Re: mysteries of regex substring matching

For completeness:

Another workaround was so trivial that I skipped it in my previous answer

You can build regexes from smaller ones with string interpolation

  DB<73> p $capt4 = '(....)' x 4
(....)(....)(....)(....)
  DB<74> x $x =~ m/ $capt4  (.+) /x
0  'AAD3'
1  4017
2  8372
3  '01D9'
4  '8AAED18778DEF993'
  DB<75>
[download]

I think that's the most readable solution.

Cheers Rolf
_{(addicted to the Perl Programming Language :)

Wikisyntax for the Monastery}

Comment on Re: mysteries of regex substring matching Download Code

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Re^2: mysteries of regex substring matching by LanX (Saint) on Jan 16, 2021 at 00:49 UTC
Same approach, but as a one liner: `DB<5> p 0123456789012345678901234567890123456789 DB<5> x / @{[ '(....)' x4 ]} (.+) /x 0 0123 1 4567 2 8901 3 2345 4 678901234567890123456789 DB<6>` [download] Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery}	[reply] [d/l]