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What is going on here?

I tweaked your code by adding a couple of extra print statements:

my $mask = hex($ARGV[0]); print $mask; print unpack 'H*', $mask; my $setbits = unpack("%32b*", $mask); printf("0x%x has %d set", $mask, $setbits);

And when I run it I get this output:

C:\test>junk 0xff 255 ## 3 bytes 323535 ## 6 nybbles with bit patterns 0011 0010 0011 0101 0011 0101 0xff has 11 set

hex converts the string '0xff' to a number which you store in $mask;

unpack expects a string, so perl helpfully converts the number stored in $mask to a string in the default decimal representation '255'.

You are counting the bits in that 3 byte string.

To count the bits in the number, you need to tell perl that you want the binary representation of that number:

#! perl -slw use strict; my $mask = hex($ARGV[0]); print $mask; print unpack 'H*', pack 'C', $mask; my $setbits = unpack("%32b*", pack 'C', $mask ); printf("0x%x has %d set", $mask, $setbits); __END__ C:\test>junk 0xff 255 ff 0xff has 8 set

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In reply to Re^4: Efficient bit counting with a twist. by BrowserUk
in thread Efficient bit counting with a twist. by BrowserUk

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