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Re^2: long integer to hexadecimal conversion

by perl_fan (Novice)
on Oct 01, 2009 at 10:27 UTC ( #798590=note: print w/replies, xml ) Need Help??

in reply to Re: long integer to hexadecimal conversion
in thread long integer to hexadecimal conversion

tried unpack same result :(
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Re^3: long integer to hexadecimal conversion
by Marshall (Abbot) on Oct 01, 2009 at 13:14 UTC
    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**4-1=15, or 'F'. 0123456789ABCDEF is all 16 possibilities with 4 bits.

    The biggest unsigned 32 bit number is 2**32-1= 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**3-1=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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