If Perl had a =>> operator, the code would look like this:

perl -le "$mask = 0x2b; print for map {$mask =>> 1?$_:()} @ARGV" and y
+ou will be pleased with the results
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... but as Perl lacks shift-and-assign operators that return the shifted-off bits, my code is uglier:

perl -le "$mask = 0x2b; print for map {my $sel = $mask & 1; $mask = $m
+ask >> 1; $sel?$_:()} @ARGV" and you will be pleased with the results
and
you
be
with
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Update: I found >>= and <<=, but they don't seem to act like I'd want, namely returning the shifted-off bits. In fact, my experiments don't make it clear to me what the operators do, as they leave $mask unchanged it seems:

perl -le "$mask = 0x2b; print for map {$mask >>= 1?$_:();} @ARGV" and 
+you will be pleased with the results
[download]

prints

43
43
43
43
43
43
43
43
[download]

... which I don't really understand.

Update 2: D'oh. Rereading the documentation, I now understand that $mask >>= 1 is equivalent to $mask = $mask >> 1, which returns the (new) value of $mask instead of returning the shifted-off bits. So it's not as usable as I'd need, for this case of golfing.