in reply to What happens with empty $1 in regular expressions? (was: Regular Expression Question)

The problem is that the $1, $2, etc. variables only get set if the regular expression matches. /(\d+)/ does NOT match "Mm", so no assignment occurs. What you need to do is to check whether there is a match before using the positional variables, like this: 
if ($sss =~ /(\d+)/) {
  # use $1 as you wish
}
else {
  # don't!
}

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Interestingly, your last code snippet will do the right thing if you remove the part after the && (and an extra opening parenthesis you seem to have there), because $1 and $2 will only be used when there is a match, and in that case they will contain either the strings they matched, or undef if their respective subexpressions didn't match anything.

--ZZamboni

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Re: Re: Regular Expression Question
by japhy (Canon) on Feb 28, 2001 at 19:28 UTC
    No, $1 will not be undef if the match fails. The only time they are undefined is when a match has not yet been done, or a match does not contain a captured pattern for that variable: 
    #!/usr/bin/perl -w

if ("abc" =~ /(\w+)/, 1) { print "abc => $1\n" }
{ 
  print "local: $1\n";
  if ("ghi" =~ /\w+/, 1) { print "ghi => $1\n" }
  if ("def" =~ /(\w+)/, 1) { print "def => $1\n" }
  if ("[=]" =~ /(\w+)/, 1) { print "[=] => $1\n" }
}
print "general: $1\n";

__END__
abc => abc
local: abc
Use of uninitialized value at regexes line 6.
ghi => 
def => def
[=] => def
general: abc

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    However, a failed match returns false, so if you removed the , 1's from each of those, you wouldn't see the line for "=".

    japhy -- Perl and Regex Hacker
[reply]
[d/l]
      I agree completely with you, and that's what I meant, that if the regex does not match, $1 is not modified in any way (neither set or unset). But your example made it way clearer than any explanation :-)

      --ZZamboni

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