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TypeScript中泛型类型的子类型推断

typescript-generics type-inference typescript

3

Peter Hudec · 技术社区 · 8 年前

我想要一个函数,它接受某个对象并返回其 x 属性需要将对象限制为泛型类型 Type<X> 我希望返回值的类型是属性的类型 x .

将输入限制为 类型(<);X> 我需要使用 T extends Type<X> 但我必须实际设置 X 对于某些类型值,如 T extends Type<string> 这不适用于 Type<number> T extends Type<any> 它将丢弃属性

<T extends Type<any>>(o: T) => T.X <T extends Type<???>>(o: T) => typeof o .

TypeScript中有什么方法可以做到这一点吗?如果是,如何?

// Suppose I have this generic interface
interface Type<X> {
  readonly x: X
}

// I create one version for foo...
const foo: Type<string> = {
  x: 'abc',
}

// ...and another one for bar
const bar: Type<number> = {
  x: 123,
}

// I want this function to restrict the type of `o` to `Type`
// and infer the type of `o.x` depending on the inferred type of `o`,
// but to restrict to `Type` I must hardcode its X to `any`, which
// makes `typeof o.x` to evaluate to `any` and the type of `o.x` is lost.
function getX<T extends Type<any>> (o: T): typeof o.x {
  return o.x
}

// These are correctly typed
const okFooX: string = getX(foo)
const okBarX: number = getX(bar)

// These should result in error but it is OK due to `Type<any>`
const errorFooX: boolean = getX(foo)
const errorBarX: boolean = getX(bar)

1 回复 | 直到 8 年前

1

2

Nitzan Tomer 8 年前

如果我理解正确,那么:

function getX<T>(o: Type<T>): T {
    return o.x;
}

然后:

const errorFooX: boolean = getX(foo); // error: Type 'string' is not assignable to type 'boolean'
const errorBarX: boolean = getX(bar); // error: Type 'number' is not assignable to type 'boolean'

( code in playground