如果将可选参数指定为特定值，则返回类型“从不" [英] Return type 'never' if an optional parameter is given to be a specific value

问题描述

我有一个函数，该函数带有一个可选的 boolean 参数，默认为 false .当参数为 false 时，该函数返回 string .当参数为 true 时，该函数应返回 never .

I have a function that takes an optional boolean argument that defaults to false. When the argument is false, the function returns a string. When the argument is true, the function should return type never.

这是我尝试过的:

function example(arg: true): never;
function example(arg = false): string {
  //...
}

这感觉像应该起作用:推断 arg 具有 boolean 类型，以及当未传递或以 false 传递时， example 返回 string .当它作为 true 传递时，重载开始，并且 example 返回 never .

This feels like it should work: arg is inferred to have a boolean type, and when it is not passed or passed as false, example returns string. When it is passed as true, the overload kicks in and example returns never.

但是，这根本不起作用.TypeScript为 arg 提供类型为 true ，并使其为必需，并使 example 始终永远返回>.

However, this doesn't work at all. TypeScript gives arg the type true and makes it required and makes example always return never.

我也尝试过这种疯狂，并且走近了:

I also tried this craziness, and got closer:

function example(arg: false): string;
function example(arg: true): never;
function example(arg: boolean): string | never;
function example(arg = false): string | never {
  //...
}

但是，编译器仍然不认为 arg 是可选的.

However the compiler still doesn't consider arg to be optional.

推荐答案

这似乎可行:

function example(arg: true): never;
function example(arg?: false): string;
function example(arg: boolean): string;
function example(arg: boolean = false): never | string {
    if (!arg) {
        return 'foo';
    }

    throw new Error();
}

const foo = example(true); // typeof foo is never
const bar = example(false); // typeof bar is string
const baz = example(); // typeof baz is string

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