in reply to /s and /m don't seem to be doing anything

Don't forget the basic debugging step of printing your things to check if they are what you think they are:

my $re1=qr/foo$/;  
my $re2=qr/foo$/m;
my $re3=qr/foo$/s;
$,="\t"; $\="\n";  # just for output
print qr/$re1/, qr/$re1/m, qr/$re1/s;
print qr/$re2/, qr/$re2/m, qr/$re2/s;
print qr/$re3/, qr/$re3/m, qr/$re3/s;
__END__
(?^:foo$)       (?^:foo$)       (?^:foo$)
(?^m:foo$)      (?^m:foo$)      (?^m:foo$)
(?^s:foo$)      (?^s:foo$)      (?^s:foo$)

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So that means...

my $re1=qr/foo$/;
print $re1, "\n";
print "foo\\nbar\\n =~ \/$re1\/ \t";
print "foo\nbar\n" =~ /$re1/ ?"true":"false"; print "\n";
print "foo\\nbar\\n =~ \/$re1\/m \t";
print "foo\nbar\n" =~ /$re1/m ?"true":"false"; print "\n";
my $re2=qr/foo$/m;
print $re2, "\n";
print "foo\\nbar\\n =~ \/$re2\/ \t";
print "foo\nbar\n" =~ /$re2/ ?"true":"false"; print "\n";
print "foo\\nbar\\n =~ \/$re2\/m \t";
print "foo\nbar\n" =~ /$re2/m ?"true":"false"; print "\n";
__END__
(?^:foo$)
foo\nbar\n =~ /(?^:foo$)/       false
foo\nbar\n =~ /(?^:foo$)/m      false
(?^m:foo$)
foo\nbar\n =~ /(?^m:foo$)/      true
foo\nbar\n =~ /(?^m:foo$)/m     true

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(I haven't yet found the relevant bit of documentation that makes this behavior clear.)

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Re^2: /s and /m don't seem to be doing anything
by Anonymous Monk on Dec 31, 2014 at 04:24 UTC
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Re^2: /s and /m don't seem to be doing anything
by Anonymous Monk on Dec 31, 2014 at 04:25 UTC

    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

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    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.

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      I can't test but I'm sure including $re as string w/o precompiling doesn't preserve any flags.

      Cheers Rolf

      (addicted to the Perl Programming Language and ☆☆☆☆ :)

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        If a regex is defined in string form, then no regex modifiers are associated with it unless they happen to be explicitly embedded within the string using a  (?adlupimsx-imsx) or  (?adluimsx-imsx:pattern) construct (see Extended Patterns in perlre). 

        c:\@Work\Perl\monks>perl -wMstrict -le
"my $rs1 = '(?:(?-s).*)';
 ;;
 my $rx1 = qr{ ($rs1) (.*) }xms;
 print qq{\$rx1: $rx1};
 ;;
 qq{foo\nBAR} =~ m{ $rx1 }xms;
 print qq{from \$rx1:  \$1 '$1'   \$2 '$2'};
 ;;
 my $rs2 = '(?:.*)';
 ;;
 my $rx2 = qr{ ($rs2) (.*) }xms;
 print qq{\$rx2: $rx2};
 ;;
 qq{wig\nWAG} =~ m{ $rx2 }xms;
 print qq{from \$rx2:  \$1 '$1'  \$2 '$2'};
"
$rx1: (?^msx: ((?:(?-s).*)) (.*) )
from $rx1:  $1 'foo'   $2 '
BAR'
$rx2: (?^msx: ((?:.*)) (.*) )
from $rx2:  $1 'wig
WAG'  $2 ''

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        Give a man a fish:  <%-(-(-(-<

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