Posted by Tim Roberts on 11/18/83 11:41

"fritz-bayer@web.de" <fritz-bayer@web.de> wrote:
>
><?php
>$a = 5543039447;
>$b = 2147483648;
>$result = ($a & $b);
>echo "$result\n" ;
>?>
>
>to perl? In perl almost the exactly same code:
>
>#!/usr/bin/perl
>$a = 5543039447;
>$b = 2147483648;
>$result = ($a & $b);
>print "$result\n" ;
>
>returns 2147483648.
>
>How do I have to modify the perl code, so that it will actually do a
>bitwise and? I don't get - any workaround? Somethings I'm not getting
>here? I'm using linux debian/sarge with an intel 32 bit prozessor.

The basic problem here is that what you are asking for is nonsense. GIGO.

Both of your numbers are too large to fit into an unsigned 32-bit value.
When that happens, both languages will automatically convert to using
floating point to store those values. $a and $b are FLOATING POINT.

However, bitwise operations do not make sense for floating point values.
Thus, both languages must convert the floats back to 32-bit integers. But,
we already know that the values are too big for that. So, there is an
overflow. It just turns out that the two languages handle this overflow
differently.

PHP handles the overflow by throwing away the upper bits, leaving you with
only the bottom 32 bits of the converted value. Perl handles the overflow
by "pegging"; it returns 2^32-1, or 4,294,967,295. That value happens to
have all 32 bit set, so the bitwise "and" returns the second value.

If you need to handle integers larger than 32-bits in Perl, you can use the
Math::BigInt module to do that.
--
- Tim Roberts, timr@probo.com
Providenza & Boekelheide, Inc.

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