TypeScript sometimes struggles to accurately deduce the precise return type of a function

I have a function called myCallback

const myCallback = (param: number) => {
  // doSomething
};

Now, I want to create another function, useMyCallback, that may or may not receive a parameter and will return myCallback either bound or unmodified:

const useMyCallback = (optionalParam?: number) => {
  return optionalParam === undefined
    ? myCallback
    : () => myCallback(optionalParam);
};

The issue is that regardless of whether I pass in a parameter or not, the return value of myCallback is always typed as (param: number) => void

const myCallback = (param: number) => {
  // doSomething
};

const useMyCallback = (optionalParam?: number) => {
  return optionalParam === undefined
    ? myCallback
    : () => myCallback(optionalParam);
};

const unboundCallback = useMyCallback(); // returns (param: number) => void as expected
const boundCallback = useMyCallback(1);  // returns (param: number) => void, when expecting () => void

Is there a way to achieve what I want without having to cast each useMyCallback call?

Answer №1

We have a few things to discuss about this topic.


First, let's consider you need to refer to this type:

type Fn = ((arg: number) => void ) | (() => void)

In JavaScript, you can pass more arguments to a function than required. However, TypeScript enforces that you provide all the necessary arguments. Since the mentioned type could be a function that needs an argument, it must be called with an argument for type safety.

This is why boundCallback expects an argument.


Secondly, TypeScript does not run your code directly, so you need to indicate explicitly when omitting an argument affects the return type. One effective way to achieve this is through the use of function overloads.

// Overload signatures
function useMyCallback(): typeof myCallback
function useMyCallback(optionalParam?: number): () => void

// Implementation
function useMyCallback(optionalParam?: number) {
  return typeof optionalParam === 'undefined'
    ? myCallback
    : () => myCallback(optionalParam);
};

These overloads inform TypeScript about the specific return types corresponding to different argument patterns. The actual implementation of the function accommodates various arguments and provides the appropriate return values accordingly.

By doing so, TypeScript can understand that changing the arguments will result in distinct return values.

This approach eliminates the issue I mentioned at the beginning of this response since the return type is always a single function type that adapts based on the supplied arguments.

Now, the following scenarios are valid:

const unboundCallback = useMyCallback(); //-> (param: number) => void
unboundCallback(0)

const boundCallback = useMyCallback(1); //-> () => void
boundCallback() 

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