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我想要一个函数，它接受一些对象并返回它的x属性。该对象需要限制为泛型类型Type<X>，我希望返回值的类型是属性的类型x。

要将输入限制为Type<X>我需要使用T extends Type<X>但我必须将其设置X为某些类型值T extends Type<string>，例如不能使用Type<number>或丢弃属性T extends Type<any>的类型信息。x

我希望做类似的事情<T extends Type<any>>(o: T) => T.Xor <T extends Type<???>>(o: T) => typeof o。

TypeScript 中有没有办法做到这一点？如果是这样，怎么做？

// 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)
4

1 回答 1

2

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'

（操场上的代码）

于 2017-08-10T14:02:42.810 回答