Accessing the [] operator from a pointer

Question

If I define a pointer to an object that defines the [] operator, is there a direct way to access this operator from a pointer?

For example, in the following code I can directly access Vec's member functions (such as empty()) by using the pointer's -> operator, but if I want to access the [] operator I need to first get a reference to the object and then call the operator.

#include <vector>  int main(int argc, char *argv[]) {     std::vector<int> Vec(1,1);     std::vector<int>* VecPtr = &Vec;  if(!VecPtr->empty())      // this is fine     return (*VecPtr)[0]; // is there some sort of ->[] operator I could use?  return 0; } 

I might very well be wrong, but it looks like doing (*VecPtr).empty() is less efficient than doing VecPtr->empty(). Which is why I was looking for an alternative to (*VecPtr)[].

Answer 1

You could do any of the following:

#include <vector>  int main () {   std::vector<int> v(1,1);   std::vector<int>* p = &v;    p->operator[](0);   (*p)[0];   p[0][0]; } 

By the way, in the particular case of std::vector, you might also choose: p->at(0), even though it has a slightly different meaning.

Answer 2

return VecPtr->operator[](0); 

...will do the trick. But really, the (*VecPtr)[0] form looks nicer, doesn't it?