Clarification about identity and scalars

Also goes towards #114

Author: JJ

doc/Language/structures.pod6

@@ -30,10 +30,30 @@ can still access its internal structure.
 
     (1,2,3, $(4,5))[3][0].say; # OUTPUT: «4␤»
 
+An interesting side effect, or maybe intended feature, is that scalarization
+conserves identity of complex structures.
+
+    for ^2 {
+         my @list =(1,1);
+         say @list.WHICH;
+    } # OUTPUT: «Array|93947995146096␤Array|93947995700032␤»
+
+Every time C<(1,1)> is assigned, the variable created is going to be different
+in the sense that C<===> will say it is; as it is shown, different values of the
+internal pointer representation are printed. However
+
+    for ^2 {
+      my $list =(1,1);
+      say $list.WHICH
+    }   # OUTPUT: «List|94674814008432␤List|94674814008432␤»
+
+In this case, C<$list> is using the Scalar sigil and thus will be an C<Scalar>. Any scalar will the same value will be exactly the same, as shown when printing the pointers.
 
 =head1 Complex data structures
 
-TBD
+Complex data structures fall in two different broad categories L<Positional|/type/Positional>, or list-like and L<Associative|/type/Associative>, or key-value pair like, according to how you access its first-level elements. In general, complex data structures, including objects, will be a combination of both, with object properties assimilated to key-value pairs.
+
+
 
 =head1 Functional structures