Why can't I make a vector of references?

Question

When I do this:

std::vector hello;

Everything works great. However, when I make it a vector of references instead:

Answer 1

yes you can, look for std::reference_wrapper, that mimics a reference but is assignable and also can be "reseated"

Answer 2

As the other comments suggest, you are confined to using pointers. But if it helps, here is one technique to avoid facing directly with pointers.

You can do something like the following:

vector<int*> iarray;
int default_item = 0; // for handling out-of-range exception

int& get_item_as_ref(unsigned int idx) {
   // handling out-of-range exception
   if(idx >= iarray.size()) 
      return default_item;
   return reinterpret_cast<int&>(*iarray[idx]);
}

Answer 3

Ion Todirel already mentioned an answer YES using std::reference_wrapper. Since C++11 we have a mechanism to retrieve object from std::vector and remove the reference by using std::remove_reference. Below is given an example compiled using g++ and clang with option
-std=c++11 and executed successfully.

#include <iostream>
#include <vector>
#include<functional>

class MyClass {
public:
    void func() {
        std::cout << "I am func \n";
    }

    MyClass(int y) : x(y) {}

    int getval()
    {
        return x;
    }

private: 
        int x;
};

int main() {
    std::vector<std::reference_wrapper<MyClass>> vec;

    MyClass obj1(2);
    MyClass obj2(3);

    MyClass& obj_ref1 = std::ref(obj1);
    MyClass& obj_ref2 = obj2;

    vec.push_back(obj_ref1);
    vec.push_back(obj_ref2);

    for (auto obj3 : vec)
    {
        std::remove_reference<MyClass&>::type(obj3).func();      
        std::cout << std::remove_reference<MyClass&>::type(obj3).getval() << "\n";
    }             
}