Re: Regexp problem

Thanks for the quick answers

I was watching this on the debugger and I think I misinterpreted it.

So now my question is: why in the debugger I get

x "24-06" =~ /\d{2}-\d{2}/ as 1,

but x "24-06" =~ /\d{2}-\d{2}(p)*/ as undef?

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Re^2: Regexp problem by linuxer (Curate) on Aug 28, 2012 at 20:35 UTC
As far as I understand the debugger, both expressions are evaluated in list context. The first expression simply returns 1 for success and ~~I assume 0~~ an empty list for failure; so it's a single boolean result in list context. The second expression has capturing parantheses, which (in list context) produces a list of captured results. As your `(p)` could not be matched, there is no captured value for `(p)`, so it returns a list with one element: `undef`. That's my attempt to explain it. I am sure there are others who can explain it more detailed and accurately and even show some insight into the internals ... `DB<4> x "24-06" =~ /\d+-\d+/ 0 1 DB<5> x "24-06" =~ /\d+-\d+failure/ empty array` [download] edit:* fixed assumption upon failed regex match, added code example, some rephrasing	[reply] [d/l] [select]

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Re^2: Regexp problem
by linuxer (Curate) on Aug 28, 2012 at 20:35 UTC

As far as I understand the debugger, both expressions are evaluated in list context.

The first expression simply returns 1 for success and ~~I assume 0~~ an empty list for failure; so it's a single boolean result in list context.

The second expression has capturing parantheses, which (in list context) produces a list of captured results. As your (p)* could not be matched, there is no captured value for (p), so it returns a list with one element: undef.

That's my attempt to explain it. I am sure there are others who can explain it more detailed and accurately and even show some insight into the internals ...

  DB<4> x "24-06" =~ /\d+-\d+/ 
0  1
  DB<5> x "24-06" =~ /\d+-\d+failure/
  empty array
[download]

edit:

fixed assumption upon failed regex match, added code example, some rephrasing

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[d/l]
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