如何将我的函数参数类型设置为以配置结构为条件？

Question

### this is my mapping data structure saying: some contexts have specific own tags
const config = {
  contexts: ["A", "B", "C"] as const,

  tags: {
    A: ["aa", "bb"] as const,
    B: ["aa1", "bb1"] as const,
    C: ["aa2", "bb2"] as const,
  },
};

# union type for contexts possible
type C = typeof config.contexts[number]; // "A" | "B" | "C"

# union type for tags specific for context U
type F<U extends C> = typeof config.tags[U][number];

const a: C = config.contexts[1]; // OK "B"

const f: F<typeof a> = "aa1"; // OK will accept only 'aa1','bb1'


# then I tried the above into a couple of function def but with NO SUCCESS

const func =  (a:C) => {
  return (g:F<typeof a>) => {}
}


func('A')(g) // ERROR g can be all of : "aa" | "bb" | "aa1" | "bb1" | "aa2" | "bb2"


const func2 =  (a:C,g:F<typeof a>) => {}


func('A',gg) // ERROR gg can be all of : "aa" | "bb" | "aa1" | "bb1" | "aa2" | "bb2"

Answer 1

使用 typoeof a 并不意味着捕获传入参数的实际类型，它只是意味着 a 的声明类型是什么，因此在本例中为 C。它适用于第一种情况，因为 TS 将使用控制流分析将 a 的类型缩小为 B，因为那是本地信息，但它不适用于函数。

要捕获调用站点信息，您需要使用类型参数：

const func =  <T extends C>(a:T) => {
  return (g:F<T>) => {}
}

func('A')("bb") // ok
func('A')("bb1") // err

Playground Link

如何将我的函数参数类型设置为以配置结构为条件？

How can I set up my function arg types to be mutually conditioned upon a config structure?

typescript

typescript-generics

typescript-typings