Wrong representation of JavaScript Array Length

Question

If I have an array in JavaScript, say:

var a = [1, 2, 3];
var b = ["Praveen", "Kumar", "Stack", "Overflow"];

If I get the lengths of the above arrays by:

a.length;
b.length;

I get the correct values. ie,

a.length;  // 3
b.length;  // 4

But, if I create another array, where I set my indices like:

c = [];
c[5] = "Five";
c[10] = "Ten";

And then if I query the length, it shows me 11 .

c.length  // 11

Is this wrong? Or is this way JavaScript interprets arrays? Please guide.

Answer 1

The .length is defined to be one greater than the value of the largest numeric index. (It's not just "numeric"; it's 32-bit integer values, but basically numbered properties.)

Conversely, setting the .length property to some numeric value (say, 6 in your example) has the effect of deleting properties whose property name is a number greater than or equal to the value you set it to.

Answer 2

The effect of

var c = [];
c[10] = "foo";

is that c will look like this:

c[0] === undefined
c[1] === undefined
c[2] === undefined
c[3] === undefined
c[4] === undefined
c[5] === undefined
c[6] === undefined
c[7] === undefined
c[8] === undefined
c[9] === undefined
c[10] === "foo"

Having elements 0 through 10, the length of the array is therefore 11.

Answer 3

It's correct. The length depends on the highest index used, not the number of actual items in the array.

Creating an array like this:

c = [];
c[5] = "Five";
c[10] = "Ten";

Has the exact same effect as this:

c = [,,,,,"Five",,,,,"Ten"];

Although the array only has two items, the length is 11 as the highest index used is 10.

The string representation of the array will be [undefined, undefined, undefined, undefined, undefined, "Five", undefined, undefined, undefined, undefined, "Ten"] as it uses the length and shows a value for all indexes up to that.

If you loop the properties in the array, you will only get indexes of the actual items:

for (n in c) console.log(n);

Output:

5
10