Perl: the Markov chain saw  
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Re^3: long integer to hexadecimal conversionby Marshall (Abbot) 
on Oct 01, 2009 at 13:14 UTC ( #798644=note: print w/replies, xml )  Need Help?? 
The basic problem appears to be that you have a 32 bit machine and you need a bigger integer than can be represented in 32 bits. If you have 4 bits, the biggest unsigned number that you can represent is with all 4 bits "on", is, "1111", or 15 in decimal, 2**41=15, or 'F'. 0123456789ABCDEF is all 16 possibilities with 4 bits. The biggest unsigned 32 bit number is 2**321= 4,294,967,295. The normal convention for representing a "signed" number is what is called 2's complement arithmetic. If you have 4 bits and have say 0001, to get "1", you complement (reverse) all bits and then add one. 1110+1=1111. Now we come to the interpretation of whether "1111" means "15" or "1". Basically the most significant bit becomes the "sign bit". 0111 with 4 bits is the max positive number or 2**31=7. 1000 is the max negative number which weirdly enough is 8. So if you have 32 bits as signed 2's complement, you get "0111 1111 1111 1111 1111 1111 1111 1111" or 2,147,483,647 as the max positive number and 2,147,483,648 as the max negative number. I think a Perl float uses more than 32 bits for the "mantissa". I think it maybe as many as 53 bits. I'm not sure. I think the "safest" way is to use these big int modules. The 32 bit integer to think about is 2 billion. When you get to that size integer number, things can get more complicated. 123,143,230,627 is a lot bigger than 2,147,483,647 and hence the problems.
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