Substituting without modifying

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

/me kneels before the Perl monks...

After asking the resident Perl geeks, I turn to you for wisdom. What is the Right Way to use a variable as a basis for an s///, without modifying it?

    $foo = "bar baz";
    (my $qux = $foo) =~ s/bar/quux/;
[download]

Here $foo is unchanged, and $qux equals "quux baz". Is there a more elegant way to do this, e.g. without first copying $foo?

(BTW, brother dep informs me that using 'my' inside of the parentheses is not a good thing. Maybe that's why I insist on doing it that way... ;] What traps can I run in to doing that?)

Comment on Substituting without modifying Download Code

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Re: Substituting without modifying by chipmunk (Parson) on May 20, 2001 at 21:19 UTC
If I understand your question... You want to perform a substitution on the value in $foo, without changing the value of $foo or copying the value to another variable first? I think that you will be able to meet any two of those three requirements. :) Here's something to consider, although it requires $foo to be a package variable rather than a lexical, and still involves copying the value: `$foo = "bar baz"; { (local $foo = $foo) =~ s/bar/quux/; # do something with the modified $foo } # now the original $foo is back again` [download] BTW, there is nothing wrong with your use of my inside the parentheses.	[reply] [d/l]
Re: Re: Substituting without modifying by sierrathedog04 (Hermit) on May 20, 2001 at 22:51 UTC
Your method works great using lexical variables as well, because the perl interpreter correctly interpolates the temporary value of $foo when that temporary value is in scope. `use strict; my $foo = "bar baz"; { (my $foo = $foo) =~ s/bar/quux/; # do something with the modified $foo $foo = "changed it"; print "The temporary value of foo is $foo.\n"; } # now the original $foo is back again print "The original value of foo is $foo.\n";` [download] Running under ActiveState 5.6.1 and Windows ME, the above code first returns the temporary value of "changed it" and then the original value of "bar baz".	[reply] [d/l]
Re: Substituting without modifying by chromatic (Archbishop) on May 21, 2001 at 00:57 UTC
That's syntactically equivalent to a two-step operation: `my $qux = $foo; $qux =~ s/bar/quux/;` [download] It just takes one line of code, so it's more idiomatic. There's absolutely nothing wrong with it, and I quite like it myself. There are only two things you might run into -- programmers who don't understand the idiom, and the looks of a my inside braces. If it makes your life easier, go for it.	[reply] [d/l]
Re: Substituting without modifying by lemming (Priest) on May 20, 2001 at 22:30 UTC
That is a method right out of the camel. You have to use the parentheses, otherwise $qux would have the number of times the pattern matched and $foo would be changed. Nothing wrong with my in my opinion	[reply]