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'.

In reply to Re^12: Using pos() inside regexp (no /e) by braveghost
in thread Using pos() inside regexp by braveghost

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