in reply to PERL regex modifiers for m//

$s = "foo\nfoot\nroot";
$s =~ /^foo/g;           # matches only the first foo
$s =~ /^foo/gm;          # matches both foo
$s =~ /f.*t/g;           # matches only foot
$s =~ /f.*t/gs;          # matches foo\nfoot\nroot
$s =~ /f.*?t/gs;         # matches foo\nfoot
$s =~ /^foot.*root$/g;   # doesn't match
$s =~ /^foot.*root$/gm;  # doesn't match
$s =~ /^foot.*root$/gs;  # doesn't match
$s =~ /^foot.*root$/gms; # matches foot\nroot

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Re^2: example of 'm / / m' related example and compare to 'm / / s'
by ww (Archbishop) on Nov 28, 2011 at 18:09 UTC
    Line 3: " # matches both foo": are you sure? 
    C:>perl -e "$s = \"foo\nfoot\nroot\";($y) = $s =~ /^foo/gm;print $y;"
foo

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[d/l]
      Yes, but not in the same pass. 
      >perl -E"say join ',', qq{foo\nfoot\nroot} =~ /^foo/gm;"
foo,foo

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