是否可以创建一个复杂的类型保护函数，根据提供的参数推断返回类型？

Question

假设我有两个工会：

type Animal = "cat" | "elephant"
type Vehicle = "some small vehicle" | "truck"

我想创建一个函数来推断它正在使用哪个联合，然后保护特定联合中的类型 - 类似于：

function isBig(thing: Animal | Vehicle): (typeof thing extends Animal) ? thing is "elephant" : thing is "truck" {
    if(isAnimal(thing)) return thing === "elephant"
    return thing === "truck"
}

这可行吗？甚至合理？

Answer 1

你可以用两个函数签名来做到这一点：

type Animal = "cat" | "elephant"
type Vehicle = "some small vehicle" | "truck"

function isBig(thing: Animal): thing is "elephant"
function isBig(thing: Vehicle): thing is "truck"
function isBig(thing: Animal | Vehicle) {
    if(isAnimal(thing)) return thing === "elephant"
    return thing === "truck"
}