Simple job queue

main.js

const queue = [];

let dispatching;
async function dispatch() {
  if (dispatching) return;
  dispatching = true;
  for (let task; (task = queue.shift()); await Promise.resolve(task()).catch(console.error));
  dispatching = false;
}

function postJob(f) {
  if (typeof f !== 'function') return;
  queue.push(f);
  dispatch();
}

const sleep = ms => new Promise(resolve => setTimeout(resolve, ms));
postJob(async () => { await sleep(1000); console.log('foo'); });
postJob(async () => { await sleep(1000); console.log('bar'); throw new Error(); });
postJob(() => console.log('baz'));
postJob(() => console.log('qux'));

dispatch 里的 for 整的阳气啊

@NoirVoider 回过头来看当时写复杂了，可以直接

let p = Promise.resolve();

const postJob = (f) => {
  if (typeof f !== "function") throw new TypeError("f is not a function");
  p = p.then(f).catch(console.error);
};