Re^3: Regular Expressions

In the first case, there is nothing stored in \1 so there shouldn't be any match (and it is an error.) In the latter case, \1 probably refers to the outer parentheses, but that hasn't taken effect yet, so there is no match and no error (perhaps it really doesn't make sense but perl tries to interpret it that way...)
If you instead do

print "Match: $1\n" if $test_string =~ /((?:f))\1/;
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the result is: Match: f
(Doing that doesn't seem especially useful, though...)
chas

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Re^4: Regular Expressions by Nkuvu (Priest) on May 17, 2005 at 21:37 UTC
Allow me to clarify a bit. What I find interesting is that for the first case perl doesn't warn about it ~~at all~~. If I have a regex that uses backreferences but has no capturing parens, I'd expect that perl (with `use warnings` enabled) would mention something about it. I'm not saying it's wrong behavior, and I understand why it's not matching. It just didn't provide the warning I'd expect. update: Or put it another way. I mean that the error message "Reference to nonexistent group in regex;" goes away if the regex is changed slightly. Consider: `print "matches\n" if $test_string =~ /(?:f)\1(.)/;` There is a valid set of capturing parens there, but after the \1 backreference is used. So this won't match -- and Perl doesn't warn about it. Surprised me a bit.	[reply] [d/l] [select]
Re^5: Regular Expressions by chas (Priest) on May 17, 2005 at 22:20 UTC
Yes, it is a bit surprising/confusing that there isn't a warning. I was just trying to guess why there isn't an error message ("nonexistent group") as in the previous case, and maybe it is for the reason I stated (i.e. \1 refers to the outer parens);I'm really not convinced, though. chas	[reply]