25 Matching Annotations
  1. Jan 2022
    1. My gut told me calling an async function from the setTimeout callback was a bad thing. Since the setTimeout machinery ignores the return value of the function, there is no way it was awaiting on it. This means that there will be an unhandled promise. An unhandled promise could mean problems if the function called in the callback takes a long time to complete or throws an error.
    1. const rejectedP = Promise.reject('-'); const finallyP = rejectedP.finally(); const result1 = rejectedP; const result2 = new Promise(resolve => { const rejectedP = Promise.reject('-'); const finallyP = rejectedP.finally(); resolve(rejectedP); }); we can see that the first snippet creates two promises (result1 and rejectedP being the same) while the second snippet creates three promises. All of these promises are rejected, but the rejectedP rejection is handled by the callbacks attached to it, both through ….finally() and resolve(…) (which internally does ….then(resolve, reject)). finallyP is the promise whose rejection is not handled in the both examples. In the second example, result2 is a promise distinct from rejectedP that is also not handled, causing the second event.
    1. You basically did var a = promise.then(…); var b = promise.catch(…); creating a branch in the chain. If promise is getting rejected now, the catch callback will be called and b will be a fulfilled promise just fine, but the a promise is getting rejected too and nobody handles that. Instead, you should use both arguments of then and write Requirement.create({id: id, data: req.body.data, deleted: false}) .then(requirement => { res.json(requirement); }, reason => { let err = {'error': reason}; res.json(err); });
  2. www.npmjs.com www.npmjs.com
  3. Oct 2020
    1. Another example:

      const expensiveOperation = async (value) => {
        // return Promise.resolve(value)
          // console.log('value:', value)
          await sleep(1000)
          console.log('expensiveOperation: value:', value, 'finished')
          return value
      }
      
      var expensiveOperationDebounce = debounce(expensiveOperation, 100);
      
      // for (let num of [1, 2]) {
      //   expensiveOperationDebounce(num).then(value => {
      //     console.log(value)
      //   })
      // }
      (async () => { await sleep(0   ); console.log(await expensiveOperationDebounce(1)) })();
      (async () => { await sleep(200 ); console.log(await expensiveOperationDebounce(2)) })();
      (async () => { await sleep(1300); console.log(await expensiveOperationDebounce(3)) })();
      // setTimeout(async () => {
      //   console.log(await expensiveOperationDebounce(3))
      // }, 1300)
      

      Outputs: 1, 2, 3

      Why, if I change it to:

      (async () => { await sleep(0   ); console.log(await expensiveOperationDebounce(1)) })();
      (async () => { await sleep(200 ); console.log(await expensiveOperationDebounce(2)) })();
      (async () => { await sleep(1100); console.log(await expensiveOperationDebounce(3)) })();
      

      Does it only output 2, 3?

  4. Sep 2020
    1. // LoginApi.js return loginDao.login(username, password).then(function (res) { store.dispatch(loginSuccess()); log.log("[loginApi.login] END"); return true; }, function (err) { store.dispatch(loginFail()); errorUtils.dispatchErrorWithTimeout(errorLogin); log.log(err); return false; }); // never supposed to reject
  5. May 2020
  6. developer.mozilla.org developer.mozilla.org