! operator in typescript after object method

Question

I have an object X with a method getY() returning an object Y with a method a(), in typescript. What does it mean an expression like this one:

X.getY()!.a()

I guess the ! operator is used to check against null, but how does it work concretely? Where is defined in the language?

Answer 1

It's called the "Non-null assertion operator" and it tells the compiler that x.getY() is not null.

It's a new typescript 2.0 feature and you can read about it in the what's new page, here's what it says:

A new ! post-fix expression operator may be used to assert that its operand is non-null and non-undefined in contexts where the type checker is unable to conclude that fact. Specifically, the operation x! produces a value of the type of x with null and undefined excluded. Similar to type assertions of the forms x and x as T, the ! non-null assertion operator is simply removed in the emitted JavaScript code.

// Compiled with --strictNullChecks
function validateEntity(e?: Entity) {
    // Throw exception if e is null or invalid entity
}

function processEntity(e?: Entity) {
    validateEntity(e);
    let s = e!.name;  // Assert that e is non-null and access name
}

---

Edit

There's an issue for documenting this feature: Document non-null assertion operator (!)

Answer 2

Non-null assertion operator: !

• You tells the TS compiler that the value of a variable is not null | undefined
• Use it when you are in possession of knowledge that the TS compiler lacks.

---

Here is a trivial example of what it does:

let nullable1: null | number;
let nullable2: undefined | string;

let foo  = nullable1! // type foo: number
let fooz = nullable2! // type fooz: string

It basically removes null | undefined from the type

---

When do I use this?

Typescript is already pretty good at inferring types for example using typeguards:

let nullable: null | number | undefined;

if (nullable) {
    const foo = nullable; // ts can infer that foo: number, since if statements checks this
}

---

However sometimes we are in a scenario which looks like the following:

type Nullable = null | number | undefined;

let nullable: Nullable;

validate(nullable);

// Here we say to ts compiler:
// I, the programmer have checked this and foo is not null or undefined
const foo = nullable!;  // foo: number

function validate(arg: Nullable) {
    // normally usually more complex validation logic
    // but now for an example
    if (!arg) {
        throw Error('validation failed')
    }
}

My personal advice is to try to avoid this operator whenever possible. Let the compiler do the job of statically checking your code. However there are scenarios especially with vendor code where using this operator is unavoidable.

Answer 3

It gets rid of the typescript compile error, when a variable can be of type null/undefined and you are performing a function on that variable which requires it be non-null/non-undefined.

function getValue(): string | undefined {
    return "MyStr"; // TS assumes getValue can return undefined as well
  //return undefined;
  //return OutsidePlugin(); // The plugin can return both string and undefined, e.g.
}
            
let value = getValue();
console.log('value length: ' + value.length); //value.length can requires non-undefined

This will generate the following Typescript compile error:

'value' is possibly 'undefined'.

Non-null assertion operator is just a quick escape hatch:

console.log('value length: ' + value!.length); //TS, I assure you it's non-undefined