JavaScript IE 9: Custom sort function

Question

In IE 9, if I type this in the console:

[1, 4, 2, 3].sort(function (a, b) { return a < b; });

The resulting array is: [1, 4, 2, 3].

If I do this in FF/Chrome, I get it, reverse-sorted: [4, 3, 2, 1].

How come this doesn't work in IE?

EDIT: Is there a function out there that abstracts these browser differences? In other words, is there a function I can pass function(a, b) { return a < b; } in and get the same result in all browsers? Any open-source stuff?

Possible duplicate of stackoverflow.com/questions/4782893/… — JaredMcAteer
– JaredMcAteer, Commented Sep 21, 2011 at 14:29
I see this answer you have linked but I don't understand how to make it work in IE like it does in FF/Chrome. — IEHaterNumber1
– IEHaterNumber1, Commented Sep 21, 2011 at 14:31

Felix Kling · Accepted Answer · 2011-09-22 17:51:02Z

7

Maybe because the function is supposed to return -1 if a is smaller, 0 if they are equal, and 1 if a is larger (or vice versa if you want reverse order). Edit: In fact, it must be zero, a positive or negative number (as @pimvdb pointed out, and that's what I make use of in the example below). Your function will never return -1 though and that might create problems.

Consider 1 and 3. Your function returns 1 for 1 < 3, which is ok, but returns 0 for 3 < 1. In one case the number are different, in the other you are saying that they are equal.

That FF/Chrome sort it in reverse order might be due to the sorting algorithm they are using.

Try:

[1, 4, 2, 3].sort(function (a, b) { return b - a; });

Update: To substantiate this, we can have a look at the specification, Section 15.4.4.11 Array.prototype.sort(comparefn), where the properties are given which have to be fulfilled by a comparison function:

A function comparefn is a consistent comparison function for a set of values S if all of the requirements below are met for all values a, b, and c (possibly the same value) in the set S: The notation a <_CF b means comparefn(a,b) < 0; a =_CF b means comparefn(a,b) = 0 (of either sign); and a >_CF b means comparefn(a,b) > 0.

Calling comparefn(a,b) always returns the same value v when given a specific pair of values a and b as its two arguments. Furthermore, Type(v) is Number, and v is not NaN. Note that this implies that exactly one of a <_CF b, a =_CF b, and a >_CF b will be true for a given pair of a and b.

Calling comparefn(a,b) does not modify the this object.

a =_CF a (reflexivity)

If a =_CF b, then b =_CF a (symmetry)

If a =_CF b and b =_CF c, then a =_CF c (transitivity of =_CF)

If a <_CF b and b <_CF c, then a <_CF c (transitivity of <_CF)

If a >_CF b and b >_CF c, then a >_CF c (transitivity of >_CF)

NOTE The above conditions are necessary and sufficient to ensure that comparefn divides the set S into equivalence classes and that these equivalence classes are totally ordered.

edited Sep 22, 2011 at 17:51

answered Sep 21, 2011 at 14:32

Felix Kling

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3 Comments

pimvdb Over a year ago

+1 Although it's rather a negative/positive/zero number. Yours doesn't always return -1, 1 or 0 either :)

c69 Over a year ago

ES5 #15.4.4.11 - sais you must return 1, 0 or -1 in custom sort function. ecma-international.org/publications/files/ECMA-ST/Ecma-262.pdf

Felix Kling Over a year ago

@c69: As far as I understood it, this is what the internal compare function returns. A function passed as parameter just has to follow these rules: The notation a <CF b means comparefn(a,b) < 0; a =CF b means comparefn(a,b) = 0 (of either sign); and a >CF b means comparefn(a,b) > 0. But thanks for the reference!

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