Future-proof non-aliasing/always-expanding of mapped/intersection/union/etc. types

Search Terms

non-aliasing types

Suggestion

I would like an "officially blessed" way to tell TS if it should give me a type alias, or give me an "expanded" type. (See examples)

Now that this issue has been fixed,
https://github.com/microsoft/TypeScript/issues/32824

the technique I've been using to force TS to not alias a type will break when I migrate to TS 3.6/3.7

Use Cases

In some cases, seeing Mapped1<T>/Intersected1<T>/Unioned1<T> is better for a developer (see examples).

In other cases, seeing the full "expanded" type is better (see examples for Mapped2<T>/Intersected2<T>/Unioned2<T>)

Examples

type Identity<T> = T;

type Mapped1<T> = { [k in keyof T]: [T[k], "mapped"] };
type Mapped2<T> = Identity<{ [k in keyof T]: [T[k], "mapped"] }>;

declare function foo1(): Mapped1<{ x: string, y: number }>;
declare function foo2(): Mapped2<{ x: string, y: number }>;

/*
Aliased, bad
const m1: Mapped1<{
    x: string;
    y: number;
}>
*/
const m1 = foo1();
/*
Non-aliased, good
const m2: {
    x: [string, "mapped"];
    y: [number, "mapped"];
}
*/
const m2 = foo2();

//==============================================================

type Intersected1<T> = T & { hi: string };
type Intersected2<T> = Identity<T & { hi: string }>;

declare function bar1(): Intersected1<{ x: string, y: number }>;
declare function bar2(): Intersected2<{ x: string, y: number }>;

/*
const i1: Intersected1<{
    x: string;
    y: number;
}>
*/
const i1 = bar1();
/*
const i2: {
    x: string;
    y: number;
} & {
    hi: string;
}
*/
const i2 = bar2();

//==============================================================

type Unioned1<T> = T | { hi: string };
type Unioned2<T> = Identity<T | { hi: string }>;

declare function baz1(): Unioned1<{ x: string, y: number }>;
declare function baz2(): Unioned2<{ x: string, y: number }>;

/*
const u1: Unioned1<{
    x: string;
    y: number;
}>
*/
const u1 = baz1();
/*
const u2: {
    hi: string;
} | {
    x: string;
    y: number;
}
*/
const u2 = baz2();

Playground

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. library functionality, non-ECMAScript syntax with JavaScript output, etc.)
• This feature would agree with the rest of TypeScript's Design Goals.

Related

https://github.com/microsoft/TypeScript/issues/32824#issuecomment-521127740

@weswigham brought up a different way to force TS to not alias a type.

I prefer my Identity<T> trick because it doesn't create a new "temporary" object type.

この issue の mapped、intersection、union の例から始め、次に issue #32824 の関連する議論と、そこからリンクされているコメントを確認してください。TypeScript が、これらの型をエイリアスのままにするか展開するかを制御する、公式にサポートされた方法を備え、示されているユースケースを維持できれば完了です。