Why does my function change the argument when it modifies an array?

This question already has an answer here:

Is JavaScript a pass-by-reference or pass-by-value language? 29 answers

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Objects, including arrays are assigned by reference in ECMAScript. So, when you modify the array in the function you are modifying the same array as you passed into the function, not a new copy of the array.

A quick way to perform a shallow copy of an array is using slice(0) (and in some modern engines you can leave out the 0 ). See this answer about copying an array.

Also see this answer for some examples.

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var foo = [1,2,3];

bar = function1(foo);
bar = function2(foo);
bar = function3(foo);


function function1(newFoo){
    return [newFoo,'a',1];
} //foo after function1 = 1,2,3


function function2(newFoo){
    var otherFoo = newFoo.slice(0);

    otherFoo[0] = 'a';
    return otherFoo;
} //foo after function2 = 1,2,3


function function3(newFoo){
    var otherFoo = newFoo.slice(0);

    otherFoo.push('4');
    return otherFoo;
} //foo after function2 = 1,2,3

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In function2 and function3, you're modifying the input variable (by changing its contents).

In function1, you're not modifying the input variable; you're merely assigning it to point to something else with = .

Objects (including arrays) are passed by reference - if you don't want to modify the input, the safest thing to do is make a copy of the input at the beginning of the function (before potentially modifying it). For arrays, you can do this with the slice function: var copy = input.slice(0)

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