从可选参数更改函数返回签名

Question

这是我的示例代码:

type base = {
a: string;
};
type base_plus = {
foo: string;
};
const get = (param?: number) => {
const newBase: base = { a: '1' };
if (param) {
return { ...newBase, foo: 'bar' } as base_plus;
}
return newBase;
};

我想要的是这个:

const a = get(1); // => a is of type base_plus
const b = get(); // => b is of type base

。E能够知道get()函数的返回类型直接从事实，如果我调用可选参数或没有

Answer 1

您想这样声明重载函数吗?

type base = {
a: string;
};
type base_plus = {
foo: string;
};
function get(param: number): base_plus;
function get():base;
function get(param?: number) {
const newBase: base = { a: '1' };
if (param) {
return { ...newBase, foo: 'bar' } as base_plus;
}
return newBase;
};
const a = get(1); // => a is of type base_plus
const b = get();

Checkout this playground