Combining this with LanX's comment about metachars gives enlightenment:

my $re1=qr/foo$/;  
my $re2=qr/foo$/m;
$,="\t"; $\="\n";  # just for output
print tf(qr/$re1/),  tf(qr/$re1/m);
print tf(qr/$re1$/), tf(qr/$re1$/m);
print tf(qr/$re2/),  tf(qr/$re2/m);
print tf(qr/$re2$/), tf(qr/$re2$/m);
sub tf { "$_[0] ".("foo\nbar\n"=~$_[0]?"T":"F") }
__END__
(?^:foo$) F          (?^:foo$) F
(?^:(?^:foo$)$) F    (?^m:(?^:foo$)$) F
(?^m:foo$) T         (?^m:foo$) T
(?^:(?^m:foo$)$) F   (?^m:(?^m:foo$)$) T

[download]

i.e. the modifiers need to be on the regex which contains the metacharacters which they affect. Makes sense but I didn't catch it right away.

perlop says this:

If a precompiled pattern is embedded in a larger pattern then the effect of "msixpluad" will be propagated appropriately.

It's possible I've missed some other documentation, but I find this statement possibly a bit misleading, since "appropriately" could be taken to mean that, for example, qr/$re1/m should propagate /m into $re1, as the OP expected.

In reply to Re^2: /s and /m don't seem to be doing anything by Anonymous Monk
in thread /s and /m don't seem to be doing anything by sciguy

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