Code still executed after return statement in catch?

Question

console.log('1');
await get()
    .then()
    .catch(() => {
        console.log('2');

        return;
    });
console.log('3');

Why did console logs 1 2 3? I thought after return statement, the code will be not executed?

Answer 1

return will only terminate the current function. Here, the function that gets terminated by the return is the .catch callback.

Since .catch returns a resolved Promise, the awaited Promise chain will resolve when the catch resolves.

If you want to stop the outer function when the catch runs, have your catch return something that you check outside, eg:

(async() => {
  console.log('1');
  const result = await Promise.reject()
    .catch(() => {
      console.log('2');

      return 2;
    });
  if (result === 2) {
    return;
  }
  console.log('3');
})();

Or have the .catch throw an error, so that the outer Promise chain rejects. Since it's being awaited, the whole outer Promise will then reject:

const fn = async () => {
  console.log('1');
  const result = await Promise.reject()
    .catch(() => {
      console.log('2');
      throw new Error();
    });
  console.log('3');
};

fn()
  .then(() => console.log('resolved'))
  .catch(() => console.log('rejected'))

If your .catch doesn't do anything substantial but you want this sort of behavior, it's usually a good idea to omit it entirely. That way, when there's an error, the awaited Promise will reject, the function will terminate, and the error can be caught by the appropriate consumer.

const fn = async () => {
  console.log('1');
  const result = await Promise.reject();
  console.log('3');
};

fn()
  .then(() => console.log('resolved'))
  .catch(() => console.log('rejected'))