I think the answer lies in that $_ is set to $test inside the regular expression statement, so it is not a numerical argument unless $test is numerical. As proof (as I was curious to find out what was going on I tried a couple of things to come to a conclusion)

1. When I changed the $_ to a numerical value inside the regex instead:

if ($test =~m#(??{substr($test, 0, 8)})#) {  }

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and the regex will work, without warning.

2. When I set the string $test to a something numerical (12,444.343,etc.) and again it will work without warning.

3. I tried setting $_ to a number outside and again it failed. 

use strict;
use warnings;

my $test = "fee fi fo fee";
$_++;

if ($test =~ m!((??{ substr($test,0,$_) }))!) { };

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So what I figured it dealt with the $_ So what I finally did was the following as final proof that I was on the right track: 

use strict;
use warnings;

my $test = "fee fi fo free";

if ($test =~ m!((??{ print "$_\n";substr($test,0,$_) }))!) {
};

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Which when it runs displays the following:
fee fi fo free
Argument "fee fi fo free" isn't numeric in substr at (re_eval 1) line 
+1.

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-enlil

In reply to Re: Troubles with m!(??{substr(...)})! by Enlil
in thread Troubles with m!(??{substr(...)})! by jens

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