in reply to please explain this script

First of all, the program can be reduced in size: 
$/=\1;while(<>){$c=($c<<1)+($c>>31)+($.>>2==3?0:ord)}printf"%x",$c

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The program calculates a checksum, it does so by iterating over each character of the file ($/ = \1; while (<>)). It keeps a running, 32 bit checksum, initially zero. For each character, it will rotate the current checksum one bit leftward (note that it's rotating, not shifting). This is the ($c << 1) + ($c >> 31) part. Then for each fourth character, the code point number will be added to the running total. ($x >> 2 == 3 ? 0: ord); $x ++). Finally, the checksum is printed as a hex number.

Note that it will display different sums depending whether you have a 32-bit or a 64-bit perl!

Abigail

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Re: Re: please explain this script
by El Linko (Beadle) on May 28, 2004 at 13:54 UTC
    Probably wouldn't be a bad idea to replace it with a md5 hash using Digest::MD5 
    use Digest::MD5;
my $hash = Digest::MD5->new();
$hash->addfile(*STDIN);
print $hash->hexdigest()."\n";

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    The resulting string is 4 times longer so its possible this would require adjustments elsewhere. You do however get the benefit of name recognition so anyone who comes across the code knows exactly what it is doing.
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[d/l]
      the binary file actually stores the calulated checksum starting at the 12 byte and is expecting 32 bits. I was just wanting to know what the script was doing.
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Re: Re: please explain this script
by arcnon (Monk) on May 28, 2004 at 15:32 UTC
    one more question.. what is ord actually doing I thought ord makes asci characters. Please excuse my ignorance. and thanks for the help.
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      ord returns the ordinal value (usually the ascii value) of a given literal character. For example, print ord('A'); prints 65.


      Dave

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[d/l]

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