Compiling regular expressions to perform substitution

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

I am writing a program which will read a bunch of substitution regular expressions from a configuration file and then apply them to the contents of other files. The problem I am running into is that the regexps don't seem to be compiling because they just aren't working. The following code illustrates what I mean:

#!/usr/bin/perl -w

$regex = "s/foo/bar/g";
$compiled = qr/$regex/;   # program works the same 
                          # whether I do this or not

$string = "foo fru foodyfoo";

$string =~ $compiled;
print "$string\n";      # no substitution! $string is still
                        # "foo fru foodyfoo" instead of
                        # "bar fru bardybar".
[download]

I know this is probably a stupid question but that's only because I am a stupid person. Sorry if my question makes you groan with disgust. I like chicken.

Comment on Compiling regular expressions to perform substitution Download Code

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Re: Compiling regular expressions to perform substitution by broquaint (Abbot) on Apr 11, 2002 at 13:11 UTC
The reason they aren't compiling properly is that `s///` is an operator. If you are going to be compiling regexps from a string, you need only compile the left-hand side of a `s///`, as the right-hand side is essentially a double quoted string. `use strict; my $re = qr/foo./; my $str = "foo bar baz"; $str =~ s/$re/FOOD /; print "$str\n"; __output__ FOOD bar baz` [download] HTH broquaint	[reply] [d/l]
Re: Re: Compiling regular expressions to perform substitution by Stegalex (Chaplain) on Apr 11, 2002 at 13:20 UTC
Well pluck my feathers and fry me in a vat! You're right. Thanks! I like chicken.	[reply]
It doesn't mean what you think it means by Fletch (Bishop) on Apr 11, 2002 at 13:14 UTC
`qr//` only compiles regular expressions, not substitution operators. What you've done is compiled a regular expression for matching `s/foo/bar/g'. You can use compiled regexen on the left hand side of an `s///` expression, but you can't use one to capture the whole expression. Now if what you're trying to do is capture an operation or three to be performed on a string that you can select programattically, you possibly could use a hash of coderef like this: `my %mungers = ( foo => sub { $_[0] =~ s/foo/bar/g }, fru => sub { $_[0] =~ s/(fru.)/uc($1)/e }, ); if( $phase_of_moon > WAXING_GIBBOUS ) { $mungers->{foo}->( $string ); } else { $mungers->{fru}->( $string ); }` [download] Update:* Feh, as was pointed out that first `->` inside the if/else blocks is extraneous. That's what I get for posting while trying to eat breakfast.	[reply] [d/l] [select]
Re: It doesn't mean what you think it means by Stegalex (Chaplain) on Apr 11, 2002 at 15:05 UTC
Slight modification needed to make this compile. Here it is: `my %mungers = ( foo => sub { $_[0] =~ s/foo/bar/g }, fru => sub { $_[0] =~ s/(fru.*)/uc($1)/e }, ); if( $phase_of_moon > WAXING_GIBBOUS ) { $mungers{foo}->( $string ); # this line changed } else { $mungers{fru}->( $string ); # this line changed }` [download] Again, thanks for everyone's help. I like chicken.	[reply] [d/l]


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