Typescript 函数返回类型基于可选参数存在而不使用函数重载 [英] Typescript function return type based on optional parameter presence without using function overloads

问题描述

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我的目标是根据可选的 condition: CONDITION" 参数的存在返回不同的类型.我正在努力在不诉诸重载的情况下实现这一目标.

My goal here is to return a different type based on the presence of a optional condition: "CONDITION" parameter. And I'm trying to accomplish this without resorting to overloads.

type TYPE_1 = "TYPE_1"
type TYPE_2 = "TYPE_2"

type CONDITION = "CONDITION"

function foo(condition?: CONDITION): TYPE_1 | TYPE_2 {
    if (condition) {
        return "TYPE_1";
    }
    else {
        return "TYPE_2";
    }
}

const shouldBeType_1 = foo("CONDITION");    // ERROR: THIS IS BEING EVALUATED AS UNION TYPE: "TYPE_1" | "TYPE_2"
const shouldBeType_2 = foo();               // ERROR: THIS IS BEING EVALUATED AS UNION TYPE: "TYPE_1" | "TYPE_2"

这很容易通过重载来实现:

This is easy to accomplish with overloads:

/* ########################################### */
/* #### THIS IS EASY TO DO WITH OVERLOADS #### */
/* ########################################### */

function foo_overloaded(): TYPE_2
function foo_overloaded(condition: "CONDITION"): TYPE_1
function foo_overloaded(condition?: "CONDITION"): TYPE_1 | TYPE_2 {
    if (condition) {
        return "TYPE_1";
    }
    else {
        return "TYPE_2";
    }
}

const overloaded_shouldBeType_1 = foo_overloaded("CONDITION");   // SUCCESS: THIS IS TYPE_1
const overloaded_shouldBeType_2 = foo_overloaded();              // SUCCESS: THIS IS TYPE_2

没有重载的正确方法是什么?还是我把它复杂化了，重载只是在这种情况下的方法?

What is the proper way of doing it without overloads? Or am I over complicating it and overloading is simply the way to go in this situation?

SO 上也有这个问题:TypeScript:基于参数的函数返回类型，无需重载

Also there is this question here on SO: TypeScript: function return type based on argument, without overloading

它建议应该使用接口作为返回类型的映射，例如:

It suggests that an interface should be used as a map to the return type, like:

interface Registry {
    A: number,
    B: string,
    C: boolean
}

function createType<K extends keyof Registry>(type: K, value: Registry[K]): Registry[K] {
    return value;
}

但我不能那样做，因为 condition 要么是 "CONDITION"|未定义.那么如何映射 undefined 类型呢?我也试过用条件类型来做.类似的东西:

But I can't do that, because condition is either "CONDITION" | undefined. So how can I map the undefined type? I also tried doing that with conditional type. Something like:

type RETURN_TYPE<T extends undefined | "CONDITION"> = T extends "CONDITION" ? TYPE_1 : TYPE_2;

但这也没有用.

推荐答案

我会说在这种情况下你会使用重载函数，你可以通过以下方式部分解决这个问题:

I would say you would go with overloaded functions in this case, you can solve this partly with the following:

function foo<T extends CONDITION | undefined>(condition?: T): T extends CONDITION ? TYPE_1 : TYPE_2 {
    if (condition) {
        //`as any` is intentional here: https://stackoverflow.com/questions/55641731/typescript-conditional-type-complains-type-not-assignable
        return "TYPE_1" as any;
    } else {
        return "TYPE_2" as any;
    }
}

有了这个，以下工作正常:

With this, the following works fine:

const shouldBeType_1 = foo("CONDITION") // It is TYPE_1;

但是当您不传递任何参数时，它将不起作用:

But when you don't pass any params it would not work:

const shouldBeType_2 = foo(); // It is TYPE_1 | TYPE_2

当然，如果你直接传递 undefined，它就可以正常工作:

Of course if you pass undefined directly, it works correctly:

const shouldBeType_2 = foo(undefined); // It is TYPE_2;

---

长话短说，现在，解决您的问题最干净的方法是使用函数重载.

---

So long stroy short, right now, the cleanest approach for your problem, is to use function overloading.

编辑正如已经指出的那样，如果我为泛型类型添加默认参数，这也适用于省略的参数.

EDIT As it has been pointed out if I add a default parameter for the generic type, this would work with omitted param as well.

function foo<T extends CONDITION | undefined = undefined>(condition?: T): T extends CONDITION ? TYPE_1 : TYPE_2 {
    if (condition) {
        return "TYPE_1" as any;
    } else {
        return "TYPE_2" as any;
    }
}

// Both work:
const shouldBeType_1 = foo("CONDITION") // It is TYPE_1;
const shouldBeType_2 = foo(); // It is TYPE_2;

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