outbreak » Javascript compare (written on May 18th, 2006 by Marko Mrdjenovic-2)

published on
May 18th, 2006
by Marko Mrdjenovic

Today I found something in Javascript that just doesn’t make any sense. Ok, maybe some other things don’t make sense either, but I think this one just beats all of them.

Let’s say we have a string and a number:

a = "1";
b = 1;

Now we want to compare these:

alert((a>b) ' : ' (aa) ' : ' (b==a));

Both will give you ‘false : false : true’.

If we change a to “2″

a = "2";
b = 1;

we get a ‘true : false : false’. We’re still OK.

Now let’s change a to “a”

a = "a";
b = 1;

The weird thing is we get ‘false : false : false’ on both alerts.

In typestrict languages you’d get an error. Or at least you should. In php you get a ‘false : true : false’ and ‘true : false : false’ which would mean they decided that all strings are smaller than numbers. This is weird – if you make b a string “1″ the return values change – but at least they made a choice.

For what I can comprehend Javascript casts both values to int and I can understand that. The core of the problem is obviously with the NaN value. What I can’t understand is how something’s not smaller, not bigger and not equal at the same time? I’m sure someone can explain this…

posted on
May 19th, 2006
by Marko

Is a dream bigger than an apple?

That’s how :)

posted on
May 19th, 2006
by Marko Mrdjenovic

I could accept that in typestrict languages. And I would even accept it here, but how come then

var a = ['a',1];
alert(a.sort().join());
var b = [1,'a'];
alert(b.sort().join());

both return ’1, a’. Doesn’t this mean that it decides that 1 is smaller than ‘a’ ?

I don’t know, it might. But my point is that if you go comparing uncomparable items, then you can reasonable expect undetermined result.

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Javascript compare

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