Don't they want to get a list "e.g. ( 1, 0, 0, 1, 1, 0 ... )" as stated in the beginning of that section,

Why not just try it out?

$vec = pack 'V', int( 2**32* rand() );;

print $bits;;
011000001111110110010000101101

@ints = (); 
{ 
    use integer; 
    push @ints, 0 if $bits =~ s/^(\d)// && $1;
    push @ints, pos $bits while $bits =~ /1/g;
};;

print @ints;;
1 2 8 9 10 11 12 13 15 16 19 24 26 27 29
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The are producing a list of the positions of the bits that are set.

I do agree that the section title and description and the sub name are woefully bad.

I'm also less than impressed with the claims for efficiency:

C:\test>perl -MMath::Random::MT=rand -minteger -MBenchmark=cmpthese
$vec = pack 'V*', map int( 2**32* rand() ), 1 ..10;;
cmpthese -1, {
 a=>q[ my @ints; my $bits = unpack 'b*', $vec; use integer; push @ints
+, 0 if $bits =~ s/^(\d)// && $1;push @ints, pos $bits while($bits =~ 
+/1/g); ],
 b=>q[ my @ints = grep{ vec( $vec, $_, 1 ) } 0 .. length( $vec ) -1; ]
+,
};;
^Z
      Rate    a    b
a  22306/s   -- -79%
b 108646/s 387%   --
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Thanks for great answers. Maybe, part of my confusion was because first "push" in faq code is unnecessary - why not replacing both lines

push @ints, 0 if $bits =~ s/^(\d)// && $1;
push @ints, pos $bits while $bits =~ /1/g;
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with

push @ints, pos($bits)-1 while $bits =~ /1/g;

And it's faster, if i'm not mistaken with how i checked it with Benchamrk module.

why not replacing both lines

If you look carefully at the first line, you'll see that in addition to pushing 0 if the first bit(byte!) is set, it also removes the first byte.

Which means that the second line operates on the string with the first byte removed.

That's because the value pos returns is the position after where the last match occurred, so by removing the first byte the positions come out correctly.

Of course, they could have just subtracted 1 from pos, then: a) that first line; b) two visits to the regex engine; and c) modifying the string; become unnecessary.

But then, if you use the tool perl provides for the job, you not only avoid the regex engine, but also the need to unpack the vector to a string of 1s & 0s:

print grep vec( pack('V',1431655765), $_, 1), 0 .. 31;;
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30

print grep vec( ~pack('V',1431655765), $_, 1), 0 .. 31;;
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31
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A very curious FAQ answer.

---

With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.

"Science is about questioning the status quo. Questioning authority".

In the absence of evidence, opinion is indistinguishable from prejudice.

The wording does seem to imply that it would return a list of bit values, but that isn't what it is doing. Rather, that routine returns a list of bit positions that are set (have a 1 in them).

If you just want the list of bits, it is probably simplest to split an unpacked string.

my @bits = split //, unpack "b*", $vec;
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To furthur demonstrate:

use integer;
my $vec = 42;
my @ints;
my $bits = unpack "b*", $vec;
my @bits = split //, unpack "b*", $vec;
push @ints, 0 if $bits =~ s/^(\d)// && $1;
push @ints, pos $bits while($bits =~ /1/g);

print "\n";
print join ', ', @ints;
print " <-- bit positions set\n";
print join ', ', @bits;
print " <-- bits \n";
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2, 4, 5, 9, 12, 13 <-- bit positions set
0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0 <-- bits