问题

Here is the translation of the provided code snippet:

如果我有

```typescript
type T1 = {
  a: () => string,
  b: () => number
}

type T2 = {
  [P in keyof T1]: ReturnType<T1[P]>
}

那么 T2 是

{
  a: string,
  b: number
}

我如何创建一个通用类型函数来映射返回类型，假设通用类型的每个键都是一个函数？

type MapReturnType<T extends Record<string, ???>> = {
  [P in keyof T]: ReturnType<T[P]>
}

<details>
<summary>英文:</summary>

If I have

type T1 = {
a: () => string,
b: () => number
}

type T2 = {
[P in keyof T1]: ReturnType<T1[P]>
}

Then T2 is

{
a: string,
b: number
}

How do I make a generic type function to map the return types, assuming each key of the generic type is a function?

type MapReturnType<T extends Record<string, ???>> = {
[P in keyof T]: ReturnType<T[P]>
}

</details>


# 答案1
**得分**: 1

你可以使用 `() => any` 作为你拥有的值，因为它是一个函数。

```typescript
type MapReturnType<T extends Record<string, () => any>> = {
  [P in keyof T]: ReturnType<T[P]>
}

type MapReturnType<T extends Record<string, () => any>> = {
  [P in keyof T]: ReturnType<T[P]>
}