PHP var becoming negative

Question

I have this code :

<?php
$integer = 33963677451;
$integer &= 0xFFFFFFFF;
echo $integer;
?>

But, when I execute it, the output is

-396060917

The same function in Python

if __name__ == '__main__':
    integer = 33963677451
    integer &= 0xFFFFFFFF
    print integer

The output is

3898906379

It seems PHP can't store big variables. Do you guys have any idea how to get the same result in PHP I have in Python ?

Thanks !

Thats _not_ the same function in Python! `0F FF FF FF FF FF != FF FF FF FF` — KingCrunch, Feb 17 '11 at 14:30
Doesn't happen in Python 3, and doesn't happen in Python 2 if you use a long (`33963677451L`). (Applies to 4 and 6 byte operand.) — , Feb 17 '11 at 14:31
Even though the two numbers are different, the essence of the question is same: php doesnt support unsigned integers out of the box — Foo Bah, Feb 17 '11 at 14:35
@Foo: Neither does Python. However, Python has bigints and uses those to avoid overflow. — , Feb 17 '11 at 14:36

score 4 · Accepted Answer · answered Feb 17 '11 at 14:34

That value is too big for an integer, python dynamically casts that into a long variable. INT_MAX defines how big your integer can be before it switches over into a double. If you want to manipulate large numbers without quirks use GMPLIB a multiple precision arithmetic library.

score 1 · Answer 2 · answered Feb 17 '11 at 14:33

1

Your 33,963,677,451 is above what a signed 32bit integer supports (max 2,147,483,647). 64bit PHP versions do support higher values.

answered Feb 17 '11 at 14:33

Marc B

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score 0 · Answer 3 · answered Feb 17 '11 at 14:30

0

php only supports signed integers. You are seeing overflow

also, the answer for the python case should be 33963677451

answered Feb 17 '11 at 14:30

Foo Bah

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score 0 · Answer 4 · answered Feb 17 '11 at 14:34

0

I don't know if it's just a typo, but you are using different values when performing the AND operation in each language. :)

You have "0xFFFFFFFFFFF" (11 chars) in your PHP version, as opposed to only 8 in the Python version.

Either way, it should be integer overflow.

answered Feb 17 '11 at 14:34

João Neves

937
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oops, you're right, it was supposed to be 8 chars :) thanks for noticing – MisterDoy Feb 17 '11 at 14:36

score 0 · Answer 5 · answered Feb 17 '11 at 14:37

0

<?php
$integer = (float) 33963677451;
$integer &= 0xFFFFFFFFFFF;
echo $integer;

try making sure that $integer is of type float.

for more: http://www.php.net/manual/de/language.types.integer.php#language.types.integer.overflow

answered Feb 17 '11 at 14:37

Daniel Kutik

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yeah i guess you have 32bit :) u might want to have a look at http://stackoverflow.com/questions/211345/working-with-large-numbers-in-php also – Daniel Kutik Feb 17 '11 at 14:53
I'm on a 64bit system... But WAMP probably doesn't support this. On my webserver, it works well :) ! – MisterDoy Feb 17 '11 at 15:20

score 0 · Answer 6 · answered Feb 17 '11 at 15:06

If you wish to see whats happening, here is a printf statement showing the different types.

$i1 = 33963677451;
$i2 = 0xFFFFFFFFFFF;
printf("%b (%d, %f) & \n%b (%d, %f)\n = \n%b (%d, %f)", 
        $i1, $i1, $i1, $i2, $i2, $i2, $i1&$i2, $i1&$i2, $i1&$i2);

PHP casts integers larger than 32 bits to floats. It seems that you can not perform this operation on floats.

PHP var becoming negative

6 Answers6