Question

169

Difference between Array.length = 0 and Array =[]?

rated 0 times [ 173] [ 4] / answers: 1 / hits: 60292 / 14 Years ago, wed, january 26, 2011, 12:00:00

Can some one explain the conceptual difference between both of them. Read somewhere that the second one creates a new array by destroying all references to the existing array and the .length=0 just empties the array. But it didn't work in my case

//Declaration 

var arr = new Array();

The below one is the looping code that executes again and again.

$(#dummy).load(something.php,function(){

   arr.length =0;// expected to empty the array

   $(div).each(function(){

       arr = arr + $(this).html();

   });

});

But if I replace the code with arr =[] in place of arr.length=0 it works fine. Can anyone explain what's happening here.

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aileent

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answered 14 Years ago mckinleyi · Accepted Answer

foo = [] creates a new array and assigns a reference to it to a variable. Any other references are unaffected and still point to the original array.

foo.length = 0 modifies the array itself. If you access it via a different variable, then you still get the modified array.

Read somewhere that the second one creates a new array by destroying all references to the existing array

That is backwards. It creates a new array and doesn't destroy other references.

var foo = [1,2,3];

var bar = [1,2,3];

var foo2 = foo;

var bar2 = bar;

foo = [];

bar.length = 0;

console.log(foo, bar, foo2, bar2);

gives:

[] [] [1, 2, 3] []

arr.length =0;// expected to empty the array

and it does empty the array, at least the first time. After the first time you do this:

arr = arr + $(this).html();

… which overwrites the array with a string.

The length property of a string is read-only, so assigning 0 to it has no effect.