typescript - 如何在 TypeScript 中区分 `(() => any) | undefined` 和 `() => any`？

Question

我有两种类型:(() => any) |未定义 和() => 任何。我想提取函数的返回类型，但前提是该值绝对不是未定义的。

我尝试过以下方法:

type IsUndefined<T> = T extends (() => any) ? "no" : "yes";

但这解析为“yes” |当类型未定义时为“no”。我想在不创建联合的情况下检测这些类型之间的差异。

请参阅this playground link举个例子。

这是短篇故事，长篇故事是我有一个如下所示的结构:

type MyStruct = {
    optionalHook?: ((...args: any) => any) | undefined,
    requiredHook: (...args: any) => any,
}

我想提取可选钩子(Hook)的返回类型，但仅限于它存在时。否则我想提取所需 Hook 的返回类型。

参见this playground link以获得更全面的示例。

Answer 1

不知道为什么缺少 optionalHook 不会将 T 推断为 createStruct 中的 undefined 但我'通过添加默认通用参数，我们设法获得了所需的结果:

// Create the generic struct
type MyStruct<T extends ((...args: any) => any) | undefined, U extends (...args: any) => any> = {
    optionalHook?: T,
    requiredHook: U,
}

// Utility function for easily creating stucts with the correct type.
// Without this I'd have to type the generic arguments manually for every struct instance.
function createStruct<T extends ((...args: any) => any) | undefined = undefined, U extends (...args: any) => any = (...args: any) => any>(struct: MyStruct<T, U>) {
    return struct;
}

const withOptional = createStruct({
    optionalHook: () => 5,
    requiredHook: () => "hello",
});

const withoutOptional = createStruct({
    requiredHook: () => true,
});


// The function for extracting the return type
type getReturnType<T> =
    T extends MyStruct<infer O, infer R> ?
        O extends (...args: any) => infer OR ?
            OR :
            R extends (...args: any) => infer RR ? RR : never :
        never;


type ResultWithOptional = getReturnType<typeof withOptional>;
type ResultWithoutOptional = getReturnType<typeof withoutOptional>;

TypeScript playground

将 optionHook 显式设置为 undefined 也可以:

TypeScript playground