Re: When is "z" != "z" ?

Martin Alterisio wrote:

2006/6/4, Rasmus Lerdorf <rasmus@lerdorf.com>:

tedd wrote:

Hi gang: Here's your opportunity to pound me again for not knowing the basics of

php.

I vaguely remember something like this being discussed a while back,

but can't find the reference.

In any event, if one uses -- for ($i="a"; $i<"z"; $i++) { echo($i); } -- it stops at "y" But, if you use -- for ($i="a"; $i<="z"; $i++) { echo($i); } -- it prints considerably more characters after "z" than what one would

normally expect -- why is that?

Just stopping at "z" would seem to make more sense, wouldn't it? After

all, when $i = "z" in the first expression, then wouldn't $i be equal to "z" in the second expression and thus halt the operation?

What am I missing here?

It's a bit of a quirk. "z"++ is "aa" and "aa" < "z". I would guess this would loop until until just before "za" which would be "yz".

What? "z"++ < "z" returns true? =S Is there a reason for this? Why aren't they handled like C chars?

Because PHP is not C. If you want them handled by their char codes, then do so with chr() calls.

It's a bit like looping through the hexadecimal characters. You would

have the same effect. However instead of being base-16 with 0-9-a-f you have base-26 using a-z.

0xF++ would be 0x10 which is greater than 0xF (0xF < 0x10). It's not the same.

Sure it is. If you treat 0x10 as a string the same way he is treating these as strings, then 0x10 is going to be smaller than 0xF. The fact is that there is just no right answer for what "z"++ should do. -Rasmus