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

TL; DR

Use std::reference_wrapper like this:

#include 
#include 
#include 
#include 

int main()
{
    std::string hello = "Hello, ";
    std::string world = "everyone!";
    typedef std::vector> vec_t;
    vec_t vec = {hello, world};
    vec[1].get() = "world!";
    std::cout << hello << world << std::endl;
    return 0;
}

Demo

Long answer

As standard suggests, for a standard container X containing objects of type T, T must be Erasable from X.

Erasable means that the following expression is well formed:

allocator_traits::destroy(m, p)

A is container's allocator type, m is allocator instance and p is a pointer of type *T. See here for Erasable definition.

By default, std::allocator is used as vector's allocator. With the default allocator, the requirement is equivalent to the validity of p->~T() (Note the T is a reference type and p is pointer to a reference). However, pointer to a reference is illegal, hence the expression is not well formed.