The special variable $+ will hold the same contents as the last matched capture. But which one is that? That is, given a set of possibilities 
(foo)|(bar)|(baz)

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I'll get something in either $1, $2, or $3. Depending on which case matched, I want to do different logic. I can do something like this: 
if (defined $1) { sub1 }
elseif (defined $2) { sub2 }
...

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But I'm thinking that something along the lines of 
$sub[$n]->();

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would be much better, if only I knew the value of n. Computing it by looking at the definedness of each capture is just as bad as doing the work directly as above, so don't bother.

It seems to me that there should be a value for this somewhere. Anybody know for sure, or know of an elegant way to find which $n is chosen by $+?

—John

In reply to What's like $+ but not gives the ordinal? by John M. Dlugosz

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