javascript closure not working with array properties?

Question

 var a = [false, 3];
 console.info('a: '+a);
     (function(item){
            var jamie = item;
        jamie [1]--;
         console.info('jamie:  '+jamie);
     })(a);
 console.info('a: '+a);

In my mind a[1] should always equal 3 in this javascript.

And when not using an array it works as expected:

 var a = 3;
 console.info('b: '+a);
     (function(item){
            var jamie = item;
        jamie--;
         console.info('jamie b:  '+jamie);
     })(a);

console.info('b: '+a);

Why does a[1] output 2 after I have ran this JS? Fiddle of said problem :O

SeanCannon · Accepted Answer · 2014-05-21 21:15:04Z

3

Because you're providing a reference to the object not passing it by value;

If you want to clone the object, here is a helpful function to allow it:

function clone(obj) {
    if (null == obj || "object" != typeof obj) return obj;
    var copy = obj.constructor();
    for (var attr in obj) {
        if (obj.hasOwnProperty(attr)) copy[attr] = obj[attr];
    }
    return copy;
}

var a = [false, 3];
console.info(a); // [false, 3];
    (function(item){
        var jamie = clone(item);
        jamie [1]--;
        console.log(jamie); // [false, 2];
    })(a);
console.info(a); // [false, 3];

edited May 21, 2014 at 21:15

answered May 21, 2014 at 21:08

SeanCannon

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7 Comments

Pointy Over a year ago

He is passing "a" by value; its value is a reference to the array object.

SeanCannon Over a year ago

Right, what Pointy said :)

Pointy Over a year ago

Sorry to be so pedantic but blurring the meaning of the word "reference" with the term "pass by reference" has been driving me slightly crazy for decades :)

Jamie Hutber Over a year ago

Boy I need to get better at JS.

Jamie Hutber Over a year ago

heh, your "backwards" if statement variables threw me off. You sly cat

|

Pointy · Accepted Answer · 2014-05-21 21:09:39Z

1

This has nothing to do with closures.

You're passing a reference to an array into a function, and the function is using that reference to alter the value of a property of the array.

When you assign an object (arrays are objects) to a variable, you're setting the value of the variable to a reference to the object. You can't manipulate objects as values directly; all you have are references.

answered May 21, 2014 at 21:09

Pointy

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4 Comments

Jamie Hutber Over a year ago

For clarity could you please explain how one might pass it by value and not by reference?

Pointy Over a year ago

@JamieHutber you are passing by value. The value of variable "a" is a reference to the array. It is not possible to pass the array itself as a value without your explicitly making a copy of the array. (In that case, you'd be passing a reference to the copy of the array.)

Jamie Hutber Over a year ago

I see, so because its an object (all objects are passed by reference) I can not pass it as a new value. Interesting, interesting that I knew this in theory. But this is the first time it has confused me. So to not alter the array's details I should put a[1] in as the value of the annoyomus function.

Pointy Over a year ago

@JamieHutber yes, or pass a.slice(0) - that'd pass a copy (a shallow copy) of the array.

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