在第一个泛型上显式但在 TS 中推断出第二个？ [复制]

Question

是否可以在第一个泛型上显式，但在第二个泛型上隐式（推断）？

例如一个pick函数：

function pick<T extends { [K: string]: any }, U extends keyof T>(obj: T, key: U): T[U] {
    return obj[key];
}

interface Obj {
    foo: string;
    bar: string;
}

const obj = {
    foo: 'foo',
    bar: 'bar',
};

// works, available keys are inferred
pick(obj, 'bar');

// error: Expected 2 type arguments, but got 1.
// Is there a way I can tell to TS to infer the 2nd generic instead of expecting it explicitly?
pick<Obj>(obj, '');

Answer 1

const pick = <T extends { [K: string]: any }>(obj: T) => <U extends keyof T>(key: U): T[U] => {
    return obj[key];
}

interface Obj {
    foo: string;
    bar: string;
}

const obj = {
    foo: 'foo',
    bar: 'bar',
};

// works, available keys are inferred
pick(obj)(bar)

// error: Expected 2 type arguments, but got 1.
// Is there a way I can tell to TS to infer the 2nd generic instead of expecting it explicitly?
pick<Obj>(obj)('foo');

你可以通过柯里化函数；让我知道这是否有帮助