是否可以制作一个复杂的typeguard函数，根据提供的参数推断返回类型

Question

假设我有两个联合：

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

我想做一个单一的函数，推断它与哪个联合工作，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

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"
}