TypeScript:基于参数的函数返回类型，无需重载 [英] TypeScript: function return type based on argument, without overloading

问题描述

是否可以根据参数返回函数的类型?

Is it possible to return a function's type based on an argument?

我看到了 基于字符串文字类型参数的变量返回类型，但它使用重载.在这里，我有 100 多种类型，所以我不想做重载.

I saw Variable return types based on string literal type argument, but it uses overloading. Here, I have 100+ types, so I do not want to do overloading.

interface Registry {
    A: number,
    B: string,
    C: boolean,
    // ... 100 more types like this
}

function createType<T = Registry[typeof myType]>(myType: keyof Registry, value: any): T {
    // do some magic
    // ...
    return value;
}

const a = createType('A', 2); // Expected type: number. Actual: error above

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推荐答案

你可以做到，但是你需要一个类型参数来捕获传入的参数.使用这个新的类型参数你可以索引到你的 Registry 得到你想要的类型:

You can do it, but you will need a type parameter to capture the argument passed in. With this new type parameter you can index into your Registry to get the type you want:

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

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

const a = createType('A', 2); // ok 
const b = createType('B', 2); // err

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