How is it possible for a TypeScript function to return something when its return type is void?

I find the book Learning JavaScript to be confusing. It delivers conflicting information regarding the use of void as a return type in functions.

const foo = (s: string): void => {
  return 1;  // ERROR
} 

At first, it states that if a function has a return type of void, it cannot return anything. But then it contradicts itself by explaining that void means the return type should be ignored, allowing functions like:

arr.forEach(foo);

This code snippet uses a function foo with a supposed return type of void, even though it does return something.

To further illustrate this contradiction, consider the following example:

type Foo = (s: string) => number;

const foo: Foo = (s) => {
  return 1;
}

function bar(cb: (s: string) => void): boolean {
  return true;
}

bar(foo);

In this scenario, the function cb is expected to have a return type of void, but it can actually return a value without any issues.

The conflicting explanations from the book leave me puzzled about how these discrepancies can be reconciled.

Answer №1

When looking at the example in TypeScript Playground, it is clear that TypeScript does not give an error. This is because a function signature like (s: string) => number actually fulfills the requirement of another function signature like (s: string) => void, as demonstrated in the proof provided in TypeScript Playground.

The key point to note here is that you can define a function with a specific return type but still use it as if it had a different return type by simply disregarding its return value.

Illustratively:

type Foo = (s: string) => number;

const foo: Foo = (s) => {
    return 1;
}

function bar(cb: (s: string) => void): boolean {
    cb('hello')
    return true;
}

bar(foo);

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