C++ Difference Between Std::ref(T) And T&?

Answer :

Well ref constructs an object of the appropriate reference_wrapper type to hold a reference to an object. Which means when you apply:

auto r = ref(x); 

This returns a reference_wrapper and not a direct reference to x (ie T&). This reference_wrapper (ie r) instead holds T&.

A reference_wrapper is very useful when you want to emulate a reference of an object which can be copied (it is both copy-constructible and copy-assignable).

In C++, once you create a reference (say y) to an object (say x), then y and x share the same base address. Furthermore, y cannot refer to any other object. Also you cannot create an array of references ie code like this will throw an error:

#include <iostream> using namespace std;  int main() {     int x=5, y=7, z=8;     int& arr[] {x,y,z};    // error: declaration of 'arr' as array of references     return 0; } 

However this is legal:

#include <iostream> #include <functional>  // for reference_wrapper using namespace std;  int main() {     int x=5, y=7, z=8;     reference_wrapper<int> arr[] {x,y,z};     for (auto a: arr)         cout << a << " ";     return 0; } /* OUTPUT: 5 7 8 */ 

Talking about your problem with cout << is_same<T&,decltype(r)>::value;, the solution is:

cout << is_same<T&,decltype(r.get())>::value;  // will yield true 

Let me show you a program:

#include <iostream> #include <type_traits> #include <functional> using namespace std;  int main() {     cout << boolalpha;     int x=5, y=7;     reference_wrapper<int> r=x;   // or auto r = ref(x);     cout << is_same<int&, decltype(r.get())>::value << "\n";     cout << (&x==&r.get()) << "\n";     r=y;     cout << (&y==&r.get()) << "\n";     r.get()=70;     cout << y;     return 0; } /* Ouput: true true true 70 */ 

See here we get to know three things:

1. A reference_wrapper object (here r) can be used to create an array of references which was not possible with T&.

2. r actually acts like a real reference (see how r.get()=70 changed the value of y).

3. r is not same as T& but r.get() is. This means that r holds T& ie as its name suggests is a wrapper around a reference T&.

I hope this answer is more than enough to explain your doubts.

std::reference_wrapper is recognized by standard facilities to be able to pass objects by reference in pass-by-value contexts.

For example, std::bind can take in the std::ref() to something, transmit it by value, and unpacks it back into a reference later on.

void print(int i) {     std::cout << i << '\n'; }  int main() {     int i = 10;      auto f1 = std::bind(print, i);     auto f2 = std::bind(print, std::ref(i));      i = 20;      f1();     f2(); } 

This snippet outputs :

10 20 

The value of i has been stored (taken by value) into f1 at the point it was initialized, but f2 has kept an std::reference_wrapper by value, and thus behaves like it took in an int&.

A reference (T& or T&&) is a special element in C++ language. It allows to manipulate an object by reference and has special use cases in the language. For example, you cannot create a standard container to hold references: vector<T&> is ill formed and generates a compilation error.

A std::reference_wrapper on the other hand is a C++ object able to hold a reference. As such, you can use it in standard containers.

std::ref is a standard function that returns a std::reference_wrapper on its argument. In the same idea, std::cref returns std::reference_wrapper to a const reference.

One interesting property of a std::reference_wrapper, is that it has an operator T& () const noexcept;. That means that even if it is a true object, it can be automatically converted to the reference that it is holding. So:

• as it is a copy assignable object, it can be used in containers or in other cases where references are not allowed
• thanks to its operator T& () const noexcept;, it can be used anywhere you could use a reference, because it will be automatically converted to it.