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### Re: Recursive substitution

by bobf (Monsignor)
 on Nov 10, 2009 at 04:27 UTC ( #806118=note: print w/replies, xml ) Need Help??

I like to take a very simple approach to such things, since the tricky methods usually confuse me when I come back to it 6 months later and have to modify it somehow. :-)

my \$s = 'ab'; my \$pat = qr/^(a{1,3})(?=b)/; while( \$s =~ m/\$pat/ ) { \$s =~ s/\$pat/\$1a/g; }

Or, if you prefer more of the 1-liner feel:

my \$s = 'ab'; my \$pat = qr/^(a{1,3})(?=b)/; do{ \$s =~ s/\$pat/\$1a/g } while \$s =~ m/\$pat/;

I'm sure it is possible to do this in the regex itself (probably with /e). I'm sure other monks will enlighten both of us.

Update: I changed approaches mid-code and forgot to remove the superfluous m//. These solutions are not very perlish.

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Re^2: Recursive substitution
by JadeNB (Chaplain) on Nov 10, 2009 at 04:32 UTC

Thanks very much. I realised shortly after posting that I hadn't explained why I didn't want the most natural answer (current explanation: because I don't, that's why), but didn't update my post in time.

By the way, I think that s/\$pat/\$sub/ is (effectively) a no-op unless m/\$pat/, so I think that it's redundant to do a separate m/\$pat/ check.

it's redundant to do a separate m/\$pat/ check
You are absolutely correct, of course. I started playing with patterns that would have required the extra check (because the ones used in s/// and m// were different), but I failed to remove the extra code after changing my approach. I added a note to my original reply but left the crufty code to avoid confusing future readers of this thread.

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