Dynamic and static Scoping program differences

Question

int x;  int main() {    x = 14;    f();     g(); }  void f() {    int x = 13;    h(); }  void g() {    int x = 12;    h(); }  void h() {    printf("%d\n",x);   } 

If static scoping is used, what is the result? If dynamic scoping is used, what is the result?

Now if I understand scoping right, the difference between static and dynamic scoping is that static makes variables local to a class. So the value x would be local to void f(), void g() and int main () and dynamic would make them globally available. I'm just not sure how to apply it to this code. If static scoping was used would it only print the last value (12 from void g()) and dynamic scoping would be using all the values of x?

I'm a little confused on how scoping actually works. I know C uses static scoping though.

Answer 1

Static scoping means that x refers to the x declared innermost scope of declaration that has one. Since h is declared inside the global scope, the innermost x is the one in the global scope(it has no access to the xs in f and g, since it was not declared inside them), so the program prints 14 twice.

Dynamic scoping means that x refers to the x declared in the most recent frame of the call-stack the has one. If C used dynamic scoping, h would use the x from either f or g - whichever one that called it - so the program would print 13 and 12.

Answer 2

C/C++ doesn't use Dynamic scoping. Your programming language will use one or the other, you don't get to choose (Unless you are using Clojure! according to Idan Arye below).

Here is a great explanation/comparison with a good example: http://msujaws.wordpress.com/2011/05/03/static-vs-dynamic-scoping/