Function Pointer - Automatic Dereferencing [duplicate]

Question

Possible Duplicate:
How does dereferencing of a function pointer happen?

  void myprint(char* x) {       printf("%s\n", x);    }    int main() {      char* s = "hello";      void (*test)(char*);      void (*test2)(char*);       test = myprint;      test2 = &myprint;       test(s);      (*test)(s);      test2(s);      (*test2)(s);    } 

Can anyone explain to me why all of the above code is valid? "hello" is printed four times. By applying the function pointer, is it implicitly derefenced? Basically I want to know how function pointers are actually stored, because the above is kind of confusing.

Answer 1

This is just a quirk of C. There's no other reason but the C standard just says that dereferencing or taking the address of a function just evaluates to a pointer to that function, and dereferencing a function pointer just evaluates back to the function pointer.

This behavior is (thus obviously) very different from how the unary & and * operators works for normal variables.

So,

test2 = myprint; test2 = &myprint; test2 = *myprint; test2 = **********myprint; 

All just do exactly the same, gives you a function pointer to myprint

Similarly,

test2(s); (*test2)(s); (***********test2)(s); 

Does the same, call the function pointer stored in test2. Because C says it does.