Can somebody explain to me how to do 32-bit operations in Perl?

I wouldn't be very comfortable doing 32-bit unsigned integer operations in perl if overflows can occur. I'd probably do it in C and access the operations from perl via Inline::C or XS.
As already mentioned, 'use integer', is probably of value though I wouldn't like to vouch for its reliability - especially on a perl version as old as 5.8.

I don't know what "TinyPerl" is - could you provide the output of perl -V (that's an uppercase "V").

But on the first line, I get -1, and on the second line I get the positive number in base 10

As you've noted, "-1" and "4294967925" are the same 32-bit integer.
On the first line you asked for it to be expressed as a signed value (ie -1). On the second line you got to see the unsigned value (4294967295).
It seems to me that since you're actually interested in the unsigned values, the first line should have specified "%u" instead of "%d".

But, of course, the value you really expected to see was 855638016, for which you need to satisfy 2 requirements:
1) That you 'use integer'
2) That the size of your perl's integer is 4 bytes (check perl -V:ivsize).

With both of those conditions satisfied I get: 
LINE 1: 855638016 == 855638016
LINE 2: 855638016 == 855638016

[download]
But if perl's integer is 8 bytes then, irrespective of whether I 'use integer', I get: 
LINE 1: 855638016 == 5150605312
LINE 2: 855638016 == 855638016

[download]
Note that, in this instance, the "%d" in "LINE 1" is providing a signed 64-bit value because the perl integer is a 64-bit type.

Cheers,
Rob

In reply to Re: perl arithmetic is killing me! HELP! by syphilis
in thread perl arithmetic is killing me! HELP! by harangzsolt33

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