Question

拥有这些嵌套函数：

Promise.all( mapDomains( x => x.close() ) )

我考虑过编写它们：

compose(Promise.all, mapDomains)( x => x.close() )

但是上面的代码在没有将Promise.all绑定到自身的情况下失败了。这是修复：

let f1 = Promise.all.bind(Promise)
compose(f1, mapDomains)( x => x.close() )

虽然我理解这是关于this关键字如何在Javascript中运行的所有内容，但我想知道：为什么.all已经不受限制了？这有什么价值吗？

Answer 1

我的猜测是这样的问题：

var Promise = {
  all: function() {
    console.log(this.text)
  },
  text: 'some text'
};

Promise.all(); //some text

function compose(f) {
  f(); //Here 'this' will not be what you expect, unless you've bound it
}

compose(Promise.all); //undefined
compose(Promise.all.bind(Promise)); //some text

Answer 2

您可能希望将其与另一个Promise实现一起使用

＆＃13;

// just a demo
// this should be another library to make sense
class MyPromise extends Promise { 
  then(onFulfilled, onRejected) {
    const wrappedOnFulfilled = function(arg) {
       console.log(`I was fulfilled with ${arg}`)
       return onFulfilled(arg)
    }
    return super.then.call(this, wrappedOnFulfilled, onRejected)
  }
}

var all = Promise.all

const myPromise = all
  .call(MyPromise, [MyPromise.resolve(1), MyPromise.resolve(2)])
  .then(([x,y]) => console.log(x, y))

console.log(myPromise instanceof MyPromise)

＆＃13;

为什么Promise.all不受Promise约束？

2 个答案: