javascript - How does function.apply.bind work in the following code? - Stack Overflow

admin管理员组

文章数量:1186688

So I get that an array of [200,599] is returned from the promise and the callback function inside spread is being passed into Function.apply.bind, but now I'm lost. How is the array of [200,599] split into x and y? How exactly does the apply.bind work?

function getY(x) {
        return new Promise( function(resolve,reject){
            setTimeout( function(){
                resolve( (3 * x) - 1 );
            }, 100 );
        } );
    }

function foo(bar,baz) {
    var x = bar * baz;

    // return both promises
    return [
        Promise.resolve( x ),
        getY( x )
    ];
}

function spread(fn) {
    return Function.apply.bind( fn, null );
}

Promise.all(
    foo( 10, 20 )
)
.then(
    spread( function(x,y){
        console.log( x, y );    // 200 599
    } )
)

So I get that an array of [200,599] is returned from the promise and the callback function inside spread is being passed into Function.apply.bind, but now I'm lost. How is the array of [200,599] split into x and y? How exactly does the apply.bind work?

function getY(x) {
        return new Promise( function(resolve,reject){
            setTimeout( function(){
                resolve( (3 * x) - 1 );
            }, 100 );
        } );
    }

function foo(bar,baz) {
    var x = bar * baz;

    // return both promises
    return [
        Promise.resolve( x ),
        getY( x )
    ];
}

function spread(fn) {
    return Function.apply.bind( fn, null );
}

Promise.all(
    foo( 10, 20 )
)
.then(
    spread( function(x,y){
        console.log( x, y );    // 200 599
    } )
)

• javascript
• promise
• ecmascript-6
• bind
• apply

Share Improve this question Follow asked Oct 6, 2016 at 22:59 Gwater17Gwater17 2,31433 gold badges2323 silver badges4545 bronze badges 1

• You know what they do on their own, right? – Bergi Commented Oct 6, 2016 at 23:24

Add a comment  | 

4 Answers 4

Sorted by: Reset to default 14

.apply() is a method on function objects. Like so:

console.log(Math.max.apply(null, [1, 2, 3])); // 3

.apply() accepts two arguments, the context (what would become this inside of the target function) and an iterable of arguments (usually an array, but the arguments array like also works).

---

.bind() is a method on function objects. Like so:

const x = {
  foo: function() {
    console.log(this.x);
  },
  x: 42
}

var myFoo = x.foo.bind({x: 5});

x.foo(); // 42
myFoo(); // 5

.bind() takes a context (what would become this), and optionally, additional arguments, and returns a new function, with the context bound, and the additional arguments locked

---

Since .apply() is a function in on itself, it can be bound with .bind(), like so:

Function.prototype.apply.bind(fn, null);

Meaning that the this of .apply() would be fn and the first argument to .apply() would be null. Meaning that it would look like this:

fn.apply(null, args)

Which would spread the parameters from an array.

Spread takes a function and binds the apply method to, partially applying the null argument. So in short,

spread(fn)

is transformed to

args => fn.apply(null, args)

which is the same as using the ES6 spread syntax

args => fn(...args)

where the function got its name from.

---

If you want the long answer, remember what bind does:

method.bind(context, ...args1)

returns a function that works like

(...args2) => method.call(context, ...args1, ...args2)

In our case, method is apply, the context is fn and the first arguments are null, so the call

Function.apply.bind( fn, null )

will create a function that works like

(...args2) => (Function.apply).call(fn, null, ...args2)

which is equivalent to the fn.apply(…) call above, given that apply is the method inherited from Function.prototype in both accesses.

The spread function is just a utility function to convert an array, into parameters passed to a function. The apply is doing the converting, and the bind is binding it to the calling function so that the "this" context is connected to same function.

To see how spread is working in a simpler form ->

spread(function (x,y){console.log(x,y);})([1,2])

You will get the answer, 1 2, as 1 is passed to x, and 2 is passed to y.

So in your example your promise.all is returning an array of resolved promises. These are then getting mapped to parameters to your function(x,y).

The reason it works is the "destructuring" nature of apply (if given an array of values, they would be provided spreaded to the function you use apply on).

Now back to your code when calling bind on apply, the value returned is a function which returns the same function provided to bind, the only thing different is when executed it would be called using apply (with an array as thisArg in your case), but it isn't going to be executed until you call it. In your case when the promise has resolved, the function provided tothen woule be executed with an array of arguments provided by Promise resolution.

function spread(fn){
   let boundedFn = fn.bind(fn)

   return function(arg){
      return boundedFn.apply(null,arg)
   }
}
   
spread((x, y, c) => console.log(x, y, c))([1,2,3])

// or

Promise.resolve([6, 5]).then(spread((a, b) => console.log(a, b)))

The reason bind is provided (in your code) with null as second param is to prevent the array provided by the caller from being given to apply as its first param, which reserved for this.

本文标签： javascriptHow does functionapplybind work in the following codeStack Overflow