定义一个通用的 function 类型来支持 TypeScript 中任意数量的任意类型的参数

Question

In typescript we can define a function type like:

type FunctionHandler = {
   (param1: string, param2: number): string
}

But I want to declare a generic type that can take any number of params with their type as we do in variable typings like following:

type Variable<Props> = {
   [p in keyof Props]: Props[p]
}

I want something like:

type FunctionHandler<Props> = {
   ([p in keyof Props]: Props[p]): void | boolean
}

But I am not able to do it, can anyone help me with that?

Answer 1

You can use a rest param like so to have the function accept any number of arguments of any type:

type FunctionHandler = {
  (...params: any): string
}

See also

• https://www.typescriptlang.org/docs/handbook/functions.html#rest-parameters

Answer 2

Try this:

type FunctionHandler<TArgs extends any[]> = (...params: TArgs) => void;

Or, an even more generic type, including the return type:

type FunctionHandler<TArgs extends any[], TReturn extends any> = (...params: TArgs) => TReturn;

Usage:

function doSomthingWithHandler(handler: FunctionHandler){
  const result = handler(123); // result will be of type `string`
}

doSomthingWithHandler((arg: number) => 'some string');