String substitution

abhay180 has asked for the wisdom of the Perl Monks concerning the following question:

Why is the following substitution of $expr (with $n_expr) in $line not working?

print "BEFORE:LINE = $line,EXPR= $expr,NEW_EXPR= $n_expr\n";
$line =~ s/$expr/$n_expr/;
print "AFTER:LINE = $line,EXPR= $expr,NEW_EXPR= $n_expr\n";
[download]

The o/p of above is:
BEFORE:LINE = input [`CLW:::0] abc;,EXPR = [`CLW:::0], NEW_EXPR= [$CLW
+:0]
AFTER:LINE = input [`CLW:::0] abc;,EXPR = [`CLW:::0], NEW_EXPR= [$CLW:
+0]
[download]

The PERL-version is : V5.8.8; OS is: x86_64-linux-thread-multi

Comment on String substitution Select or Download Code

Replies are listed 'Best First'.
Re: String substitution by Corion (Patriarch) on Jul 12, 2010 at 16:55 UTC
Your search string is interpreted as a regular expression, and it contains regular expression meta characters. See perlre. Maybe you wanted to use quotemeta or \Q\|? `$line =~ s/\Q$expr/$n_expr/;` [download]	[reply] [d/l]
Re^2: String substitution by abhay180 (Sexton) on Jul 12, 2010 at 17:03 UTC
Thanks a lot. That helps. Indeed i seem to have forgotten the basic rules of substitution :-)	[reply]
Re: String substitution by moritz (Cardinal) on Jul 12, 2010 at 16:54 UTC
Because `s///` treats the first string is a regex, and not all regexes match their own textual representation. Hint: `[...]` has a special meaning, see perlretut. You can avoid that with `\Q$EXPR\E`. Perl 6 - links to (nearly) everything that is Perl 6.	[reply] [d/l] [select]