如何基于继承泛型的参数推导TypeScript返回类型？

我定义了一个返回函数的类型：

type OuterFunction<OuterParams extends [...any], InnerParams extends [...any]> = 
  (outerParams: [...OuterParams]) => (innerParams: [...InnerParams]) => Promise<boolean>

我尝试创建另一个函数，接收OuterFunction和OuterParams并返回内部函数：

type OuterParamsOf<T> = T extends OuterFunction<infer Deps, any[]> ? Deps : never

function someFunction<T extends OuterFunction<any[], any[]>>(func: T, deps: [...OuterParamsOf<T>]): ReturnType<T> {
  return func(...deps)
}

但出现错误：

Type '(...args: any[]) => Promise' is not assignable to type 'ReturnType'

忽略错误调用someFunction能正常工作，请问：

• 为何会出现这个错误？
• 该如何基于func参数正确设置返回类型？

TypeScript无法精准推断func(...deps)的类型与ReturnType<T>完全匹配。虽然T继承自OuterFunction<any[], any[]>，但ReturnType<T>会被推断为对应特定InnerParams的具体函数类型，而func(...deps)的返回值被宽泛判定为(...args: any[]) => Promise<boolean>，两者类型约束程度不一致，导致兼容性报错。

方案一：显式约束泛型参数 #

直接将OuterParams和InnerParams作为泛型参数，让TypeScript精准追踪类型信息，避免通过T间接约束：

type OuterFunction<OuterParams extends [...any], InnerParams extends [...any]> = 
  (outerParams: [...OuterParams]) => (innerParams: [...InnerParams]) => Promise<boolean>

function someFunction<O extends [...any], I extends [...any]>(
  func: OuterFunction<O, I>, 
  deps: [...O]
): ReturnType<OuterFunction<O, I>> {
  return func(...deps)
}

方案二：提取InnerParams类型 #

如果要保留T作为泛型参数，可以新增条件类型提取InnerParams，显式指定返回类型的结构：

type OuterFunction<OuterParams extends [...any], InnerParams extends [...any]> = 
  (outerParams: [...OuterParams]) => (innerParams: [...InnerParams]) => Promise<boolean>

type OuterParamsOf<T> = T extends OuterFunction<infer Deps, any[]> ? Deps : never
type InnerParamsOf<T> = T extends OuterFunction<any[], infer Params> ? Params : never

function someFunction<T extends OuterFunction<any[], any[]>>(
  func: T, 
  deps: [...OuterParamsOf<T>]
): (innerParams: [...InnerParamsOf<T>]) => Promise<boolean> {
  return func(...deps)
}

两种方案都能让TypeScript准确识别返回类型，消除报错。

内容的提问来源于stack exchange，提问作者Naoric