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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";

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.

In reply to Re^2: Strange regex to test for newlines: /.*\z/ by bloonix
in thread Strange regex to test for newlines: /.*\z/ by betterworld

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