Bug #13550 Updated: intval() does not round off doubles correctly

ID: 13550 Updated by: jeroen Reported By: mlemos@acm.org Status: Bogus Bug Type: *Programming Data Structures Operating System: Linux 2.2.18 PHP Version: 4.0.6 New Comment: Manuel ment that intval($double)==intval(round($double)) should be true in the context of $double == round($double) So his problem wasn't with mixing up round and intval, but his problem was that he didn't realize that (string)$double == (string)round($double) does not necessarily imply $double == round($double) (and which is not the case in this case, because the difference being to small to show up in the string representation) Previous Comments: ------------------------------------------------------------------------ [2001-10-05 17:00:03] rasmus@php.net Ok, you could have stated this to begin with: intval($double)!=intval(round($double)) Where are you getting the idea that intval does a round()? intval(3.99) would be 3 round(3.99) would be 4 Therefore intval(round(3.99)) would be 4 and thus often intval($double) != intval(round($double)) ------------------------------------------------------------------------ [2001-10-05 16:54:14] jeroen@php.net <quote> This is a real PHP bug that needs to be fixed, because if: [1]echo $double; prints the same as [2]echo round($double); then [3]echo intval($double); should print the same as [4]echo intval(round($double)); </quote> [1] and [2] _print_ the same, but they aren't. It's because of the precision setting in php.ini (14 significant numbers, IIRC) This is inherently to floating point numbers. Simply don't trust them to the last digit. ------------------------------------------------------------------------ [2001-10-05 16:45:13] jeroen@php.net > Basically the bug is that in PHP: > > intval($double)!=intval(round($double)) converting to integer will round DOWN, while rounding first rounds towards the nearst integer. intval($float) and intval(floor($float)) should be the same for nonnegative $float's, but in the case of round that's not true. So not a bug ------------------------------------------------------------------------ [2001-10-05 16:40:50] manuel@php.net I don't think you are understanding the problem. Basically the bug is that in PHP: intval($double)!=intval(round($double)) round($double) is still a floating point number. The problem is that currently in PHP the intval() floating point rounding rule is not the same as round() rule. This is a real PHP bug that needs to be fixed, because if: echo $double; prints the same as echo round($double); then echo intval($double); should print the same as echo intval(round($double)); ------------------------------------------------------------------------ [2001-10-05 15:56:41] jeroen@php.net No, floating point numbers are ALWAYS inprecise, and you shouldn't trust on their exact values. See the warning on http://www.php.net/manual/en/language.types.float.php The problem is that $scaled is 877.99999999999999 or something, and because of the precision setting in php.ini it is printed as 878, but when using (int) cast, or intval(), it will be rounded down (or twowards zero, don't reacall). By the way, pow(int,int) will yield integer values when possible since 4.0.7, which nukes one possible source of floating point problems (unless you use 10.0 of course, but that doesn't make sense in PHP) BTW: this is a FAQ, will add an entry. ------------------------------------------------------------------------ The remainder of the comments for this report are too long. To view the rest of the comments, please view the bug report online at http://bugs.php.net/?id=13550 Edit this bug report at http://bugs.php.net/?id=13550&edit=1