TypeScript中泛型的子类型的推断

Question

我想要一个函数，它需要一些对象并返回它的x属性。该对象需要被限制为通用类型Type<X>，我希望返回值的类型是属性x的类型。

要限制输入Type<X>我需要使用T extends Type<X>但我必须实际设置X像T extends Type<string>某种类型的值不会Type<number>或T extends Type<any>其丢弃x属性的类型信息的工作。

我希望能做一些像<T extends Type<any>>(o: T) => T.X或<T extends Type<???>>(o: T) => typeof o。

在TypeScript中有办法做到这一点吗？如果是这样，怎么样？

// 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) 

Answer 1

如果我理解你正确，则：

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）