Allow method type guards

[vagarenko]

TypeScript Version: 2.0.3

Code

Why I can have function type guard:

function hasValue<T>(value: T | undefined): value is T { return value !== undefined; }

but not method type guard:

export class Maybe<T> {
    constructor(public value: T | undefined) {}

    hasValue(): this.value is T { return this.value !== undefined; }
    //              ^ '{' or ';' expected.
}

?

[kitsonk]

Semi related to a random thought here in that more complicated guards could express codependency of properties:

interface Array<T> {
    length: number;
    pop(): this.length > 0 ? T : undefined;
}

const arr = [ 1, 2, 3 ];

if (arr.length > 0) {
    const val = arr.pop(); // val is number, not number | undefined under strictNullChecks
}

[aluanhaddad]

This would be interesting... expressing path dependent types.
Read the OP backward... too early.

[mattmazzola]

Would this also fix issues with Map?
In the case where you're using map to add memoization to functions there is a similar need to use the ! operator.

const cache = new Map<number, string[]>()
if (cache.has(n)) {
  return cache.get(n)!  // The if statement above should make return type `string[]` instead of `string[] | undefined`
}

[DanielRosenwasser]

What about using this type guards instead?

export class Maybe<T> {
    constructor(public value: T | undefined) {}

    hasValue(): this is this & { value: T } { return this.value !== undefined; }
}

///////

declare var x: Maybe<string>;
if (x.hasValue()) {
    x.value.toLowerCase();
}

@mattmazzola I've used the above technique to describe a way that Map#has could work similarly, but it can get a bit messy.

[mattmazzola]

Ok, I'll have to learn/explore more about the this type guards. In my Map example this type definition is pre-defined by the lib.d.ts since they're part of ES2016 I believe and I didn't want to re-write my own type which wraps the native one.

[masaeedu]

@DanielRosenwasser Any way to make that approach work for private properties?