Now I'm starting to wonder if your assertions can be taken at face value. However, in fairness, maybe I'm just missing the point(s).

On my "unix-like" system (Ubuntu 10.4, perl 5.10.1), :

~$ perl -e '$_="qwerty"; s/r/print pos(). "\n";/e; print $_;'
3
qwe1ty
[download]

And your latest code (third block in ), modified (a) to actually do something with $*& (i.e. print it); (b) to be more explicit; and (c) to produce readable output under Ubuntu, looks like line 1 below and produces the output in lines 2-4:

perl -e "\$_='qwerty'; s/r/print pos(). \"\n\";print \$&.\"\n\";/e; pr
+int \$_ . \"\n\";"
3
r
qwe1ty
~$
[download]

(See also ikegami's Re^4: Using pos() inside regexp (no /e).)

or, looking at the issue through the mechanism of a script, executed in the Gnome terminal OR bash:

#!/usr/bin/perl
use strict;
use warnings;
use feature  qw(say);

# perl -le "\$_='qwerty';s/r/print pos()/e;"

my $str ='qwerty'; 
$str =~ s/[rq]/say pos($str); say $&;/eg; 
print "\t\$str: $str \n\n";

my $str0 ='qwerty'; 
$str0 =~ s/(r|q)/say pos($str0); say $&;/eg; 
print "\t\$str0: $str0 \n\n";

my $str1 ='qwerty'; 
$str1 =~ s/[rz]/say pos($str0); say $&;/eg; 
print "\t\$str1: $str1 \n\n";


my $str2 ='qwerty_erk'; 
$str2 =~ s/r/say pos($str2); say $&;/eg; 
print "\t\$str2: $str2 \n\n";

my $var='four'; 
$var =~ s/[fu]/say pos($var); say $&;/eg; 
print $var . "\n\n";
[download]

I don't think execution of the code above supports your interpretation of the appearance of the number 1 in the output:

~/pl_test$ perl braveghost.pl
0
q
3
r
    $str: 1we1ty 

0
q
3
r
    $str0: 1we1ty 

Use of uninitialized value in say at braveghost.pl line 17.

r
    $str1: qwe1ty 

3
r
8
r
    $str2: qwe1ty_e1k 

0
f
2
u
1o1r

~/pl_test$
[download]

Does your statement that "1(one) is returned by print function and s/// insert it because it is last return at executed section" mean you believe that invoking print in the regex somehow assigns its return value to pos()?

BTW, re "macos" -- "Mac OS" is the Apple TM style.

hmmm whats for are you modified my code?

executed part of regexp works as function and it return value which will used by regexp to make replace, i.e. result of last executed command will be a string to replace:

example

# perl -e "\$_='qwerty';\$v='_MATCH_'; s/r/print \$v/; print \"\n\".\$
+_;";

qweprint _MATCH_ty

# perl -e "\$_='qwerty';\$v='_MATCH_'; s/r/print \$v/e; print \"\n\".\
+$_;";
_MATCH_
qwe1ty

# perl -e "\$_='qwerty';\$v='_MATCH_'; s/r/print \$v;\$v/e; print \"\n
+\".\$_;";
_MATCH_
qwe_MATCH_ty
[download]

1(one) is result of print

perldoc -f print
       print FILEHANDLE LIST
       print LIST
       print   Prints a string or a list of strings.  Returns true if
               successful.  ...
[download]

Does your statement that "1(one) is returned by print function and s/// insert it because it is last return at executed section" mean you believe that invoking print in the regex somehow assigns its return value to pos()?

No, I mean that: was 'qwerty' became 'qwe1ty' because 'print pos()' return '1'.