Question

I am using PHP 5.3.

On my 32-bit system the size of an INT:

print "PHP_INT_MAX: " . PHP_INT_MAX . "\n";
print "PHP_INT_SIZE: " . PHP_INT_SIZE . " bytes (" . (PHP_INT_SIZE * 8) . " bits)\n";

• PHP_INT_MAX: 2147483647
• PHP_INT_SIZE: 4 bytes (32 bits)

However, part of an an encoding algorithm I am using relies on the fact that an int is the above size (4 bytes). When I run the code on my web host’s server, it is a 64-bit system and the int size is twice as large.

Is there a way to force “(int)” cast to use the 32-bit size?

For example, assume the following code:

$test = 4170266799;
print $test;                     
print (int) $test;

On my 32-bit system, the output is:

4170266799
-124700497

On my 64-bit system, the output is:

4170266799
4170266799

Is it possible to force the value of an INT to be 4 bytes, even when the architecture changes from 32-bit to 64-bit?

Answer 1

Referencing this question on Stack Overflow, I have found a solution that seems to work in my early tests:

function thirtyTwoBitIntval($value)
{                  
    if ($value < -2147483648)
    {
    return -(-($value) & 0xFFFFFFFF);
    }
    elseif ($value > 2147483647)
    {
        return ($value & 0xFFFFFFFF);
    }
    return $value;
}


$test = 4170266799;
print $test;                     
print (int) $test;
print thirtyTwoBitIntval($test);

And the output on the 32-bit system is:

4170266799  # $test = 4170266799
-124700497  # (int) $test
-124700497  # thirtyTwoBitIntval($test);

And the output on the 64-bit system is:

4170266799  # $test = 4170266799
4170266799  # (int) $test
-124700497  # thirtyTwoBitIntval($test);