# Set the path variable.

 $path = "\\default\\main\\Anand\\toipcs\\Tutorials\\internet";
 
 # Make a copy of it
 
 $curPath = $path;
 
 # Extract (what I assume to be) your user/home directory ("Anand").  
 # However this line won't work because the substitution operator is missing the 's'
 
 $curPath =~ !\default\main\(.*)\Tutorials\internet!$1!;
 #          ^ There should be an 's' here
 
 # Make a copy of the extracted directory name ("Anand") and print it out
 
 $parentdir = $curPath;
  print "parent Directory : $curPath\n";
  
  # This line doesn't doesn't make any sense at all??.
  # It's obviously trying to extract 2 different parts of the path froom something
  # But 
  # a) The is no variable to operate on! It would need to look something like:
  #       ($Pdir, $Sdir) = $path =~ m!(.*)\\(.*)!; !!!NOTE  the doubled backslash (\\)!
  # b) Quite what the intent of the '$1,$2' at the  end was meant to do.
  
 ($Pdir,$Sdir) =~ m!(.*)\(.*)$!$1,$2;
 
 # Nothing will be printed as the line above did nothing!

 print "parent branch : $Pdir \t sub branch $Sdir";