Catching an exception from the setTimeout function executed inside an instance of a Promise object [duplicate]

Answer 1

The timeout code is executed outside the promise context. What you want to do is call 'reject' with the error:

const TIMEOUT = 1000;

const mySecondPromise = new Promise((resolve, reject) => {
  setTimeout(() => {
    reject(new Error(`Error in asynchronous function`));
  },
  TIMEOUT
  );
});

mySecondPromise
  .catch(() => console.log(`This code WILL be executed`));

Answer 2

After some discussions and comments, my task was solved by this approach:

const someFunction = timeout => new Promise(resolve => setTimeout(resolve, timeout));

someFunction(timeout).then(() => doSomething()).catch(() => console.log(`This code will not be executed`));

If an exception occurs in an asynchronous function:

someFunction(timeout).then(ErrorEvent).catch(() => console.log(`This code will not be executed`));

Such solutions are still possible:

const testF = () => { throw new Error(`Упс! Бросили исключение.`) }

const genPromise = () => new Promise((resolve, reject) => {
    setTimeout(() => {
        try {
            const res = testF()
            resolve(res)
        } catch (e) { reject(e) }
    }, 1e3)
})

t1: {
    const log = console.log.bind(console, 't1:')
    const _then = log.bind(console, 'then:')
    const _catch = log.bind(console, 'catch:')

    genPromise()
        .then(_then)
        .catch(_catch)
        .then(_then)
}

t2: {
    const log = console.log.bind(console, 't2:')
    const _then = log.bind(console, 'then:')
    const _catch = log.bind(console, 'catch:')

    void async function () {
        let pRes = null
        try {
            pRes = await genPromise()
        } catch (e) {
            _catch(e.message)
        }
        _then(pRes)
    }()
}