javascript array.sort with undefined values

How does Array.prototype.sort handle undefined values in an array?

var array = [1,undefined,2,undefined,3,undefined,4];
var array2 = [];
array2[0] = 1;array2[2] = 2;array2[4] = 3;array2[6] = 4;

When calling array.sort(function(l,r) { ... }); The values undefined are never passed in as l or r.

Can I guarantee that all the undefined values will always go to the end of the array for all browsers?

Would the following loop handle all the non undefined data in an array

array.sort();
for (var i = 0; array[i] !== undefined; i++) {
    // handle array
}

You may assume that no-one declared undefined as a variable.

Answers


Yes, you can safely assume undefined will get moved to the end of the array.

From MDC:

In JavaScript 1.2, this method no longer converts undefined elements to null; instead it sorts them to the high end of the array

From the spec, 15.4.4.11 :

Because non-existent property values always compare greater than undefined property values, and undefined always compares greater than any other value, undefined property values always sort to the end of the result, followed by non-existent property values.


It appears that the undefined values will fall to the bottom of the list. Here's some sample code to show you what happens:

var a = [1,undefined,2,undefined,3,undefined,4];
  a = a.sort();
  for( i = 0 ; i < a.length ; i++ )
  {
    alert( a[i] );
  }

This is, of course, the default behavior of JavaScript. If you overwrite the default behavior then only you will know.


all undefined values will go to the end of the array regardless of their order in declaration. (at least this is how it works in chrome's js engine)


Safari just deletes undefined property when calling sort(); method


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