How to correctly use std::reference_wrappers

Question

I am trying to understand std::reference_wrapper.

The following code shows that the reference wrapper does not behave exactly like a reference.

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

int main()
{
    std::vector<int> numbers = {1, 3, 0, -8, 5, 3, 1};

    auto referenceWrapper = std::ref(numbers);
    std::vector<int>& reference = numbers;

    std::cout << reference[3]              << std::endl;
    std::cout << referenceWrapper.get()[3] << std::endl; 
              // I need to use get ^
              // otherwise does not compile.
    return 0;
}

If I understand it correctly, the implicit conversion does not apply to calling member functions. Is this an inherent limitation? Do I need to use the std::reference_wrapper::get so often?

Another case is this:

#include <iostream>
#include <functional>

int main()
{
    int a = 3;
    int b = 4;
    auto refa = std::ref(a);
    auto refb = std::ref(b);
    if (refa < refb)
        std::cout << "success" << std::endl;

    return 0;
}

This works fine, but when I add this above the main definition:

template <typename T>
bool operator < (T left, T right)
{
    return left.someMember();
}

The compiler tries to instantiate the template and forgets about implicit conversion and the built in operator.

Is this behavior inherent or am I misunderstanding something crucial about the std::reference_wrapper?

Answer 1

Class std::reference_wrapper<T> implements an implicit converting operator to T&:

operator T& () const noexcept;

and a more explicit getter:

T& get() const noexcept;

The implicit operator is called when a T (or T&) is required. For instance

void f(some_type x);
// ...
std::reference_wrapper<some_type> x;
some_type y = x; // the implicit operator is called
f(x);            // the implicit operator is called and the result goes to f.

However, sometimes a T is not necessarily expected and, in this case, you must use get. This happens, mostly, in automatic type deduction contexts. For instance,

template <typename U>
g(U x);
// ...
std::reference_wrapper<some_type> x;
auto y = x; // the type of y is std::reference_wrapper<some_type>
g(x);       // U = std::reference_wrapper<some_type>

To get some_type instead of std::reference_wrapper<some_type> above you should do

auto y = x.get(); // the type of y is some_type
g(x.get());       // U = some_type

Alternativelly the last line above could be replaced by g<some_type>(x);. However, for templatized operators (e.g. ostream::operator <<()) I believe you can't explicit the type.