in reply to RE: RE: Selective substitution: not in Perl?
in thread Selective substitution: not in Perl?

Black magic! I'm really impressed, I didn' know you can put that all into the substitution pattern. /me bows to Nuance ...

As for my solution, I was focused on the sample input given, I didn't think of the more general problem. But you're right about this.

As for your script, it works with a small alteration:

$string =~ s/($pattern)/$i++ == $n ? $better : $1/ge;
If you increment $i in a seperate statement, the whole block evaluates to the value of $i und you get
abacab 0b1c2b
Ok, now we're having fun :-)...

Andreas (waiting for the chinese food delivery service ...)

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RE: RE: RE: RE: Selective substitution: not in Perl?
by merlyn (Sage) on Aug 16, 2000 at 15:33 UTC
    To make it even clearer, remember that the right side of the s///ge can be an arbitrary block. Here's the "Perl Bowling" style written for clarity:
    { my $count = 0; $string =~ s{($pattern)}{ my $source = $1; $count++; if ($count == $target) { $source = $replacement; # replace! } $source; # either original or replacement now }ge; };
    And yes, you can actually write it like that, just as you see. If you're doing anything odd in the right hand side in a s///e, I recommend expanding it out like that.

    -- Randal L. Schwartz, Perl hacker