Typescript: How to define type for a function callback (as any function type, not universal any) used in a method parameter

Question

Currently I have type definition as:

interface Param {
    title: string;
    callback: any;
}

I need something like:

inter         


        

Answer 1

Here's an example of a function that accepts a callback

const sqk = (x: number, callback: ((_: number) => number)): number => {
  // callback will receive a number and expected to return a number
  return callback (x * x);
}

// here our callback will receive a number
sqk(5, function(x) {
  console.log(x); // 25
  return x;       // we must return a number here
});

If you don't care about the return values of callbacks (most people don't know how to utilize them in any effective way), you can use void

const sqk = (x: number, callback: ((_: number) => void)): void => {
  // callback will receive a number, we don't care what it returns
  callback (x * x);
}

// here our callback will receive a number
sqk(5, function(x) {
  console.log(x); // 25
  // void
});

Note, the signature I used for the callback parameter ...

const sqk = (x: number, callback: ((_: number) => number)): number

I would say this is a TypeScript deficiency because we are expected to provide a name for the callback parameters. In this case I used _ because it's not usable inside the sqk function.

However, if you do this

// danger!! don't do this
const sqk = (x: number, callback: ((number) => number)): number

It's valid TypeScript, but it will interpreted as ...

// watch out! typescript will think it means ...
const sqk = (x: number, callback: ((number: any) => number)): number

Ie, TypeScript will think the parameter name is number and the implied type is any. This is obviously not what we intended, but alas, that is how TypeScript works.

So don't forget to provide the parameter names when typing your function parameters... stupid as it might seem.