Typescript 是否可以创建一个复杂的typeguard函数，根据提供的参数推断返回类型？

假设我有两个工会：

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

我想用一个函数来推断它使用的是哪个联合，然后保护特定联合中的类型，比如：

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？甚至是合理的？

您可以使用两个函数签名：

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"
}