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:

class User {
  username: string;
}

const user = new User();

// TypeError: Cannot read property 'toLowerCase' of undefined
const username = user.username.toLowerCase();

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:

class User {
  // Type error: Property 'username' has no initializer
  // and is not definitely assigned in the constructor
  username: string;
}

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:

class User {
  username: string | undefined;
}

const user = new User();

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:

// OK
const username =
  typeof user.username === "string" ? user.username.toLowerCase() : "n/a";

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:

// OK
const username = user.username?.toLowerCase() ?? "n/a";

#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:

class User {
  username = "n/a";
}

const user = new User();

// OK
const username = user.username.toLowerCase();

#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:

class User {
  username: string;

  constructor(username: string) {
    this.username = username;
  }
}

const user = new User("mariusschulz");

// OK
const username = user.username.toLowerCase();

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:

class User {
  constructor(public username: string) {}
}

const user = new User("mariusschulz");

// OK
const username = user.username.toLowerCase();

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:

class User {
  // Type error: Property 'username' has no initializer
  // and is not definitely assigned in the constructor.
  username: string;

  constructor(username: string) {
    if (Math.random() < 0.5) {
      this.username = username;
    }
  }
}

#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:

class User {
  username!: string;

  constructor(username: string) {
    this.initialize(username);
  }

  private initialize(username: string) {
    this.username = username;
  }
}

const user = new User("mariusschulz");

// OK
const username = user.username.toLowerCase();

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.