Re^2: Why \n matches but not $^? (weight)

If you suspect that a regex isn't parsed the way you wanted it, you can use re 'debug'; to find out:

$ perl -Mre=debug -wle 'm/C$\n^D/'
Freeing REx: `","'
Omitting $` $& $' support.

EXECUTING...

Compiling REx `C
n^D'
size 6 Got 52 bytes for offset annotations.
first at 1
rarest char D at 3
   1: EXACT <C\nn>(3)
   3: BOL(4)
   4: EXACT <D>(6)
   6: END(0)
anchored "C
...
[download]

You don't have to understand everything to notice that the EXACT <C\nn> isn't what you were after. The literal <c>n has to come from the ...\n in the regex, so the thing before it (a newline) has to come from the previous token in the regex.

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Re^3: Why \n matches but not $^? (dump) by tye (Sage) on Oct 14, 2008 at 00:02 UTC
Thanks. That reminded of one thing I had been trying to remember to include: `$ perl -e'print qr/C$\n^D/,$/' (?-xism:Cn^D) $ perl -le'print qr/C$\n^D/' (?-xism:C n^D)` [download] which is a lower-tech way to notice that your regex wasn't parsed the way you expected. (It also demonstrates why I use `-l`: so I don't have a append newlines to each of my print statements.) - tye	[reply] [d/l]
Re^3: Why \n matches but not $^? (weight) by tel2 (Pilgrim) on Oct 13, 2008 at 22:53 UTC
Thanks moritz, That's very helpful. I didn't know about that re debug module. BTW: Why are you and tye using Perl's -l switch? Terry	[reply]
Re^4: Why \n matches but not $^? (weight) by moritz (Cardinal) on Oct 14, 2008 at 05:57 UTC
BTW: Why are you and tye using Perl's -l switch? It adds a newline to every print statement. See perlrun for more information. If you have perl 5.10.0 or newer, you can say `perl -wE 'say ...'` instead.	[reply] [d/l]