It doesn't return 1 for success. It returns the number of matches (or substitutions for s/// or tr///). If you're not using the g option, it stops on the first match and returns 1. I'm not really up on perl internals but I would guess undef is represented as a null pointer whereas a zero would require creation of an SV.

---
print map { my ($m)=1<<hex($_)&11?' ':'';
$m.=substr('AHJPacehklnorstu',hex($_),1) }
split //,'2fde0abe76c36c914586c';
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right you are. i know that s/// & tr/// return the number of matched; never occured to me that m// behaves the same.

i goofed when i wrote that number 1 (be it as string or numeric) was being returned on a regex match. it just happens that only one thing was being matched.

thanks for the clue.

m// returns true or false in scalar context. Perl defines:

For evidence, see the source code module perl.c where PL_sv_yes and PL_sv_no are initialized. Also see the pp_match code in pp_hot.c where RETPUSHYES and RETPUSHNO are used to return a boolean value when invoked in scalar context without /g.

ah, so that is why bitwise AND operation w/ numeric** 1 yields 0 whenever following fails...

  (defined $var && $var =~ m/regex/) & 1
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...which otherwise would have produced an empty string. thanks mark.

** per perlop, an numeric operand implies numeric bitwise operation; per the table mark posted, false value is zero in numeric context.

$_ = 123;
print 1 + m/^\d+$/, "\n"; # get 2
print "1" . m/^\d+$/, "\n"; #get 11
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