题目描述
不使用 ReturnType 实现 TypeScript 的 ReturnType<T> 泛型。
例如：
const fn = (v: boolean) => {
if (v)
return 1
else
return 2
}
type a = MyReturnType<typeof fn> // 应推导出 "1 | 2"
题解
// ============= Test Cases =============
import type { Equal, Expect } from './test-utils'
type cases = [
Expect<Equal<string, MyReturnType<() => string>>>,
Expect<Equal<123, MyReturnType<() => 123>>>,
Expect<Equal<ComplexObject, MyReturnType<() => ComplexObject>>>,
Expect<Equal<Promise<boolean>, MyReturnType<() => Promise<boolean>>>>,
Expect<Equal<() => 'foo', MyReturnType<() => () => 'foo'>>>,
Expect<Equal<1 | 2, MyReturnType<typeof fn>>>,
Expect<Equal<1 | 2, MyReturnType<typeof fn1>>>,
]
type ComplexObject = {
a: [12, 'foo']
bar: 'hello'
prev(): number
}
const fn = (v: boolean) => v ? 1 : 2
const fn1 = (v: boolean, w: any) => v ? 1 : 2
// ============= Your Code Here =============
type MyReturnType<T extends CallableFunction> = T extends (...args: any[]) => infer R ? R : never
使用T extends CallableFunction对传入的类型参数T进行约束，确保T是一个可调用的函数类型
通过T extends (...args: any[]) => infer R条件表达式来推断函数的返回值类型。其中，infer R从函数T中推断出返回值的类型，并存储在R中
当T是可调用函数时，返回推断出的返回值类型R，否则返回never
