flowtype：将函数的返回类型更改为void

Question

I need to change the return type of a function to void.

For example:

import { func_one } from './examples';

// FuncOneType is a function type with the same signature
// as func_one.
type FuncOneType = typeof func_one;

// FuncOneTypeWithoutReturn is similar to FuncOneType but 
// the return type is void. This works.
const func_one_void = (...args) => {func_one(...args)};
type FuncOneTypeVoidReturn = typeof func_one_void;

// I need something like to avoid the intermediate
// function func_one_void. How can I achieve this?
type FuncOneTypeVoidReturn = $Arguments<func_one> => void;

Answer 1

We chatted about this a bit on a GitHub Issue, but wanted to put the answer here just in case someone lands on this question via Google. Flow currently has no way to represent "the type of all the arguments to a function" or to type a higher-order function that takes another function and preserves its argument types but returns a different type.

(Also note that the func_one_void example above doesn't really work as you would like. The arguments to that function no longer have any types but are instead just any .)