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Re: Strange regex to test for newlines: /.*\z/

by avar (Beadle)
on May 21, 2007 at 23:13 UTC ( #616664=note: print w/replies, xml ) Need Help??


in reply to Strange regex to test for newlines: /.*\z/

To elaborate a bit what should probably be happening is that the regex engine should always backtrack on .* into the equivalent of NOTHING, but instead it commits to the match and fails under /m once it reaches \n.
perl5.9.5 -Mre=debug -e 'my @re = (qr/.*\z/, qr/.?\z/, qr/(|.+)\z/); "\n" ~~ $_ for @re'

Compiling REx ".*\z"
Final program:
   1: STAR (3)
   2:   REG_ANY (0)
   3: EOS (4)
   4: END (0)
floating ""$ at 0..2147483647 (checking floating) anchored(MBOL) implicit minlen 0 
Compiling REx ".?\z"
Final program:
   1: CURLY {0,1} (4)
   3:   REG_ANY (0)
   4: EOS (5)
   5: END (0)
floating ""$ at 0..1 (checking floating) minlen 0 
Compiling REx "(|.+)\z"
Final program:
   1: OPEN1 (3)
   3:   BRANCH (5)
   4:     NOTHING (8)
   5:   BRANCH (FAIL)
   6:     PLUS (8)
   7:       REG_ANY (0)
   8: CLOSE1 (10)
  10: EOS (11)
  11: END (0)
floating ""$ at 0..2147483647 (checking floating) minlen 0 
Guessing start of match in sv for REx ".*\z" against "%n"
Found floating substr ""$ at offset 0...
Position at offset 0 does not contradict /^/m...
Guessed: match at offset 0
Matching REx ".*\z" against "%n"
   0 <> <%n>                 |  1:STAR(3)
                                  REG_ANY can match 0 times out of 2147483647...
   0 <> <%n>                 |  3:  EOS(4)
                                    failed...
                                  failed...
Match failed
Guessing start of match in sv for REx ".?\z" against "%n"
Found floating substr ""$ at offset 0...
Guessed: match at offset 0
Matching REx ".?\z" against "%n"
   0 <> <%n>                 |  1:CURLY {0,1}(4)
                                  REG_ANY can match 0 times out of 1...
   0 <> <%n>                 |  4:  EOS(5)
                                    failed...
                                  failed...
   1 <%n> <>                 |  1:CURLY {0,1}(4)
                                  REG_ANY can match 0 times out of 1...
   1 <%n> <>                 |  4:  EOS(5)
   1 <%n> <>                 |  5:  END(0)
Match successful!
Guessing start of match in sv for REx "(|.+)\z" against "%n"
Found floating substr ""$ at offset 0...
Guessed: match at offset 0
Matching REx "(|.+)\z" against "%n"
   0 <> <%n>                 |  1:OPEN1(3)
   0 <> <%n>                 |  3:BRANCH(5)
   0 <> <%n>                 |  4:  NOTHING(8)
   0 <> <%n>                 |  8:  CLOSE1(10)
   0 <> <%n>                 | 10:  EOS(11)
                                    failed...
   0 <> <%n>                 |  5:BRANCH(8)
   0 <> <%n>                 |  6:PLUS(8)
                                  REG_ANY can match 0 times out of 2147483647...
                                  failed...
   1 <%n> <>                 |  1:OPEN1(3)
   1 <%n> <>                 |  3:BRANCH(5)
   1 <%n> <>                 |  4:  NOTHING(8)
   1 <%n> <>                 |  8:  CLOSE1(10)
   1 <%n> <>                 | 10:  EOS(11)
   1 <%n> <>                 | 11:  END(0)
Match successful!
Freeing REx: ".*\z"
Freeing REx: ".?\z"
Freeing REx: "(|.+)\z"
  • Comment on Re: Strange regex to test for newlines: /.*\z/

Replies are listed 'Best First'.
Re^2: Strange regex to test for newlines: /.*\z/
by bloonix (Monk) on May 22, 2007 at 09:39 UTC
    I am not quite agreeable to the statement about what '.*' should match.

    For my understanding '.' should ignore newlines always but if the operator /s is used. That means that '.+' and '.*' are just multiple searches of '.' and should still ignore newlines.

    Now I understand $ and \z as the following... $ means to matches both the end and the newline before - quote perldoc - and \z only the end but not the newline.
    print "foo matched\n"         if "foo\n"     =~  /^foo$/;
    print "bar matched\n"         if "bar\n"     =~  /^bar$ \n/x;   # $ before end or newline
    print "baz doesn't matched\n" if "baz\n"     !~  /^baz\z/;
    print "foobar matched\n"      if "foobar\n"  =~  /^foobar\n\z/; # \z after newline
    
    print "match foo\n"         if "foo\n" =~ /.*$/;     # .* ignore newline and $  is before newline
    print "doesn't match bar\n" if "bar\n" !~ /.*\z/;    # .* ignore newline and \z is after  newline
    print "match baz\n"         if "baz\n" =~ /.?\z/;    # but what the hell happends here?
    
    for ( qr/(.?)\n\z/, qr/(.?)\z/ ) {
       "hello world\n" =~ $_;
       print "-$1-\n";
    }
    
    -d-
    --
    
    It seems that '.?' ignore the newline as expected and search on after the newline with '.?\z', because it searches _until_ '\z'. Also it seems that '.*' matches until the newline and not between '\n' and '\z'. '.*' is greedy, '.?' not. Maybe I missunderstand it.
      $ and \Z work pretty much the same in normal mode, both match the end of search string or before a string-ending newline. the difference between them lies in the multiline mode when you issue an 'm' modifier.

      \z means the real end of string even after the string-ending newline.

      If you use an 's' modifier, then things become more different but that's mainly coz of the '.' which changes its behaviors, not the three end-of-string anchors..

      check the following snippets:
      perl -e 'print "match\n" if "foo\n" =~ /.+$/' # ok # perl -e 'print "match\n" if "foo\n" =~ /.+\z/' perl -e 'print "match\n" if "foo\n" =~ /.+\Z/' # ok # perl -e 'print "match\n" if "foo\n\n\n" =~ /.+\Z/' perl -e 'print "match\n" if "foo\n\n\n" =~ /.+\z/' perl -e 'print "match\n" if "foo\n\n\n" =~ /.+$/' perl -e 'print "match\n" if "foo\n\n\n" =~ /.+$/m' # ok # perl -e 'print "match\n" if "foo\n\n\n" =~ /.+\z/m' perl -e 'print "match\n" if "foo\n\n\n" =~ /.+\Z/m' perl -e 'print "match\n" if "foo\n\n\n" =~ /.+\Z/s' # ok # perl -e 'print "match\n" if "foo\n\n\n" =~ /.+\z/s' # ok # perl -e 'print "match\n" if "foo\n\n\n" =~ /.+$/s' # ok #
      BTW. When comparing between \z, \Z and $, it's probably better to avoid using .* or .? quanifiers the ways in your examples.

      BTW. my previous statement about \Z had some error and I have updated that post.

      Regards,
      Xicheng

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