Support simple check in if for discriminated unions with true/false discriminant

[WiseBird]

Search Terms

discriminated true false if

Suggestion

Use Cases

Having discriminated union with true/false discriminant:

interface Square {
    kind: true;
    size: number;
}
interface Rectangle {
    kind: false;
    width: number;
    height: number;
}

type Shape = Square | Rectangle;

we can treat them differently using verbose comparison in if:

let s: Shape = null;

if (s.kind === true) {
    console.log(s.size);
} else {
    console.log(s.height);
}

With simple check:

let s: Shape = null;

if (s.kind) {
    console.log(s.size);
} else {
    console.log(s.height);
}

ts doesn't recognize Rectangle in else branch: Property 'height' does not exist on type 'Square'

Examples

Would be nice to be able to do simple if(s.kind) check.

Checklist

My suggestion meets these guidelines:

• This wouldn't be a breaking change in existing TypeScript / JavaScript code
• This wouldn't change the runtime behavior of existing JavaScript code
• This could be implemented without emitting different JS based on the types of the expressions
• This isn't a runtime feature (e.g. new expression-level syntax)

[MartinJohns]

This works already if you enable strictNullChecks. If you don't enable strictNullChecks, then you need to explicitly check for false in your else block (aka else if (s.kind === false), because the value can be undefined and null too (tho theoretically TS could infer that it's not undefined / null because you accessed a property before).

[WiseBird]

But this former check works without strictNullChecks:

if (s.kind === true) {
    console.log(s.size);
} else {
    console.log(s.height);
}

here too in else the value can be undefined or null.

[DanielRosenwasser]

I think this is a duplicate of #10564.

[WiseBird]

Duplicate