Re^2: What's the "...0" mean?

If .. or ... in used scalar context, and one of the arguments is a numerical literal (as is the case here), there's an implicit comparison against $. is made. So, in the given examples, after matching the left hand side of ..., $. equals 1. So, before the right hand side is tested, $. equals 2. Hence, ...0 or ...1 will never match, but ...2 will.

Comment on Re^2: What's the "...0" mean?

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Re^3: What's the "...0" mean? by tel2 (Pilgrim) on Oct 15, 2010 at 22:42 UTC
Thanks JavaFan (brave name for a Perl forum), > Hence, ...0 or ...1 will never match, but ...2 will. But "...0" and "...1" seem to have matched the 1st line of the input file, where they changed "zabcz" to "abcz", right? Why did they both match the 1st line, but "...2" matched the 3rd line?	[reply]
Re^4: What's the "...0" mean? by lidden (Curate) on Oct 15, 2010 at 22:56 UTC
The part before the dots turns the flip flop operator to true, the part after sets it to false when it matches. An example of how the flip flop operator works. `my @array = qw- 1 2 3 on 4 5 off 6 7 8 on 9 10 11 12 off 13 14 15 -; for (@array){ say if /on/ ... /off/; }` [download] Update: Added example.	[reply] [d/l]
Re^5: What's the "...0" mean? by morgon (Priest) on Oct 15, 2010 at 23:24 UTC
That is a good example. But in the original question `perl -ne 'print if s/.*?(?=abc)//...2' input.txt` [download] the flip-flop is true on the first line (the substitution) and remains to be true on the second line (the comparison with line number). But should it then not remain false for the rest of the file resulting in no more lines being printed?	[reply] [d/l]