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Re^3: Efficient bit counting with a twist.

by Anonymous Monk
on Jun 01, 2013 at 00:41 UTC ( #1036355=note: print w/ replies, xml ) Need Help??


in reply to Re^2: Efficient bit counting with a twist.
in thread Efficient bit counting with a twist.

Maybe I'm crazy, but this doesn't seem to work:

#!/usr/bin/perl use strict; use warnings; my $mask = hex($ARGV[0]); my $setbits = unpack("%32b*", $mask); printf("0x%x has %d set\n", $mask, $setbits);

Then I run it:
./bittest 0xff
0xff has 11 set

Or worse:
./bittest 0x1001
0x1001 has 14 set

Even the simple case:
./bittest 0
0x0 has 2 set

What is going on here?


Comment on Re^3: Efficient bit counting with a twist.
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Re^4: Efficient bit counting with a twist.
by BrowserUk (Pope) on Jun 01, 2013 at 09:49 UTC
    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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