perl code in regex replacement

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

What I want to be able to do:

$x = '123'; $y = $x; $y =~ s/(2)/?{$1+1}/x; print $y
[download]

with the result:

133
[download]

But instead I get the following:

1?{2+1}3
[download]

Is there a regex way to that? The best non-regex way?

Comment on perl code in regex replacement Select or Download Code

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Re: perl code in regex replacement by toolic (Bishop) on Mar 12, 2011 at 18:16 UTC
The best non-regex way? The best way to add 10 to a number is to use `+` `use warnings; use strict; my $x = 123; my $y = $x + 10; print $y;` [download] But, I suspect your example is a much-reduced version of a larger problem you are trying to solve.	[reply] [d/l] [select]
Re: perl code in regex replacement by ww (Archbishop) on Mar 12, 2011 at 17:29 UTC
Assuming (dangerous) that I've interpreted your problem statement as you intended, one way of doing the job (in the problem you specified; adding 10 to a 3 digit decimal value) and without resort to executing code inside a regex is: `#/usr/bin/perl use strict; use warnings; # 892837 my $x = 123; # no quotes my $y = $x; $y =~ /(\d{2})(\d)/; # regex, but see also perlretut re executing c +ode in a regex $y = ($1+1).$2; print $y;` [download]	[reply] [d/l]
Re: perl code in regex replacement by Anonymous Monk on Mar 12, 2011 at 17:17 UTC
See perlre/perlretut, and use /e if you want code execution	[reply]
Re: perl code in regex replacement by wind (Priest) on Mar 12, 2011 at 18:15 UTC
`my $x = '123'; (my $y = $x) =~ s/(2)/$1+1/e; print $y;` [download] Or `my $x = '123'; (my $y = $x) =~ tr/2/3/; print $y` [download] Or even ;) `my $x = '123'; my $y = $x + 10; print $y` [download]	[reply] [d/l] [select]