求解关于promise的一道面试题？

Question

题目：
实现一个retry函数，使得该defer函数可以在执行5次以后，输出成功与否

const defer = num => new Promise(resolve => setTimeout(() => { console.log(`defer函数：${num}`); resolve() }, 1000));

retry(defer, 5).then(() => { console.log('success') }, () => { console.log('reject') })

我目前的实现：

function retry(func, count) {
  const index = count - 1
  console.log(`retry函数：${index}`)
  if (count === 0) {
    return Promise.resolve()
  }
  const next = () => retry(func.bind(null, index), index)
  return func(index).then(next)
}

请问为什么在能打印如下，defer函数中的值为什么一直都是4，而不是递减到0 ？
另外，请问我的实现是否正确呢？

retry函数：4
defer函数：4
retry函数：3
defer函数：4
retry函数：2
defer函数：4
retry函数：1
defer函数：4
retry函数：0
defer函数：4
retry函数：-1
success

Answer 1

function testA(number) {
   console.log(number);
}

testA(-1) // -1
const testB = testA.bind(null, 5);
testB(-2) // 5
const testC = testB.bind(null, 4);
testC(-3) // 5
const testD = testC.bind(null, 3);
testD(-4) // 5

因为 bind 过后参数就绑定了，再 bind 并不会改变绑定的值

const defer = (num) =>
  new Promise((resolve, reject) =>
    setTimeout(() => {
      console.log(`defer函数：${num}`);
      const value = Math.random();
      return value > 0.5 ? resolve(value) : reject(value);
    }, 1000)
  );

async function retry(fn, count) {
  try {
    console.log(`retry函数：${count}`);
    const result = await fn(count);
    return result;
  } catch (e) {
    if (count <= 0) throw e;
    return retry(fn, count - 1);
  }
}

retry(defer, 5).then(
  (v) => {
    console.log("success", v);
  },
  (e) => {
    console.log("reject", e);
  }
);

Answer 2

const defer = num => new Promise(resolve => setTimeout(() => { console.log(`defer函数：${num}`); resolve() }, 1000));

            function retry(fun, num) {
                if (num == 0){
                    return Promise.resolve();
                }
                return fun(num).then(() => {
                    return retry(fun, --num)
                })
            }

            retry(defer, 5).then(() => { console.log('success') }, () => { console.log('reject') })

这样？

Answer 3

// 实现一个retry函数，使得该defer函数可以在执行5次以后，输出成功与否
const defer = num => new Promise(resolve => setTimeout(() => { console.log(`defer函数：${num}`); resolve() }, 1000));
async function retry(deferFun, num) {
  await deferFun(num) // 等待函数执行完
  --num // 执行完后数字减1
  if (num <= 0) {
    return // 见底就退出
  } else {
    return retry(deferFun, num) // 没见底继续打印
  }
}
retry(defer, 5).then(() => { console.log('success') }, () => { console.log('reject') })

Answer 4

const retry = (handler, count) => {
    var sequence = Promise.resolve();
    for(let i=0; i<count; i++) {
        sequence = sequence.then(() => handler(i));
    }
    return sequence;
}

Answer 5

const retry = async (fun, num) => {
    while (num) {
        await fun(num);
        num--;
    }
};

Answer 6

const defer = num => new Promise(resolve => setTimeout(() => { console.log(`defer函数：${num}`); resolve() }, 100));
 async function retry(func, count) {
     try{
           for(let i=count;i>0;i--){
              let res=await func(i)
            
            }
          return Promise.resolve()
     }
     catch(err){
         console.log(err)
          return Promise.reject()
     }

}
retry(defer, 5).then(() => { console.log('success') }, () => { console.log('reject') })