I was simply wishing to test for overflow on an integer, such as in C (well, if it were just over integer max anyway). When I looked to see if PHP was actually doing what I told it to, it seems it fails for some reason. Here are my tests of the problem:
define('INT_MAX', 0x7FFFFFFF);
print "In decimal: " . hexdec(INT_MAX) . "<br/>";
print "In decimal: " . hexdec(0x7FFFFFFE) . "<br/>"; //Under int_max
print "In hex: " . dechex(hexdec(INT_MAX)) . "<br/>";
print "Float: " . ((bool)is_float(INT_MAX)?'true':'false') . "<br/>";
Results being:
In decimal: 142929835591
In decimal: 142929835590
In hex: 47483647
Float: false
As I saw on the manual, it will cast to float if overthrown, but it seems to not and is clearly way higher. Am I being insane and missing something here, or is there some odd problem I should really need to know about when working with hexidecimal in PHP?
define('INT_MAX', 0x7FFFFFFF);
This defines INT_MAX to be integer 2147483647. It is unnecessary to interpret it as a hexadecimal number. If you really want to use INT_MAX as a literal hexadecimal value, then you need to declare it as '7FFFFFFF' (inside a string); then the hexdec function will interpret the hexadecimal notation and convert it to a decimal value.
print(dechex(INT_MAX) . "
");
This prints "7fffffff".
Your program makes no sense, because you are taking 0x7FFFFFFF which is 2147483647, and then treating it like 0x2147483647, which is 142929835591 in decimal.
Anyway, PHP already has a constant that you can use:
var_dump(PHP_INT_MAX + 1); // converted to float