Function returning pointer to itself?

Question

Is it possible in C++ to write a function that returns a pointer to itself?

If no, suggest some other solution to make the following syntax work:

some_type f () {     static int cnt = 1;     std::cout << cnt++ << std::endl; } int main () {     f()()()...(); // n calls } 

This must print all the numbers from 1 to n.

Answer 1

struct function {    function operator () ()    {         //do stuff;        return function();    } };  int main() {    function f;    f()()()()()(); } 

You can choose to return a reference to function if needed and return *this;

Update: Of course, it is syntactically impossible for a function of type T to return T* or T&

Update2:

Of course, if you want one to preserve your syntax... that is

some_type f() { } 

Then here's an Idea

struct functor; functor f(); struct functor {    functor operator()()    {       return f();    } };  functor f() {       return functor(); }  int main() {     f()()()()(); } 

Answer 2

No, you can't, because the return type has to include the return type of the function, which is recursive. You can of course return function objects or something like that which can do this.