Perl Cookbook

Copying and Substituting Simultaneously

Problem

You’re tired of constantly using two separate statements with redundant information, one to copy and another to substitute.

Solution

Instead of:

$dst = $src;
$dst =~ s/this/that/;

use:

($dst = $src) =~ s/this/that/;

Discussion

Sometimes what you wish you could have is the new string, but you don’t care to write it in two steps.

For example:

# strip to basename
($progname = $0)        =~ s!^.*/!!;

# Make All Words Title-Cased
($capword  = $word)     =~ s/(\w+)/\u\L$1/g;

# /usr/man/man3/foo.1 changes to /usr/man/cat3/foo.1
($catpage  = $manpage)  =~ s/man(?=\d)/cat/;

You can even use this technique on an entire array:

@bindirs = qw( /usr/bin /bin /usr/local/bin );
for (@libdirs = @bindirs) { s/bin/lib/ }
print "@libdirs\n";

                  /usr/lib /lib /usr/local/lib

The parentheses are required when combining an assignment if you wish to change the result in the leftmost variable. Normally, the result of a substitution is its success: either "" for failure, or the number of times the substitution was done. Contrast this with the preceding examples where the parentheses surround the assignment itself. For example:

($a =  $b) =~ s/x/y/g;      # copy $b and then change $a
 $a = ($b  =~ s/x/y/g);     # change $b, count goes in $a
               
               
               

See Also

The “Variables” section of Chapter 2 of Programming Perl, and the “Assignment Operators” section of perlop(1)