Regex Substitution, Interpolating on the RHS?

Cody Pendant has asked for the wisdom of the Perl Monks concerning the following question:

Replies are listed 'Best First'.
Re: Regex Substitution, Interpolating on the RHS? by dave_the_m (Monsignor) on Jul 18, 2005 at 11:52 UTC
`my $repl = '"foo$1bar"'; $a = '12345'; $a =~ s/(23)/$repl/ee; print $a,"\n"; $ ./perl /tmp/p 1foo23bar45 $` [download] Dave.	[reply] [d/l]
Re^2: Regex Substitution, Interpolating on the RHS? by Cody Pendant (Prior) on Jul 18, 2005 at 12:04 UTC
Wow. Interesting stuff. Double quotes, double-e... Thank you. ($_='kkvvttuu bbooppuuiiffss qqffssmm iibbddllffss') =~y~b-v~a-z~s; print	[reply]
Re^3: Regex Substitution, Interpolating on the RHS? by diotalevi (Canon) on Jul 18, 2005 at 13:04 UTC
You should realize this means you're building source code on the fly and evaling it. Its a serious security and correctness issue. The following is equivalent and makes the string-eval more visible than just the second /e would. The first /e is completely different in character than the second /e. `s<$regex->[0]>{ eval qq["$regex->[1]"]; }e` [download]	[reply] [d/l]
Re: Regex Substitution, Interpolating on the RHS? by Roy Johnson (Monsignor) on Jul 18, 2005 at 13:07 UTC
Another approach to delayed evaluation is to use closures (update: ok, not really closures, but subroutines). To get a sub to be evaluated in the double-quotish context of a s/// replacement, it needs to be part of a block. The only block interpolated in double quotes is a deref. So you make the closure return a ref to the string you want, and deref it in the replacement. It will be evaluated at replacement time, and you get the substitution you're looking for. `my $regex = ['(foo)',sub {\"bar$1"}]; my $string = 'blah foo blah'; $string =~ s/$regex->[0]/${$regex->[1]->()}/; print "S=$string\n";` [download] Caution: Contents may have been coded under pressure.	[reply] [d/l]
Re^2: Regex Substitution, Interpolating on the RHS? by Cody Pendant (Prior) on Jul 18, 2005 at 23:30 UTC
That's very smart and I'm very grateful. Can I spoil all that for asking how I get a generalised solution where I can have the same code run both evaluated and non-evaluated regexes? How do I code, for instance, so that $regex may sometimes be: `my $regex = ['(foo)',sub {\"bar$1"}];` [download] but at other times just `my $regex = ['foo','bar'];` [download] I'm guessing I'll need to use `ref()` or something? ($_='kkvvttuu bbooppuuiiffss qqffssmm iibbddllffss') =~y~b-v~a-z~s; print	[reply] [d/l] [select]
Re^3: Regex Substitution, Interpolating on the RHS? by Roy Johnson (Monsignor) on Jul 19, 2005 at 13:20 UTC
Yes, you would check `ref($regex->[1])`. It will be `'CODE'` for a coderef and false (empty) for a non-ref. You could put the whole test/selection into the block like so: `my $regex = ['(foo)',sub {\"bar$1"}]; my $string = 'blah foo blah'; $string =~ s/$regex->[0]/${(ref($regex->[1]) eq 'CODE' ? $regex->[1]-> +() : \$regex->[1])}/; print "S=$string\n";` [download] Caution: Contents may have been coded under pressure.	[reply] [d/l] [select]
Re: Regex Substitution, Interpolating on the RHS? by jbrugger (Parson) on Jul 18, 2005 at 11:23 UTC
~~try `"bar$1"`?~~ update: Won't give an error but doesn't work... I assume you need the outpur `blah barfoo blah` like the regex: `s/(foo)/bar$1/`does... this created `foo bar blah`... weird thing is that the parentheses are evaluated correctly in the regex: this works: `my $regex = ['(foo)','bar']; $string =~ s/$regex->[0]/$regex->[1]$1/;` [download] update2: dave_the_m was quicker and right, it's `#!/usr/bin/perl use strict; my $regex = ['(foo)', '"bar$1"' ]; my $string = 'blah foo blah'; $string =~ s/$regex->[0]/$regex->[1]/ee; print "$string\n";` [download] "We all agree on the necessity of compromise. We just can't agree on when it's necessary to compromise." - Larry Wall.	[reply] [d/l] [select]
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