Typescript推断泛型中的泛型

问题描述：

假设我有一个界面 Wrapped :

interface Wrapped<T> {
  data: T
}

我想定义一个这样的函数:

And I want to define a function like this:

function f<T>(arg: any): T {
  const obj: Wrapped<T> = doSomethingAndGetWrappedObject<T>(arg)
  return obj.data
}

// Don't pay attention to the argument, it is not important for the question
const n: number = f<number>(/* ... */)

问题是，在我的应用程序中，传递 number 作为类型参数非常不方便，我想传递 Wrapped< number> ，即调用 f 像这样:

The problem is, in my application it is very inconvenient to pass number as type argument, I would like to pass Wrapped<number> instead, i.e. to call f like this:

const n: number = f<Wrapped<number>>(/* ... */)

问题是:如何键入 f 使其成为可能?

The question is: how to type f to make it possible?

function f<T extends Wrapped<V>, V>(arg: any) {
  // ...
}
// Now this works, but it is very annoying to write the second type argument
const n: number = f<Wrapped<number>, number>() 
// I would like to do this, but it produces an error
// Typescript accepts either no type arguments or all of them
const n: number = f<Wrapped<number>>()

// This just works in an unpredictable way
function f<T extends Wrapped<any>>(
  arg: any
): T extends Wrapped<infer V> ? V : any {
  /* ... */
}

答

您可以使用 infer 关键字来创建用于提取通用类型的帮助程序类型.

You could create helper type for extracting the generic type by using infer keyword.

interface Wrapped<T> {
  data: T
}

type ExtractGeneric<T> = T extends Wrapped<infer X> ? X : never

function f<T extends Wrapped<any>>(): ExtractGeneric<T> {
  ....
}

const n = f<Wrapped<number>>()

Typescript推断泛型中的泛型

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