In Perl, how can I match a number N and the next N characters?

What I'm trying to do seems deceptively simple, but I can't find a way to make it work.

I want to match a number, capture it, and then match the following N characters in the string. Naively, I thought something like this would work:

$myString = "1abc3cdf\n"; 

# Capture the number, and use a back-reference in the {} to define how many characters to match
$myString =~ s/(\d+).{\1}//g;

print $myString;

I expect to get bc, but it just returns the original string 1abc3cdf, ie no substitution has taken place.

I tried with an extended regex (ie s///ge) but that didn't help. Any suggestions?

Solution

You may use the (??{...}) inline code block:

$myString =~ s/(\d+)(??{ ".{$^N}" })//g;

See the Perl demo.

Here is some reference:

This is a "postponed" regular subexpression. It behaves in exactly the same way as a (?{ code }) code block as described above, except that its return value, rather than being assigned to $^R, is treated as a pattern, compiled if it's a string (or used as-is if its a qr// object), then matched as if it were inserted instead of this construct.

During the matching of this sub-pattern, it has its own set of captures which are valid during the sub-match, but are discarded once control returns to the main pattern.

And here is more about the whole regex:

(\d+) - Capturing group #1 that matches any one or more digits
(??{ ".{$^N}" }) - a code block that represents "a "postponed" regular subexpression" whose return value is "treated as a pattern, compiled if it's a string". The "$^N contains whatever was matched by the most-recently closed group (submatch)" (see the perlre reference). In other words, if (\d+) captures 45, the (??{ ".{$^N}" }) turns into .{45} and thus matches any 45 characters other than line break characters.