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I wonder what's wrong with this

$(function() {
var arr1=new Array('A','B','C','D','E','F','G');
var arr2=new Array('F','D','B');
var arr3=arr1;
    for(x=0; x<arr3.length; x++) {
        if(jQuery.inArray(arr3[x],arr2) == -1) {arr3.splice(x, 1);}

    }
    alert(arr1.join(','));
    alert(arr3.join(','));
});

I thought arr1 should still be Array('A','B','C','D','E','F','G'), but after this operation, arr1 becomes arr3. It doesn't make sense to me since the entire operation doesn't touch arr1 at all.

Found the answer after posting this. See duplicating arrays javascript splicing

Community
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Vern
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3 Answers3

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Assignment like var arr3=arr1; are done by reference i.e. arr3 variable now points to same location where arr1 variable points to. The underlined data (in this case array) is the same. Essentially arr1 and arr3 are same. So any change done via one is visible via other. If you want to duplicate the array of simple types use slice(0)

var arr3 = arr1.slice(0);
amit_g
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var x = [3, 4, 5]
var y = [3, 4, 5]
var z = y

x === y // false
y === z // true (y and z reference the same object)

y.push(6)
y // [3, 4, 5, 6]
z // [3, 4, 5, 6]

var x2 = x.slice(0)
x2 === x // false
x2.push('foo')
x2 // [3, 4, 5, 'foo']
x // [3, 4, 5]

Use Array.prototype.slice to copy an array.

davidchambers
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arr1 is a Array Object. when you copy arr1 like var arr3 = arr1, we just get a new reference to the original object. so , when you change the arr3 , arr1 also change.

you should search "how to clone a object", this may help you.

Jin Huang
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