基于条件类型的可选参数 [英] Optional parameters based on conditional types

问题描述

是否可以根据 TypeScript 中的条件类型使函数具有强制或可选参数?

Is it possible to make a function have either mandatory or optional parameters based on conditional types in TypeScript?

这是我目前得到的:

const foo = <T extends string | number>(
    first: T,
    second: T extends string ? boolean : undefined
) => undefined;

foo('foo', true); // ok, as intended
foo(2, true); // not ok, as intended
foo(2, undefined); // ok, as intended
foo(2); // compiler error! I want this to be ok

推荐答案

您可以在 3.1 中使用 其余参数和传播表达式中的元组

You can do this in 3.1 using Tuples in rest parameters and spread expressions

const foo = <T extends string | number>(
  first: T, 
  ...a: (T extends string ? [boolean] : [undefined?])
) => undefined;

foo('foo', true); // ok, as intended
foo(2, true); // not ok, as intended
foo(2, undefined); // ok, as intended
foo(2); // ok

但更好的方法是使用重载.

But the better way is to use overloads.

function foo2(first: string, second: boolean) : undefined
function foo2(first: number, second?: undefined): undefined
function foo2<T>(first: T, second?: boolean): undefined{
  return undefined
}

foo2('foo', true); // ok, as intended
foo2(2, true); // not ok, as intended
foo2(2, undefined); // ok, as intended
foo2(2); // ok

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