泛型类型为 function 返回类型

Question

Say I have some code as below:

function add<T extends boolean>(n1: number, n2: number, showResult: T): L<T> {
  if (showResult) {
    console.log(n1 + n2)
    return undefined // Type 'undefined' is not assignable to type 'L<T>'.(2322)
  } else {
    return n1 + n2 // Type 'number' is not assignable to type 'L<T>'.(2322)
  }
}

type L<G extends boolean> = G extends true ? undefined : number

type K = L<true> // K: undefined

const result = add(1, 2, false) // result: number

The add function returns type depends on showResult , so I introduce a generic type parameter T , and a generic type L . I'm a little confused about why there are two errors that indicate neither undefined nor number can assign to type L<T> .

How can I eliminate these errors while still have the benefit of inferring result type?

Thanks in advance.

Answer 1

Your type signature indicates that the caller can pass any type that extends boolean, and your function will return L parameterized of that type . undefined is not " L parameterized by whatever type the caller passed, which extends boolean." (It is true that boolean is a bit magical here, and it may not be possible in practice to extend it, but your signature indicates that you expect it to be.)

While I wouldn't particularly recommend this function at all (passing a boolean is often a sign of a design problem, and you really should just have two functions), if I were going to implement it I'd recommend overloads:

function add(n1: number, n2: number, showResult: true): undefined
function add(n1: number, n2: number, showResult: false): number
function add(n1: number, n2: number, showResult: boolean): number | undefined
{
    // Implementation
}

This says if you pass a boolean, it will return a number or undefined, but if it's known at compile time to be true, it will be undefined, and if it's known to be false, it will be a number.