operater xor (^)

[peeratep]

on windows->
-2694274578 ^ 1136121015 = 484406617

on linux->
-2694274578 ^ 1136121015 = -1011362633

i want value 484406617 on linux..help me please!
Thank you...

[redarrow]

what is it then a time stamp or what?

your are blunt sorry.

[effigy]

How many bit systems are they? 32? 64?

[Barand]

intersting relationship between the second integer and the linux result

[code]<?php
$a = 1136121015;
$b = -1011362633;
echo '<pre>';
printf ('%032s %12d<br>%032s %12d linux<br>',
decbin($a), $a, decbin($b), $b);
echo '</pre>';
?>[/code]

-->[pre]
01000011101101111101010010110111  1136121015
11000011101101111101010010110111  -1011362633 linux
[/pre]

[effigy]

From wiki:

[quote]
The standard Unix time_t (data type representing a point in time) is a signed integer data type, traditionally of 32 bits (but see below), directly encoding the Unix time number as described in the preceding section. Being integer means that it has a resolution of one second; many Unix applications therefore handle time only to that resolution. [b]Being 32 bits (of which one bit is the sign bit)[/b] means that it covers a range of about 68 years in total.
[/quote]

How does Windows handle signed integers?

[Barand]

Does this clarify ?

[code]
<?php
$a = 31;
$b = -31;
echo '<pre>';
printf ('%032s %12d<br>%032s %12d<br>',
decbin($a), $a, decbin($b), $b);
echo '</pre>';
?>[/code]
-->[pre]
00000000000000000000000000011111          31
11111111111111111111111111100001          -31
[/pre]

[effigy]

From the guys (ikegami) over at www.perlmonks.net:

[quote]
Unix and Windows handle signs the same way: 2's complement.
The problem is that -2147483648 is the largest negative that can be stored in 32 bits.
Well, the problem is that -2694274578 is larger than -2147483648, but you are using 32 bit numbers.
[/quote]