infer

• 表示待推断类型

type ParamType<T> = T extends (...args: infer P) => any ? P : T;

• 如果T能赋值给(...args: infer P) => any, 结果是(...args: infer P) => any 否则返回T

interface User {
  name: string;
  age: number;
}

type Func = (user: User) => void;

type Param = ParamType<Func>; // Param = User
type AA = ParamType<string>; // string