Re: Regex Subexpressions

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@matches = m/(((a)b)c)/ and print "@matches\n";
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Comment on Re: Regex Subexpressions Download Code

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Re^2: Regex Subexpressions by liverpole (Monsignor) on Sep 10, 2005 at 03:02 UTC
Wow! Now that's a great answer! Let me just add the following, since it took me a few minutes to see exactly what you were doing, as well as making it conform to the original requirements for the output: `$_ = "abc"; # Regex match of /.../ uses $_ when =~ is unspecified # Nicely solves the problem stated, and demonstrates list assignment # to a regex capture. # @matches = m/(((a)b)c)/ and print join "\n", reverse @matches;` [download] I realized a "reverse" was necessary, and then immediately fell into the trap of putting it before the "join". I like your answer a lot for its succinct approach!	[reply] [d/l]
Re^3: Regex Subexpressions by ikegami (Patriarch) on Sep 10, 2005 at 03:52 UTC
Does your solution have to find "ab" in "abd"? If so, the above won't work.	[reply]