如何键入函数的参数但将返回类型保留"inferred"？

Question

下面我有两个函数originalhelloWorld，它是非类型的，helloWorld是有类型的。您可以看到，类型o的返回返回"0"；推断的"；返回类型(这个名称是什么(，并且类型x返回"0"；任何"；。

如何让ExampleFunction键入函数参数，而不推断返回类型？我尝试了几种泛型组合，但似乎都不起作用。

打字游戏场

const originalhelloWorld = (greeting: string | boolean) => {
if (typeof greeting === 'boolean') return greeting
return `hello ${greeting}`
}
type o = ReturnType<typeof originalhelloWorld>
//  ^? type o = string | boolean
/* ------------------------------------ */
type ExampleFunction = (greeting: string | boolean) => any
const helloWorld: ExampleFunction = (greeting) => {
if (typeof greeting === 'boolean') return greeting
return `hello ${greeting}`
}
type x = ReturnType<typeof helloWorld>
//  ^? type x = any

Answer 1

typescript 4.9中的新satisfies运算符起作用：

游乐场链接：

const originalhelloWorld = (greeting: string | boolean) => {
if (typeof greeting === 'boolean') return greeting
return `hello ${greeting}`
}
type o = ReturnType<typeof originalhelloWorld>
//  ^?
/* ------------------------------------ */
type ExampleFunction = (greeting: string | boolean) => any
const helloWorld = ((greeting) => {
if (typeof greeting === 'boolean') return greeting
return `hello ${greeting}`
}) satisfies ExampleFunction
type x = ReturnType<typeof helloWorld>
//  ^?