Allow method type guards

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
}

[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.