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【Q491】如何实现一个 Promise.all

如何实现一个 Promise.all

更多描述

await Promise.all([1, 2, 3]);
//-> [1, 2, 3]
 
await Promise.all([1, Promise.resolve(2), 3]);
//-> [1, 2, 3]
 
await Promise.all([1, Promise.resolve(2)]);
//-> [1, 2]
 
await Promise.all([1, Promise.reject(2)]);
//-> Throw Error: 2

Issue 欢迎在 Gtihub Issue 中回答此问题: Issue 500 (opens in a new tab)

Author 回答者: shfshanyue (opens in a new tab)

代码可在线调试: 如何实现一个 Promise.all - codepen (opens in a new tab)

有一次头条面试,一道手写题目是:如何手写实现 promise.all

我从来没有想过要手写实现 promise.all 函数,稍微一想,大概就是维护一个数组,把所有 promise 给 resolve 了之后都扔进去,这有啥子好问的。没想到,一上手还稍微有点棘手。

先来看一个示例吧:

await Promise.all([1, Promise.resolve(2)]);
//-> [1, 2]
 
await Promise.all([1, Promise.reject(2)]);
//-> Throw Error: 2
  1. 传入一个 Iterable,但大部分情况下是数组,以下以数组代替
  2. 传入一个数组,其中可包含 Promise,也可包含普通数据
  3. 数组中 Prmise 并行执行
  4. 但凡有一个 Promise 被 Reject 掉,Promise.all 失败
  5. 保持输出数组位置与输入数组一致
  6. 所有数据 resolve 之后,返回结果
function pAll(_promises) {
  return new Promise((resolve, reject) => {
    // Iterable => Array
    const promises = Array.from(_promises);
    // 结果用一个数组维护
    const r = [];
    const len = promises.length;
    let count = 0;
    for (let i = 0; i < len; i++) {
      // Promise.resolve 确保把所有数据都转化为 Promise
      Promise.resolve(promises[i])
        .then((o) => {
          // 因为 promise 是异步的,保持数组一一对应
          r[i] = o;
 
          // 如果数组中所有 promise 都完成,则返回结果数组
          if (++count === len) {
            resolve(r);
          }
          // 当发生异常时,直接 reject
        })
        .catch((e) => reject(e));
    }
  });
}

为了测试,实现一个 sleep 函数

const sleep = (seconds) =>
  new Promise((resolve) => setTimeout(() => resolve(seconds), seconds));

以下示例进行测试,没有问题

pAll([1, 2, 3]).then((o) => console.log(o));
pAll([sleep(3000), sleep(2000), sleep(1000)]).then((o) => console.log(o));
pAll([sleep(3000), sleep(2000), sleep(1000), Promise.reject(10000)])
  .then((o) => console.log(o))
  .catch((e) => console.log(e, "<- Error"));

Author 回答者: haotie1990 (opens in a new tab)

Promise.all = function (promises) {
  const len = promises.length;
  const result = new Array(len);
  let countDone = 0;
  return new Promise((resolve, reject) => {
    if (len === 0) {
      resolve(result);
    }
    for (let i = 0; i < len; i++) {
      const promise = promises[i];
      Promise.resolve(promise).then(
        (data) => {
          result[i] = data;
          countDone++;
          if (countDone === len) {
            resolve(result);
          }
        },
        (error) => {
          reject(error);
        },
      );
    }
  });
};

Author 回答者: Vi-jay (opens in a new tab)

function all(arr) {
  return new Promise((resolve, reject) => {
    const results = [];
    arr.forEach(async (val) => {
      Promise.resolve(val)
        .then((data) => {
          results.push(data);
          if (results.length === arr.length) return resolve(results);
        })
        .catch((e) => reject(e));
    });
  });
}

Author 回答者: voezy (opens in a new tab)

function all(arr) {
  return new Promise((resolve, reject) => {
    const results = [];
    arr.forEach(async (val) => {
      Promise.resolve(val)
        .then((data) => {
          results.push(data);
          if (results.length === arr.length) return resolve(results);
        })
        .catch((e) => reject(e));
    });
  });
}

@Vi-jay 这个写法里,每个 promise 结束时间可能不一样,直接用 push 而不是数组循环索引去记录的话感觉可能会有问题。

Author 回答者: justorez (opens in a new tab)

Promise.all = function (promises = []) {
  return new Promise((resolve, reject) => {
    let count = 0;
    let values = new Array(promises.length);
    const collect = (index) => (value) => {
      values[index] = value;
      ++count === promises.length && resolve(values);
    };
    promises.forEach((promise, i) => {
      if (promise instanceof Promise) {
        promise.then(collect(i), reject);
      } else {
        collect(i)(promise);
      }
    });
  });
};