foo function returns an object. typescript is know about the typing that return prop1 and prop2 wrap in promise:

const foo: ({ a, b }: {
    a: any;
    b: any;
}) => Promise<{
    prop1: string;
    prop2: string;
}>

But when I wrap foo function with fn function, then b variable that create by the callback is losing the typing and become const b: (a?: any, b?: any, c?: any) => Promise<any>.

My question is how to fix that so typescript can know about the typing? what I try to get is b variable will be the typeof the function I pass. in this case is foo function. Like that:

const b: Promise<{
    prop1: string;
    prop2: string;
}>

I try to use T: const fn = <T>(cb: T) => { but it throw an error on the callback and not sure what is mean. I don't know if I correct about this.

stackblitz

const foo = async ({ a, b }) => {
  console.log({ a, b });

  return {
    prop1: 'prop1',
    prop2: 'prop2',
  };
};

const fn = (cb) => {
  return async (a = null, b = null, c = null) => {
    const results = cb({ a, b });
    if (results) {
      return c();
    }
  };
};

const bar = () => fn(foo);

const b = bar(); // const b: (a?: any, b?: any, c?: any) => Promise<any>

CodePudding user response：

if I understand you question correctly you can do something like this

const fn = <T>(cb: ({a, b}: any) => Promise<T>) => {
  return async (a = null, b = null, c:() => T | null = null) => {
    const results =  cb({ a, b });
    if (results) {
      return c();
    }
  };
};