javascript - Code still executed after return statement in catch? - Stack Overflow

admin管理员组

文章数量:1026989

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?

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?

• javascript
• node.js
• typescript

Share Improve this question Follow asked Mar 12, 2020 at 3:38 iamhuynqiamhuynq 5,54922 gold badges1616 silver badges4040 bronze badges

Add a ment  | 

2 Answers 2

Sorted by: Reset to default 6

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'))

It's because the return is in a separate function (the catch handler).

The return only applies to the function it is inside of. It will not affect the outer scope.

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?

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?

• javascript
• node.js
• typescript

Share Improve this question Follow asked Mar 12, 2020 at 3:38 iamhuynqiamhuynq 5,54922 gold badges1616 silver badges4040 bronze badges

Add a ment  | 

2 Answers 2

Sorted by: Reset to default 6

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'))

It's because the return is in a separate function (the catch handler).

The return only applies to the function it is inside of. It will not affect the outer scope.

本文标签： javascriptCode still executed after return statement in catchStack Overflow