Re: Regex Substitution, Interpolating on the RHS?

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";
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Caution: Contents may have been coded under pressure.

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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]