承诺中的异步函数抛出错误并且不拒绝[重复]

Question

如果有一个带有异步函数的 Promise，并且如果在异步函数中发生错误，则 Promise 不会捕获但会引发错误并使应用程序崩溃，我不明白。

显然我想处理这个错误，你知道为什么承诺会这样吗？有什么办法可以解决它？

谢谢

// this promise will have an error since param is not defined,
// and the promise won't be caught
function randomPromise(param) {
    return new Promise((resolve, reject) => {
        setTimeout(() => {
            param[0] = 11;
        }, 2000);
    });
}

randomPromise()
.then(() => {
    console.log('nothing');
})
.catch((e) => {
    console.log('with set timeout or any async function in the promise, the error caused by \'param[0] = 11;\' wont bring the control here into the catch block just throws an error and crashes the application');
    console.log(e);
});

// this promise will have an error since param is not defined
// but the promise will be caught
function randomPromiseGoesToCatchBlock(param) {
    return new Promise((resolve, reject) => {
        param[0] = 11;
    });
}

randomPromiseGoesToCatchBlock()
.then(() => {
    console.log('nothing');
})
.catch((e) => {
    console.log('without the setTimeout function or any async function the error caused by \'param[0] = 11;\' brings the control here into the catch block');
    console.log(e);
});

Answer 1

Promise 构造函数中抛出的错误异步需要显式try/catched 以便可以调用reject，以便Promise 控制流量可以转移到 Promise 的catch。例如：

// this promise will have an error since param is not defined, and the promise wont be catched
function randomPromise(param) {
  return new Promise((resolve, reject) => {
    setTimeout(() => {
      try {
        param[0] = 11;
      } catch(e) {
        reject(e);
      }
    }, 2000);
  });
}

randomPromise()
  .catch((e) => {
    console.log(e.message);
  });

否则，resolve 和 reject 都不会被调用，并且错误是异步的，因此创建 Promise 的线程已经结束，因此解释器不知道抛出的错误应该拒绝该 Promise没有你明确告诉它。

相比之下，Promise 构造函数中同步抛出的错误将自动导致构造的 Promise 立即拒绝。