typescript - 获取函数/类构造函数的参数类型

Question

我正在尝试做一些我不确定在 TypeScript 中是否可行的事情:从函数中推断参数类型/返回类型。

例如:

function foo(a: string, b: number) {
  return `${a}, ${b}`;
}

type typeA = <insert magic here> foo; // Somehow, typeA should be string;
type typeB = <insert magic here> foo; // Somehow, typeB should be number;

我的用例是尝试创建一个包含构造函数和参数的配置对象。

例如:

interface IConfigObject<T> {
    // Need a way to compute type U based off of T.
    TypeConstructor: new(a: U): T;
    constructorOptions: U;
}

// In an ideal world, could infer all of this from TypeConstructor

class fizz {
    constructor(a: number) {}
}

const configObj : IConfigObj = {
    TypeConstructor: fizz;
    constructorOptions: 13; // This should be fine
}

const configObj2 : IConfigObj = {
    TypeConstructor: fizz;
    constructorOptions: 'buzz'; // Should be a type error, since fizz takes in a number
}

Answer 1

Typescript 现在有 ConstructorParameters 内置函数，类似于 Parameters 内置函数。确保传递的是类类型，而不是实例:

ConstructorParameters<typeof SomeClass>

ConstructorParameter Official Doc

Parameters Official Doc