JavaScript "arguments" -passing a function with other parameters

Question

Function multiply below is passed a callback function "addOne". Function multiply returns [3,5,7].

Since the callback function addOne is one of the arguments of function multiply, and since the arguments of function multiply are all multiplied by 2, why doesnt the callback function itself (i.e. addOne) get multiplied by 2? In other words, instead of function multiply returning [3,5,7], I would have expected it to return [3,5,7, NAN] since the function is not a number?

Does JavaScript interpreter just somehow know not to multiply it by 2?

function addOne(a) {
return a + 1;
}

function multiply(a,b,c, callback) {
    var i, ar = [];
    for (i = 0; i < 3; i++) {
        ar[i] = callback(arguments[i] * 2);
    }
    return ar;
}



myarr = multiply(1,2,3, addOne);
myarr;

alex · Accepted Answer · 2011-04-14 11:25:41Z

4

Because your loop's condition is <3 (hehe) which means it won't subscript the callback (the last element).

You should consider making the callback the first argument always, and splitting the arguments like so...

var argumentsArray = Array.prototype.slice.call(arguments),
    callback = argumentsArray.shift();

jsFiddle.

Then, callback has your callback which you can call with call(), apply() or plain (), and argumentsArray has the remainder of your arguments as a proper array.

edited Apr 14, 2011 at 11:25

answered Apr 14, 2011 at 6:38

alex

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4 Comments

mjmitche Over a year ago

I know generally what "call" does, but can you explain how it is working exactly in the example code you wrote. Thank you.

alex Over a year ago

@mjmitche The arguments variable is an object, not an array, although it exposes some array like properties. To be able to call shift() on it, we need it as a real array. To do that, we can call the Array's slice() on the arguments array, using call() to pass the object as this. When using slice() with no arguments, it returns a copy of the object as an array (or a copy of an array if used on a standard array).

mjmitche Over a year ago

thanks, so if you need arguments as an array to call shift() on it, and if slice() returns a copy of the object (i.e. arguments) as an array, I don

t understand why you need to use "call() to pass the object as this." Slice turns it into an array, and shift works on the array. Why can

t it be that simple. If you have time to clarify, please do. Thanks again.

alex Over a year ago

@mjmitche Because when you do slice() normally, the array is this. call() lets you call a function and set the this variable with its first argument.

JohnP · Accepted Answer · 2011-04-14 06:40:11Z

0

This line for (i = 0; i < 3; i++) { is protecting you.

You stop before it hits the callback argument

answered Apr 14, 2011 at 6:40

JohnP

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Comments

Yasser · Accepted Answer · 2011-04-14 06:41:02Z

0

Because you are running the the loop for the first 3 args only. i < 3 runs for i=0, i=1,i=2

answered Apr 14, 2011 at 6:41

Yasser

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