To give users the same runtime behavior as Babel or Webpack, TypeScript provides a new --esModuleInterop flag when emitting to

legacy module formats.

Under the new --esModuleInterop flag, these callable CommonJS modules must be imported as default imports like so:

We strongly suggest that Node.js users leverage this flag with a module target of commonjs for libraries like express, which export a callable/constructable module.

Webpack users may want to use this as well; however, your code should target esnext modules with a moduleResolution strategy of node . Using esnext modules with --esModuleInterop really only has the effect of turning on --allowSyntheticDefaultImports .

TypeScript 2.7 understands ECMAScript symbols more deeply, allowing you to use them more flexibly.

One highly-demanded use-case is being able to declare well-typed properties with symbols. For an example, take the following:

As you can see, TypeScript can keep track of the fact that x has properties declared using the symbols Foo and Bar since both Foo and Bar were declared as constants. TypeScript leverages this fact and gives both Foo and Bar a new kind of type: unique symbol s.

unique symbol s are subtype of symbol , and are produced only from calling Symbol() or Symbol.for() , or from explicit type annotations. They can only occur on const declarations and readonly static properties, and in order to reference an existing unique symbol type, you’ll have to use the typeof operator. Each reference to a unique symbol implies a completely unique identity that’s tied to a given declaration.

Because each unique symbol has a completely separate identity, no two unique symbol types are assignable or comparable to each other.

Other potential use-cases include using symbols for tagged unions.

TypeScript’s --watch mode now clears the screen after a re-compilation is requested. This can make it much easier to read messages from the current compilation. We’d like to thank both Philipp Kretzschmar and Joshua Goldberg for helping out on this feature!

TypeScript’s --pretty flag can make error messages easier to read and manage. We have two main improvements in this functionality. First, thanks to a pull request from Joshua Goldberg --pretty now uses colors for file names, diagnostic codes, and line numbers. Second, thanks to a separate pull request by Orta Therox , file names and positions are formatted in such a way that common terminals (including the one embedded in Visual Studio Code) can allow a Ctrl+Click, Cmd+Click, Alt+Click, etc. to jump to the appropriate location in your editor

TypeScript 2.7 introduces support for ECMAScript’s numeric separators proposal. This feature allows users to place underscores ( _ ) in between digits to visually distinguish groups of digits ( much like how commas and periods are often used to group numbers ).

These separators are also useful for binary and hexadecimal numbers.

Note that, perhaps counterintuitively, numbers in JavaScript are not well-suited to represent credit card and telephone numbers. Strings will act as better representations in such cases.

In TypeScript 2.6 and earlier, [number, string, string] was considered a subtype of [number, string] . This was motivated by TypeScript’s structural nature; the first and second elements of a [number, string, string] are respectively subtypes of the first and second elements of [number, string] , and the “trailing” string type is assignable to the union of element types from [number, string] . However, after examining real world usage of tuples, we noticed that most situations in which this was permitted was typically undesirable.

Thanks to a pull request from Kiara Grouwstra , tuple types now encode their arity into the type of their respective length property, and tuples of different arities are no longer assignable to each other. This is accomplished by leveraging numeric literal types, which now allow tuples to be distinct from tuples of different arities.

Conceptually, you might consider the type [number, string] to be equivalent to the following declaration of NumStrTuple :

Note that this is a breaking change. If you need to resort to the original behavior in which tuples only enforce a minimum size, you can use a similar declaration that does not explicitly define a length property, falling back to number .

TypeScript 2.7 brings two new changes to type narrowing – the ability to get a more specific type for a value by running certain types of checks called “type guards”.

First, the instanceof operator is now leverages the inheritance chain instead of

relying on structural compatibility, more accurately reflecting whether how instanceof may behave at runtime. This can help avoid certain complex issues when instanceof narrows from structurally similar (but unrelated) types.

Second, thanks to GitHub user IdeaHunter , the in operator now acts as a type guard , narrowing out types that don’t explicitly declare properties of a given name.

There are certain patterns in JavaScript where users will omit properties so that all uses of those properties are effectively undefined .

TypeScript used seek the best common supertype between { value: number } and {} , ending up with {} . While being technically correct, this wasn’t very useful.

Starting with version 2.7, TypeScript now “normalizes” each object literal’s type to account for every property, inserting an optional property of type undefined on each object type, and unioning them together.

From the above example, the new type of foo would be { value: number } | { value?: undefined } . Combined with the ways TypeScript can narrow, this lets us write expressive code that TypeScript can still understand. As another example, take the following:

Here, TypeScript is able to narrow the type of bar based on checking its b property, allowing us to access the bData property.

Breaking Changes

This release brings some minor breaking changes:

  • Tuples now have fixed numeric length properties.
  • instanceof and in now have slightly different narrowing behavior.
  • Inferences from generic signatures now use base constraint types of type parameters instead of any .
  • The setSelectionRange API now only accepts "forward" | "backward" | "none" .
  • allowSyntheticDefaultImports no longer synthesizes default imports from TypeScript implementation files (i.e. .ts and .tsx ).

Additionally, as mentioned above, users with the --strict setting on will automatically be opted in to --strictPropertyInitialization which errors on properties which are not directly initialized on their declarations or in constructor bodies. While easy to opt out of by explicitly turning this check off, your code may be impacted.

You can get a detailed look from our list of breaking changes issues for TypeScript 2.7 on GitHub, or keep track of general breaking changes on our Breaking Changes wiki page .

What’s next?

We always try to bring more joy to the TypeScript experience. We hope this release continues the tradition of making you more productive and expressive to bring that joy to the core experience.

Now that 2.7 is out, we’ve got some fantastic things in mind for TypeScript 2.8, including conditional types ! While plans are still currently in flux, you can keep a close watch on our roadmap to get an idea of what’s on the TypeScript horizon.

Let us know what you think of this release over on Twitter or in the comments below, and feel free to report issues and suggestions filing a GitHub issue .

Happy Hacking!

add definite assignment assertion

Daniel Rosenwasser Senior Program Manager, TypeScript

add definite assignment assertion

Comments are closed. Login to edit/delete your existing comments

light-theme-icon

Intermediate TypeScript, v2

Intermediate TypeScript, v2 Definite Assignment Assertion

Mike North

Mike discusses the definite assignment operator and explains how to use the exclamation mark (!) to assert that a variable will be assigned a value before it is used. A question about using the declare keyword instead of the exclamation mark is also covered in this segment.

Transcript from the "Definite Assignment Assertion" Lesson

[00:00:00] >> The next thing we're gonna talk about is definite assignment assertion. It also involves an exclamation mark and to do this part, this exercise, we're going to need to go into our tsconfig for this notes project and we're gonna turn on strictPropertyInitialization and set it to true. You should see some errors pop up on the left, and we do, perfect.

[00:00:26] So make sure you turn that on, we're gonna talk about what that is, and how we deal with it. So it's strictPropertyInitialization. So I have a class here and it's called ThingWithAsyncSetup and I've done something that's a little odd here, I have something async inside of a constructor and some might argue I'm already doing something a bit weird.

[00:00:53] But I might pushback against that and say, well look, I'm kinda just kicking off the promise. I'm doing nothing in the constructor that depends on the promise being resolved, and it looks like I have some state I can have in place where I can from the outside observe whether whatever this setup promise does I can observe that it is successfully, has successfully completed, right?

[00:01:21] The dot then will be called and then I'll see this flip. And I have here isSetup as a class field set to a boolean but I'm getting an error, and the error is that is set up has no initializer, and it's not definitely assigned in the constructor. So if I were to take this one line here and move it up into the constructor, the error goes away, and that's because TypeScript analyzes the constructor, it looks at all the class fields that don't have an initializer which would look like this, right?

[00:01:54] I mean, you might argue in this scenario, I should do this, but this is a good example here. The problem here is Typescript saying, I don't know, it doesn't seem like, at least in the general case isSetup, like by the time we return an instance, is that actually gonna be a Boolean?

[00:02:16] It seems like it might be undefined. And why do we think that it's saying that? Does anyone have any hypotheses? Any guesses? I mean, I'm setting it in here. >> Is that run after the constructor? >> Typescript sees a callback here, and it's like, callbacks. I don't know when promise invokes this callback, I don't know if, by the time this promise function, the constructor for promise, by the time it returns, how do I know that this callback has been completed?

[00:02:58] It doesn't know effectively that once we advance beyond line 49, whether isSetup will be set, and in the general case, that is a perfectly valid thing to be concerned about, right? You could get an instance of thing with AsyncSetup, and it claims to have a boolean value for setup, but it could be undefined, except this callback here, we would call this like the promise executor, right?

[00:03:29] It's the thing you use to, it's the function that represents sort of the work or the beginning of the work that could be async, right? This is invoked synchronously within the promise constructor. So I know that by the time we assign the promise to setupPromise, this function's already done, it has been completed.

[00:03:55] I know, definitely, that isSetup, this line of code on line 50, that will have been run, I know for a fact. Now, I also know that TypeScript, there's no way to articulate that in TypeScript and to say, this is a synchronous function, you should assume that this is completed before, like chill out TypeScript, this has been run, there's no way to say that.

[00:04:19] But I can say, look, I know about this particular callback, and I'm gonna say, isSetup is definitely going to have been assigned by the time the constructor completes. Now what I've given you here, it's actually a very, this is a true example, and it does actually guarantee that by the time you get an instance of ThingWithAsyncSetup, it's gonna have a boolean value there, that is true.

[00:04:53] Sometimes you might know something else, like you might know that look, this object is gonna be created, but no one's gonna be working with it or using it until this thing is done to it. Maybe you're initializing objects and you're collecting them, and then after they're already initialized, you set some stuff up on them, and you want downstream consumers to not have to deal with the possibility that a bunch of things could be undefined.

[00:05:23] And so you might blur the lines here and say, okay, technically TypeScript, you're correct, these fields might not have been initialized, but I'll take care of initializing them myself from the outside. And the thing I wanna be sure of is that whoever is working with isSetup, they don't have to deal with this type here, they don't have to deal with the truly honest state of where it is after instantiation.

[00:05:51] You might wanna say, look, right after I create it, I'm doing this thing to it, and I'm always doing that, and I'll take responsibility for making sure that that's true and that no one gets a hold of these things before. So an example of where you might see this is like framework level code, where objects are created, not by invoking the constructor, but by some other process like they're created as a side effect of doing something.

[00:06:17] And then you can be assured that, whenever we go through this pathway where these things are created. I'm grabbing that instance and I'm doing a bunch of stuff to it, and then when I hand it off, I want it to be represented as if the stuff was always there.

[00:06:35] You're taking on risk there, but sometimes it's a valid pattern. Sometimes it's preferable over modeling an interface, or a class that represents, the partially set up thing, and then you have another class that you have to deserialize and serialize into so you get the fully set up thing, it can get complicated if you try to stick very rigidly to the rules.

[00:07:00] Those are two use cases for definitely assignment operator. >> Is it a good practice to use the non-knowledge assertion operator in production code? >> Non-null assertion operator. >> The exclamation point. >> The exclamation point, so what I said before, and I'll stick to it here. I use this all the time in my tests.

[00:07:23] I try to not use it in applications, in libraries, in any code that's actually running and the reason is the mode of failure is usually unacceptable to me, it'll just throw your codes interrupted, and it'll fail however, it's gonna fail. And I usually like to, like I would say, if you want this equivalent thing in production, what you should really do is this.

[00:08:03] Do this. And then write something explicit, like how to handle, the case where it's not there, just use a type card. In tests, this will get very messy cuz I might, on every single line, I get a big JSON object like back that has tons of properties and things on it.

[00:08:27] My test would be three quarters, like by line count. Type guards and safely finding if things are there and throwing very specific errors, it's not just that the outer object wasn't there, but the inner object too and the inner inner inner object. And that's where using this non-null assertion operator, it's valuable cuz all I want is to throw the, as I try to probe into this object, drill into this object, and this is a very concise way of doing that, it's easier to maintain, your test remains a lot more readable, but I don't use this.

[00:09:10] I'm not gonna speak in absolutes here, it would have to be exceptional case for me to use this in production code, not even a production code, any code that's outside of the testing sphere. Cuz you can always do this and this is much more graceful error handling, this is okay when throwing is the desired outcome, that is an absolute statement I'll make because that's exactly what I'm near it'll throw >> Is there a difference between [COUGH] using the exclamation point on 53 and using the declare keyword?

[00:09:55] >> Like this? That's interesting. I wouldn't necessarily do this. I'm actually surprised that this works. I use declare mostly in ambient type information, or when I'm making a mod, like a type declaration, which we'll see later, where you can do that, you can augment types in other modules or in the global scope, but I generally don't use this.

[00:10:30] It does appear to work, it appears to work. I would be kind of, if I saw this in a poll request, I would comment on it and say, you're asking people to understand that you can do this and there's an equally effective way to do this and it's much more common and people will know it when they come across it.

[00:10:53] But that's interesting, you taught me something here. I guess we're saying what we're doing there with declare, if I had to guess, I don't know this for a fact, but I'll explain how my mental model is processing that. When we say declare, we might be stating that, this is a boolean, it's of type boolean, and it's sort of a different thing then checking for definite assignment, right?

[00:11:24] You're really, yeah, that's the best way I can articulate it. I'd recommend using this, I don't think I've ever seen the use of declare before field unless it's in a declaration file, like a DTS file. >> I've ended up having to use it like in a number of projects where I'm injecting, setting that to a bad service, using that decorator.

[00:11:50] And it's probably cause like my TS config wasn't set up right or something, but it would complain about the exclamation point there but it wouldn't complain about the declares. >> I see like a collision of syntax. Yeah, I would use declare there given that it works, but man, is that uncommon and please file a bug report because I think that's odd.

[00:12:13] It's odd to the point where I'm not sure it will continue to work that way, it's that unusual. It'll work in declaration files, all of definitelytyped like that's declared everywhere but in this space, I've never seen that, interesting. I believe you though, there are all sorts of typescript syntax things that sort of collide with other things like this, this is a when you mix this with JSX, it's quite interesting here, this is why we don't write arrays like, This is popular for a long time, but this makes it complicated in JSX, where the parsers get overwhelmed with the same kind of syntax, meaning different things in different spaces.

Learn TypeScript 3 by Building Web Applications by Sebastien Dubois, Alexis Georges

Get full access to Learn TypeScript 3 by Building Web Applications and 60K+ other titles, with a free 10-day trial of O'Reilly.

There are also live events, courses curated by job role, and more.

Late binding and TypeScript definite assignment assertions

If you take a close look at the code for the previous UserSettings example component, you may notice a weird exclamation mark:

Now, what is it and why did we need it?

As the official TypeScript documentation ( https://www.typescriptlang.org/docs/handbook/release-notes/typescript-2-7.html#definite-assignment-assertions ) states, definite assignment assertions can be used to tell the compiler that variables will indeed be assigned, even if TypeScript can't determine this on its own .

The definite assignment assertion is used through the ! character, and is defined after a property or variable declaration.

If we consider ...

Get Learn TypeScript 3 by Building Web Applications now with the O’Reilly learning platform.

O’Reilly members experience books, live events, courses curated by job role, and more from O’Reilly and nearly 200 top publishers.

Don’t leave empty-handed

Get Mark Richards’s Software Architecture Patterns ebook to better understand how to design components—and how they should interact.

It’s yours, free.

Cover of Software Architecture Patterns

Check it out now on O’Reilly

Dive in for free with a 10-day trial of the O’Reilly learning platform—then explore all the other resources our members count on to build skills and solve problems every day.

add definite assignment assertion

The art of simplicity

Search this blog, typescript 2.7–definite assignment assertions.

While looking through some TypeScript 2.7 samples, I noticed the following syntax:

  • has a type that includes undefined ,
  • has an explicit initializer
  • is initialized within the constructor

This is problematic if you want to initialize a property within a helper method or have a dependency injection framework do it for you. To solve this situation you can use a definite assignment assertion using the exclamation mark: (!) . Now the type checker will no longer complain.

Remark: By using a definite assignment assertion it becomes your responsibility to make sure the property gets set.

Popular posts from this blog

Devtoys–a swiss army knife for developers, help i accidently enabled hsts–on localhost, azure devops/ github emoji.

DEV Community

DEV Community

RamaDevsign

Posted on Aug 17, 2022 • Originally published at blog.ramadevsign.com

What Does "!" Symbol Mean in Typescript?

In this short guide, I will introduce you to the Definite Assignment Assertion feature in typescript and demonstrate to you an interesting use case to implement this feature.

Prerequisites 

This guide assumes you have basic knowledge of Typescript , you have a basic understanding of the Stripe API , and you have installed the Typescript error translator extension.

The Problem

environment variables returns an error in typescript environment

The Solution

The solution in this case is to use the "!" symbol which is a feature in Typescript known as the definite assignment assertion. 

As defined in the official Typescript Documentation , the definite assignment assertion feature is used to tell the compiler that the variable assigned is indeed valid for all intents and purposes, even if Typescript's analyses cannot detect so. 

environment variables error fixed with ! symbol (definite assignment assertion)

With the definite assignment assertion, we can assert that indeed in this case we have environment variables ( process.env.STRIPE_SECRET_KEY ) in string format and there is nothing to worry about.

This workaround still leaves us with weaker type safety since any misconfiguration (in our case the environment variables) could lead to errors that cannot be caught at compile time. These errors can only be caught at runtime which beats the logic of having a strict and functional type checking system. With this in mind it's your responsibility to ensure correctness.

It's also good to note that previous Typescript versions used the "!" symbol as a non-null assertion operator. You can read more about this in the Typescript Documentation about Non-null assertion operator .

Thank you for reading this far and happy coding!

Top comments (1)

pic

Templates let you quickly answer FAQs or store snippets for re-use.

jaber-said profile image

The thing that strikes me the most is that you use like the colors I use in vs

Are you sure you want to hide this comment? It will become hidden in your post, but will still be visible via the comment's permalink .

Hide child comments as well

For further actions, you may consider blocking this person and/or reporting abuse

isaacdlyman profile image

Solving the CIA Kryptos Code (Part 1)

Isaac Lyman - Jan 26

lazypro profile image

Generating Avro Schemas from Go types

ChunTing Wu - Jan 15

nhannguyendevjs profile image

Beginner's TypeScript #23

Nhan Nguyen - Jan 15

sre_panchanan profile image

Building a CD Pipeline with GitOps and ArgoCD 🛠️

Panchanan Panigrahi - Feb 6

Once suspended, orama254 will not be able to comment or publish posts until their suspension is removed.

Once unsuspended, orama254 will be able to comment and publish posts again.

Once unpublished, all posts by orama254 will become hidden and only accessible to themselves.

If orama254 is not suspended, they can still re-publish their posts from their dashboard.

Once unpublished, this post will become invisible to the public and only accessible to RamaDevsign.

They can still re-publish the post if they are not suspended.

Thanks for keeping DEV Community safe. Here is what you can do to flag orama254:

orama254 consistently posts content that violates DEV Community's code of conduct because it is harassing, offensive or spammy.

Unflagging orama254 will restore default visibility to their posts.

DEV Community

We're a place where coders share, stay up-to-date and grow their careers.

Property '...' has no initializer and is not definitely assigned in the constructor error fix in Angular

Fixing Property ‘…’ has no initializer and is not definitely assigned in the constructor error

If you are using latest version of Angular application, you might have encountered Property '...' has no initializer and is not definitely assigned in the constructor error, when you declare a variable inside a class or component.

Consider below example.

It’s because of Strict Class Initialization flag introduced in TypeScript 2.7 version.

So If the --strictPropertyInitialization typescript flag is enabled, the compiler checks if each property declared inside a class or angular component

  • has an undefined type
  • has an explicit initializer
  • or checks if it’s assigned in the constructor or not.

If not it throws Property '...' has no initializer and is not definitely assigned in the constructor error.

The flag is enabled by default if --strict flag in enabled in typescript compiler options.

To fix Property '...' has no initializer and is not definitely assigned in the constructor error in Angular applications use the below methods.

  • Disable strictPropertyInitialization flag
  • Adding undefined type to the property
  • Add definite assignment assertion to property
  • Add initializer to property
  • Assignment in the Constructor

Solution 1: Disable strictPropertyInitialization flag

The simple way to fix this error in Angular applications is to disable --strictPropertyInitialization flag in typescript compiler options in tsconfig.json file.

If not there other ways we can by pass this error.

Solution 2: Adding undefined type to the property

It’s ok to have an undefined property.

So while declaring variable add undefined type to the property.

But while using employees property we have to check for undefined value to avoid errors in run time.

Solution 3: Add definite assignment assertion to property

If you know that we will assign the property in later point in time.

It’s better to add definite assignment assertion to the property. i.e., employees.

To add the definite assignment assertion we have to add Exclamation mark(!) symbol after the variable name.

Be careful, we are just bypassing the compiler error, so it’s our responsibility to make sure the property is definitely assigned before using it. otherwise we will get TypeError at run time.

Solution 4: Add initializer to property

Another way to make this type error go away is to add an explicit initializer to the property.

Solution 5: Assignment in the Constructor

Otherwise, we can assign some default value to the property in the constructor.

Arunkumar Gudelli

Was this page helpful?

Background Reading: Classes (MDN)

TypeScript offers full support for the class keyword introduced in ES2015.

As with other JavaScript language features, TypeScript adds type annotations and other syntax to allow you to express relationships between classes and other types.

Class Members

Here’s the most basic class - an empty one:

This class isn’t very useful yet, so let’s start adding some members.

A field declaration creates a public writeable property on a class:

As with other locations, the type annotation is optional, but will be an implicit any if not specified.

Fields can also have initializers ; these will run automatically when the class is instantiated:

Just like with const , let , and var , the initializer of a class property will be used to infer its type:

--strictPropertyInitialization

The strictPropertyInitialization setting controls whether class fields need to be initialized in the constructor.

Note that the field needs to be initialized in the constructor itself . TypeScript does not analyze methods you invoke from the constructor to detect initializations, because a derived class might override those methods and fail to initialize the members.

If you intend to definitely initialize a field through means other than the constructor (for example, maybe an external library is filling in part of your class for you), you can use the definite assignment assertion operator , ! :

Fields may be prefixed with the readonly modifier. This prevents assignments to the field outside of the constructor.

Constructors

Background Reading: Constructor (MDN)

Class constructors are very similar to functions. You can add parameters with type annotations, default values, and overloads:

There are just a few differences between class constructor signatures and function signatures:

  • Constructors can’t have type parameters - these belong on the outer class declaration, which we’ll learn about later
  • Constructors can’t have return type annotations - the class instance type is always what’s returned

Super Calls

Just as in JavaScript, if you have a base class, you’ll need to call super(); in your constructor body before using any this. members:

Forgetting to call super is an easy mistake to make in JavaScript, but TypeScript will tell you when it’s necessary.

Background Reading: Method definitions

A function property on a class is called a method . Methods can use all the same type annotations as functions and constructors:

Other than the standard type annotations, TypeScript doesn’t add anything else new to methods.

Note that inside a method body, it is still mandatory to access fields and other methods via this. . An unqualified name in a method body will always refer to something in the enclosing scope:

Getters / Setters

Classes can also have accessors :

Note that a field-backed get/set pair with no extra logic is very rarely useful in JavaScript. It’s fine to expose public fields if you don’t need to add additional logic during the get/set operations.

TypeScript has some special inference rules for accessors:

  • If get exists but no set , the property is automatically readonly
  • If the type of the setter parameter is not specified, it is inferred from the return type of the getter
  • Getters and setters must have the same Member Visibility

Since TypeScript 4.3 , it is possible to have accessors with different types for getting and setting.

Index Signatures

Classes can declare index signatures; these work the same as Index Signatures for other object types :

Because the index signature type needs to also capture the types of methods, it’s not easy to usefully use these types. Generally it’s better to store indexed data in another place instead of on the class instance itself.

Class Heritage

Like other languages with object-oriented features, classes in JavaScript can inherit from base classes.

implements Clauses

You can use an implements clause to check that a class satisfies a particular interface . An error will be issued if a class fails to correctly implement it:

Classes may also implement multiple interfaces, e.g. class C implements A, B { .

It’s important to understand that an implements clause is only a check that the class can be treated as the interface type. It doesn’t change the type of the class or its methods at all . A common source of error is to assume that an implements clause will change the class type - it doesn’t!

In this example, we perhaps expected that s ’s type would be influenced by the name: string parameter of check . It is not - implements clauses don’t change how the class body is checked or its type inferred.

Similarly, implementing an interface with an optional property doesn’t create that property:

extends Clauses

Background Reading: extends keyword (MDN)

Classes may extend from a base class. A derived class has all the properties and methods of its base class, and can also define additional members.

Overriding Methods

Background Reading: super keyword (MDN)

A derived class can also override a base class field or property. You can use the super. syntax to access base class methods. Note that because JavaScript classes are a simple lookup object, there is no notion of a “super field”.

TypeScript enforces that a derived class is always a subtype of its base class.

For example, here’s a legal way to override a method:

It’s important that a derived class follow its base class contract. Remember that it’s very common (and always legal!) to refer to a derived class instance through a base class reference:

What if Derived didn’t follow Base ’s contract?

If we compiled this code despite the error, this sample would then crash:

Type-only Field Declarations

When target >= ES2022 or useDefineForClassFields is true , class fields are initialized after the parent class constructor completes, overwriting any value set by the parent class. This can be a problem when you only want to re-declare a more accurate type for an inherited field. To handle these cases, you can write declare to indicate to TypeScript that there should be no runtime effect for this field declaration.

Initialization Order

The order that JavaScript classes initialize can be surprising in some cases. Let’s consider this code:

What happened here?

The order of class initialization, as defined by JavaScript, is:

  • The base class fields are initialized
  • The base class constructor runs
  • The derived class fields are initialized
  • The derived class constructor runs

This means that the base class constructor saw its own value for name during its own constructor, because the derived class field initializations hadn’t run yet.

Inheriting Built-in Types

Note: If you don’t plan to inherit from built-in types like Array , Error , Map , etc. or your compilation target is explicitly set to ES6 / ES2015 or above, you may skip this section

In ES2015, constructors which return an object implicitly substitute the value of this for any callers of super(...) . It is necessary for generated constructor code to capture any potential return value of super(...) and replace it with this .

As a result, subclassing Error , Array , and others may no longer work as expected. This is due to the fact that constructor functions for Error , Array , and the like use ECMAScript 6’s new.target to adjust the prototype chain; however, there is no way to ensure a value for new.target when invoking a constructor in ECMAScript 5. Other downlevel compilers generally have the same limitation by default.

For a subclass like the following:

you may find that:

  • methods may be undefined on objects returned by constructing these subclasses, so calling sayHello will result in an error.
  • instanceof will be broken between instances of the subclass and their instances, so (new MsgError()) instanceof MsgError will return false .

As a recommendation, you can manually adjust the prototype immediately after any super(...) calls.

However, any subclass of MsgError will have to manually set the prototype as well. For runtimes that don’t support Object.setPrototypeOf , you may instead be able to use __proto__ .

Unfortunately, these workarounds will not work on Internet Explorer 10 and prior . One can manually copy methods from the prototype onto the instance itself (i.e. MsgError.prototype onto this ), but the prototype chain itself cannot be fixed.

Member Visibility

You can use TypeScript to control whether certain methods or properties are visible to code outside the class.

The default visibility of class members is public . A public member can be accessed anywhere:

Because public is already the default visibility modifier, you don’t ever need to write it on a class member, but might choose to do so for style/readability reasons.

protected members are only visible to subclasses of the class they’re declared in.

Exposure of protected members

Derived classes need to follow their base class contracts, but may choose to expose a subtype of base class with more capabilities. This includes making protected members public :

Note that Derived was already able to freely read and write m , so this doesn’t meaningfully alter the “security” of this situation. The main thing to note here is that in the derived class, we need to be careful to repeat the protected modifier if this exposure isn’t intentional.

Cross-hierarchy protected access

Different OOP languages disagree about whether it’s legal to access a protected member through a base class reference:

Java, for example, considers this to be legal. On the other hand, C# and C++ chose that this code should be illegal.

TypeScript sides with C# and C++ here, because accessing x in Derived2 should only be legal from Derived2 ’s subclasses, and Derived1 isn’t one of them. Moreover, if accessing x through a Derived1 reference is illegal (which it certainly should be!), then accessing it through a base class reference should never improve the situation.

See also Why Can’t I Access A Protected Member From A Derived Class? which explains more of C#‘s reasoning.

private is like protected , but doesn’t allow access to the member even from subclasses:

Because private members aren’t visible to derived classes, a derived class can’t increase their visibility:

Cross-instance private access

Different OOP languages disagree about whether different instances of the same class may access each others’ private members. While languages like Java, C#, C++, Swift, and PHP allow this, Ruby does not.

TypeScript does allow cross-instance private access:

Like other aspects of TypeScript’s type system, private and protected are only enforced during type checking .

This means that JavaScript runtime constructs like in or simple property lookup can still access a private or protected member:

private also allows access using bracket notation during type checking. This makes private -declared fields potentially easier to access for things like unit tests, with the drawback that these fields are soft private and don’t strictly enforce privacy.

Unlike TypeScripts’s private , JavaScript’s private fields ( # ) remain private after compilation and do not provide the previously mentioned escape hatches like bracket notation access, making them hard private .

When compiling to ES2021 or less, TypeScript will use WeakMaps in place of # .

If you need to protect values in your class from malicious actors, you should use mechanisms that offer hard runtime privacy, such as closures, WeakMaps, or private fields. Note that these added privacy checks during runtime could affect performance.

Static Members

Background Reading: Static Members (MDN)

Classes may have static members. These members aren’t associated with a particular instance of the class. They can be accessed through the class constructor object itself:

Static members can also use the same public , protected , and private visibility modifiers:

Static members are also inherited:

Special Static Names

It’s generally not safe/possible to overwrite properties from the Function prototype. Because classes are themselves functions that can be invoked with new , certain static names can’t be used. Function properties like name , length , and call aren’t valid to define as static members:

Why No Static Classes?

TypeScript (and JavaScript) don’t have a construct called static class the same way as, for example, C# does.

Those constructs only exist because those languages force all data and functions to be inside a class; because that restriction doesn’t exist in TypeScript, there’s no need for them. A class with only a single instance is typically just represented as a normal object in JavaScript/TypeScript.

For example, we don’t need a “static class” syntax in TypeScript because a regular object (or even top-level function) will do the job just as well:

static Blocks in Classes

Static blocks allow you to write a sequence of statements with their own scope that can access private fields within the containing class. This means that we can write initialization code with all the capabilities of writing statements, no leakage of variables, and full access to our class’s internals.

Generic Classes

Classes, much like interfaces, can be generic. When a generic class is instantiated with new , its type parameters are inferred the same way as in a function call:

Classes can use generic constraints and defaults the same way as interfaces.

Type Parameters in Static Members

This code isn’t legal, and it may not be obvious why:

Remember that types are always fully erased! At runtime, there’s only one Box.defaultValue property slot. This means that setting Box<string>.defaultValue (if that were possible) would also change Box<number>.defaultValue - not good. The static members of a generic class can never refer to the class’s type parameters.

this at Runtime in Classes

Background Reading: this keyword (MDN)

It’s important to remember that TypeScript doesn’t change the runtime behavior of JavaScript, and that JavaScript is somewhat famous for having some peculiar runtime behaviors.

JavaScript’s handling of this is indeed unusual:

Long story short, by default, the value of this inside a function depends on how the function was called . In this example, because the function was called through the obj reference, its value of this was obj rather than the class instance.

This is rarely what you want to happen! TypeScript provides some ways to mitigate or prevent this kind of error.

Arrow Functions

Background Reading: Arrow functions (MDN)

If you have a function that will often be called in a way that loses its this context, it can make sense to use an arrow function property instead of a method definition:

This has some trade-offs:

  • The this value is guaranteed to be correct at runtime, even for code not checked with TypeScript
  • This will use more memory, because each class instance will have its own copy of each function defined this way
  • You can’t use super.getName in a derived class, because there’s no entry in the prototype chain to fetch the base class method from

this parameters

In a method or function definition, an initial parameter named this has special meaning in TypeScript. These parameters are erased during compilation:

TypeScript checks that calling a function with a this parameter is done so with a correct context. Instead of using an arrow function, we can add a this parameter to method definitions to statically enforce that the method is called correctly:

This method makes the opposite trade-offs of the arrow function approach:

  • JavaScript callers might still use the class method incorrectly without realizing it
  • Only one function per class definition gets allocated, rather than one per class instance
  • Base method definitions can still be called via super .

In classes, a special type called this refers dynamically to the type of the current class. Let’s see how this is useful:

Here, TypeScript inferred the return type of set to be this , rather than Box . Now let’s make a subclass of Box :

You can also use this in a parameter type annotation:

This is different from writing other: Box — if you have a derived class, its sameAs method will now only accept other instances of that same derived class:

this -based type guards

You can use this is Type in the return position for methods in classes and interfaces. When mixed with a type narrowing (e.g. if statements) the type of the target object would be narrowed to the specified Type .

A common use-case for a this-based type guard is to allow for lazy validation of a particular field. For example, this case removes an undefined from the value held inside box when hasValue has been verified to be true:

Parameter Properties

TypeScript offers special syntax for turning a constructor parameter into a class property with the same name and value. These are called parameter properties and are created by prefixing a constructor argument with one of the visibility modifiers public , private , protected , or readonly . The resulting field gets those modifier(s):

Class Expressions

Background Reading: Class expressions (MDN)

Class expressions are very similar to class declarations. The only real difference is that class expressions don’t need a name, though we can refer to them via whatever identifier they ended up bound to:

Constructor Signatures

JavaScript classes are instantiated with the new operator. Given the type of a class itself, the InstanceType utility type models this operation.

abstract Classes and Members

Classes, methods, and fields in TypeScript may be abstract .

An abstract method or abstract field is one that hasn’t had an implementation provided. These members must exist inside an abstract class , which cannot be directly instantiated.

The role of abstract classes is to serve as a base class for subclasses which do implement all the abstract members. When a class doesn’t have any abstract members, it is said to be concrete .

Let’s look at an example:

We can’t instantiate Base with new because it’s abstract. Instead, we need to make a derived class and implement the abstract members:

Notice that if we forget to implement the base class’s abstract members, we’ll get an error:

Abstract Construct Signatures

Sometimes you want to accept some class constructor function that produces an instance of a class which derives from some abstract class.

For example, you might want to write this code:

TypeScript is correctly telling you that you’re trying to instantiate an abstract class. After all, given the definition of greet , it’s perfectly legal to write this code, which would end up constructing an abstract class:

Instead, you want to write a function that accepts something with a construct signature:

Now TypeScript correctly tells you about which class constructor functions can be invoked - Derived can because it’s concrete, but Base cannot.

Relationships Between Classes

In most cases, classes in TypeScript are compared structurally, the same as other types.

For example, these two classes can be used in place of each other because they’re identical:

Similarly, subtype relationships between classes exist even if there’s no explicit inheritance:

This sounds straightforward, but there are a few cases that seem stranger than others.

Empty classes have no members. In a structural type system, a type with no members is generally a supertype of anything else. So if you write an empty class (don’t!), anything can be used in place of it:

How JavaScript handles communicating across file boundaries.

The TypeScript docs are an open source project. Help us improve these pages by sending a Pull Request ❤

Ryan Cavanaugh  (60)

Last updated: Feb 08, 2024  

Search code, repositories, users, issues, pull requests...

Provide feedback.

We read every piece of feedback, and take your input very seriously.

Saved searches

Use saved searches to filter your results more quickly.

To see all available qualifiers, see our documentation .

  • Notifications

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement . We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Add a note about definite assignment assertion of Typescript 2.7 in docs #221

@skegel13

skegel13 commented Feb 8, 2018

  • 👍 2 reactions

@ktsn

ktsn commented Feb 8, 2018

Sorry, something went wrong.

@ktsn

304NotModified commented Sep 17, 2018

@adimascio

ktsn commented Feb 10, 2020

Successfully merging a pull request may close this issue.

@skegel13

  • Manage your subscription
  • Manage payment method
  • Renew your subscription

Turn recurring billing on or off

  • When subscription expires
  • Cancel Microsoft 365
  • Share Microsoft 365 Family
  • Stop sharing Microsoft 365 Family
  • You received an invitation to share
  • Switch between Microsoft 365 subscriptions
  • Switch to a business subscription
  • Transfer to a different Microsoft account
  • About accounts
  • Sign in to Microsoft 365
  • Why you need to sign in
  • Forgot account or password
  • Get started at Microsoft 365.com
  • Meet the Microsoft 365 app launcher
  • Check version
  • Microsoft 365 for home or business
  • What business product do I have?
  • Difference between Microsoft 365 and Office 2021
  • Difference between home and business plans
  • Difference between Microsoft 365 and free web apps
  • Can't install or manage Office or Microsoft 365
  • Parental permission when signing in to Office or Microsoft 365

add definite assignment assertion

Manage your Microsoft 365 subscription or Office product

For most Microsoft 365 subscriptions and versions of Office (2013 and later), you need to associate an account with your product. This is the account you use to sign in to Microsoft365.com and what you use to install or reinstall the apps, or to manage your subscription. Depending on your product, this account can be a personal Microsoft account (such as Hotmail.com, Outlook.com, Live.com), or a work or school account assigned by someone in your organization.

Note:  Some products purchased through an employee  Microsoft Workplace Discount Program  (formerly known as Home Use Program) benefit or volume license versions (managed by an organization's IT department) might not require an account. The information below doesn't apply to these Office versions or Microsoft 365 subscriptions.

Sign in to the dashboard for your account

Open a desktop app like Word or Excel, or go to Microsoft365.com .

Tip:  You might be prompted to sign in. Make sure to sign in with the account associated with Microsoft 365 or Office. If you can’t remember which email address you associated with your subscription or Office product, see I can't remember the Microsoft account I use with Microsoft 365 .

From the header, select your Account manager .

Depending on your account type, select  My Microsoft account or  View  account .

Your account dashboard is displayed and what you can do next depends on if you're signed in with a Microsoft account , or a work or school account .

Select the tab below for the type of account you're signed in with.

If you selected My Microsoft account , the Microsoft account dashboard will open. This is where you manage your Microsoft account and any Microsoft products associated with this account.

On the Microsoft account dashboard, select  Services & subscriptions to view all Microsoft products associated with this account. 

For non-subscription versions of Office (such as Office 2013 and later): Find your Office product and select Install . Follow the prompts to install or reinstall the desktop apps.

For Microsoft 365 Family or Personal subscriptions: Select Install premium Microsoft 365 apps and follow the prompts to install or reinstall the desktop apps.

On the Microsoft 365 subscription tab, select Manage . From here you can:

Renew your subscription with a prepaid code or card

Cancel a subscription

For Microsoft 365 Family subscriptions, you can start sharing your subscription, and add or remove people you're already sharing with. If you're not the subscription owner, you can see who's sharing Microsoft 365 with you, or choose to leave the subscription.

If you selected View account , the  My Account dashboard for your work or school account will open. Here you can:

Install and manage your apps. On the Office apps card, select Manage , and select Install to install or reinstall the latest desktop apps, or other apps such as Project or Visio. (An install option is available as long as your admin assigned you a license and gave you permission to self-install.)

Manage your devices.

View your subscriptions, and any other licenses assigned to you.

Note:  For Microsoft 365 admins only     If you're the Microsoft 365 admin in your organization, you control what you want your users to have access to. Go to the Microsoft 365 Admin help center for more information about setting up users.

Facebook

Need more help?

Want more options.

Explore subscription benefits, browse training courses, learn how to secure your device, and more.

add definite assignment assertion

Microsoft 365 subscription benefits

add definite assignment assertion

Microsoft 365 training

add definite assignment assertion

Microsoft security

add definite assignment assertion

Accessibility center

Communities help you ask and answer questions, give feedback, and hear from experts with rich knowledge.

add definite assignment assertion

Ask the Microsoft Community

add definite assignment assertion

Microsoft Tech Community

add definite assignment assertion

Windows Insiders

Microsoft 365 Insiders

Find solutions to common problems or get help from a support agent.

add definite assignment assertion

Online support

Was this information helpful?

Thank you for your feedback.

IMAGES

  1. Using The "Definite Assignment Assertion" To Define Required Input

    add definite assignment assertion

  2. PPT

    add definite assignment assertion

  3. PPT

    add definite assignment assertion

  4. PPT

    add definite assignment assertion

  5. Assertion ( assert Statement ) in Java

    add definite assignment assertion

  6. Types of Assertions in TestNG with Examples (Tutorial #1)

    add definite assignment assertion

VIDEO

  1. 🔵 Assert Meaning

  2. JMeter Assertions from scratch

  3. Assertion Meaning with Examples

  4. JMeter Tutorial 08

  5. 40 Definite Assignment Assertion

  6. How to Add Assertions?

COMMENTS

  1. What's the difference between definite assignment assertion and ambient

    Definite assignment assertion: class Person { name!: string; } Ambient declaration: class Person { declare name: string; } I can't see the difference between these two techniques. They both cure the error, they both don't emit code, and they both don't allow initializers.

  2. TypeScript definite assignment assertions

    The correct fix is probably to assign this.wordsPerMinute = wordsPerMinute in the constructor - but in some cases, you may be doing something funky where the dependency will be resolved in a way the compiler is unable to determine. When you need to allow a property with no definite assignment, you can use the definite assignment assertion.

  3. TypeScript: Documentation

    In our example, we knew that all uses of x would be initialized so it makes more sense to use definite assignment assertions than non-null assertions.. Fixed Length Tuples. In TypeScript 2.6 and earlier, [number, string, string] was considered a subtype of [number, string].This was motivated by TypeScript's structural nature; the first and second elements of a [number, string, string] are ...

  4. Definite Assignment Assertions (!)

    The Definite Assignment Assertions or also called non-null assertion operator tells the TypeScript compiler that a value typed cannot be null or undefined which is a way to override the compiler's analysis and inform it that a variable will be assigned a value before it is used. type Person = { name: string; }; const printName = (person?:

  5. Using The "Definite Assignment Assertion" To Define Required Input

    To fix this workflow, Angular internals have started to use the "Definite Assignment Assertion". This is a property name followed by an exclamation point. For example: public user!:

  6. Strict Property Initialization in TypeScript

    In later versions of TypeScript there's the concept of a Strict Class Property Initializer, or Definite Assignment Assertion. This feature all of a sudden produced type errors on previously working code, so let's explore why it's here and why you should be using it. "Property … has no initializer and is not definitely assigned in the constructor."

  7. Strict Property Initialization in TypeScript

    Perhaps the most useful solution is to add a username parameter to the constructor, which is then assigned to the username property. This way, whenever an instance of the User class is constructed, the caller has to provide the username as an argument:

  8. Announcing TypeScript 2.7

    Definite assignment assertions. As expressive as we try to make the type system, we understand there are certain times you might know better than TypeScript. As we mentioned above, definite assignment assertions are a new syntax you can use to convince TypeScript that a property will definitely get assigned. But in addition to working on class ...

  9. Definite Assignment Assertion

    The "Definite Assignment Assertion" Lesson is part of the full, Intermediate TypeScript, v2 course featured in this preview video. Here's what you'd learn in this lesson: Mike discusses the definite assignment operator and explains how to use the exclamation mark (!) to assert that a variable will be assigned a value before it is used.

  10. Late binding and TypeScript definite assignment assertions

    As the official TypeScript documentation ( https://www.typescriptlang.org/docs/handbook/release-notes/typescript-2-7.html#definite-assignment-assertions) states, definite assignment assertions can be used to tell the compiler that variables will indeed be assigned, even if TypeScript can't determine this on its own.

  11. TypeScript 2.7-Definite Assignment Assertions

    To solve this situation you can use a definite assignment assertion using the exclamation mark: (!) . Now the type checker will no longer complain. Now the type checker will no longer complain. Remark: By using a definite assignment assertion it becomes your responsibility to make sure the property gets set.

  12. What Does "!" Symbol Mean in Typescript?

    The solution in this case is to use the "!" symbol which is a feature in Typescript known as the definite assignment assertion. As defined in the official Typescript Documentation, the definite assignment assertion feature is used to tell the compiler that the variable assigned is indeed valid for all intents and purposes, even if Typescript's ...

  13. Property '...' has no initializer and is not definitely assigned in the

    Solution 3: Add definite assignment assertion to property. If you know that we will assign the property in later point in time. It's better to add definite assignment assertion to the property. i.e., employees. employees!: Employee[]; To add the definite assignment assertion we have to add Exclamation mark(!) symbol after the variable name.

  14. Quick-fix duplicates comments on "Add definite assignment assertion to

    and offer a quick fix: Add definite assignment assertion to property 'myField: string;' If you accepts the quick fix, you will get the comment repeated twice (and that's not expected): class MyClass { myField!: string; // some comment // some comment }

  15. Typescript: fields defined with definite assignment assertion are

    Expected behavior I think I get the decision behind stripping out fields defined using definite assignment assertion, so conceptually, they are no different than fields defined using declare.However, for my use case, I would really like babel to be in sycn with Typescript here for the case where useDefineForClassFields is set to true, that is I want to fields to be present and initialized to ...

  16. TypeScript 2.7 introduces the definite property assignment assertion

    TypeScript 2.7 introduces the definite property assignment assertion operator (!) Jan 4, 2018. nicolo-ribaudo added the area: typescript label Jan 4, 2018. andy-ms mentioned this issue Jan 5, 2018. typescript: Support definite assignment assertion #7159. Merged Copy link Member. hzoo ...

  17. TypeScript: Documentation

    Type-only Field Declarations. When target >= ES2022 or useDefineForClassFields is true, class fields are initialized after the parent class constructor completes, overwriting any value set by the parent class.This can be a problem when you only want to re-declare a more accurate type for an inherited field. To handle these cases, you can write declare to indicate to TypeScript that there ...

  18. Add a note about definite assignment assertion of Typescript 2.7 in

    Typescript 2.7 introduces a new flag called --strictPropertyInitialization which requires class properties to be property of a class gets initialized in the constructor body, or by a property initializer. If I don't initialize props and ...

  19. Angular @input property ending with exclamation point?

    A definite assignment assertion '!' is not permitted in this context.ts (1255) So I don't think that it ever makes sense to inlude the ! operator on an @Input property. Here's a screenshot: javascript angular typescript ecmascript-6 Share Improve this question Follow edited Nov 28, 2019 at 21:43 asked Nov 28, 2019 at 17:32 Ole 43.6k 62 206 390

  20. Create a form in Word that users can complete or print

    Add content to the form. Go to Developer, and then choose the controls that you want to add to the document or form. To remove a content control, select the control and press Delete. You can set Options on controls once inserted. From Options, you can add entry and exit macros to run when users interact with the controls, as well as list items ...

  21. Manage your Microsoft 365 subscription or Office product

    If you selected My Microsoft account, the Microsoft account dashboard will open.This is where you manage your Microsoft account and any Microsoft products associated with this account. On the Microsoft account dashboard, select Services & subscriptions to view all Microsoft products associated with this account. For non-subscription versions of Office (such as Office 2013 and later):

  22. In TypeScript, why an exclamation mark on a class field?

    The exclamation on a class field is called the definite assignment assertion operator. It tells the TypeScript compiler not to generate this error: ... Add a comment | Your Answer Reminder: Answers generated by artificial intelligence tools are not allowed on Stack Overflow. Learn more. Thanks for contributing an answer to Stack Overflow! ...