C++11 and above

How std forward( ) converts to r-value in c++?

January 19, 2021
Vaibhav
12,781 Comments
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Just like, std::move( ), std::forward( ) also casts its argument (to R-value), however, unlike std::move( ) which is an unconditional cast, std::forward( ) casts under certain conditions. The forward will do casting to R-value only when the argument is bound to an R-value.

Considering only a pure technical perspective, a forward( ) can do all and the move( ) isn’t necessary. However, use of move( ) function is more convenient. Basically, std::move( ) needs only single thing, i.e. function’s argument value. Whereas the forward( ) function needs 2 things – firstly, the function’s argument value and secondly a template type argument in < > (angular) brackets.

Move:

int && izvar = std::move(4);

Forward:

int &&ipvar = std::forward<int>(3);

Universal Reference Pointer :

Generally, std::forward is used with a universal reference parameter, which can bound either to L-value or R-value. The forward will do cast to R-value only when the universal reference parameter is bound to R-value.

#include <iostream> //main header
using namespace std; //for namespace

void funda(int&& avar)    //R-value overload
{
    cout << "R-Value overload called" << endl;
}

void funda(int& avar)     //L-value overload
{
    cout << "L-Value overload called" << endl;
}

template<typename T>
void call_funda(T&& parg) //Argument Universal Reference
{
    funda(std::forward<T>(parg));
}

int main()
{
    int mvar=5;
    call_funda(mvar);    // Calls L-value overload
    call_funda(4);       // Calls R-value overload
    return 0;
}

When call_funda is called with L-value and R-value respectively, the corresponding overloads are called

Basic implementation of forward( ):

template <typename T>
T&& forward(typename std::remove_reference<T>::type& param)
{
    return static_cast<T&&>(param);
}

The std::remove_reference template class is part of C++ library, however, this also can be easily implemented (see below).

template <class T>
struct remove_reference { typedef T type; };

template <class T>
struct remove_reference<const T> { typedef const T type; };

template <class T>
struct remove_reference<T&> { typedef T type; };

template <class T>
struct remove_reference<const T&> { typedef const T type; };

template <class T>
struct remove_reference<T&&> { typedef T type; };

template <class T>
struct remove_reference<const T&&> { typedef const T type; };

How it all works ?

The concept of forward works using reference collapsing rules. When the forward function is called with an argument which is L-value, then T will deduce to become L-value type.

Case1: if the argument is int&, then T will also become T&

Therefore, the code will become

int& && forward(typename std::remove_reference<int&>::type& param)
{
    return static_cast<int& &&>(param);
}

The reference collapsing rules will make “&” of “&&” to become “&”, hence the forward( ) code will become as shown below:

int& forward(int& param)
{
    return static_cast<int&>(param);
}

Clearly, the forward casts the param to L-value type.

Case2: if the argument is int&&, then T will also become T&&

Therefore, the code will become

int&& && forward(typename std::remove_reference<int&&>::type& param)
{
    return static_cast<int&& &&>(param);
}

The reference collapsing rules will make “&&” of “&&” to become “&&”, hence the forward( ) code will become as shown below:

int&& forward(int& param)
{
    return static_cast<int&&>(param);
}

Clearly, the forward casts the param to R-value type.

Demonstration of our basic implementation of forward( )

#include <iostream> //main header
using namespace std;//for namespace

void funda(int&& avar)
{
    cout << "R-Value overload called" << endl;
}

void funda(int& avar)
{
    cout << "L-Value overload called" << endl;
}

template <typename T>
T&& my_forward(typename std::remove_reference<T>::type& param)
{
    return static_cast<T&&>(param);
}

template<typename T>
void call_funda(T&& parg)
{
    funda(my_forward<T>(parg));
}


int main()
{
    int mvar=5;
    call_funda(mvar);     // Calls L-value overload
    call_funda(4);       // Calls R-value overload
        
    return 0;  
}

Main Funda: std::forward( ) do not forward anything, but it casts its argument to R-value during compile time on the basis of reference collapsing rules

Related Topics:

 What are the drawbacks of using enum ?
Which member functions are generated by compiler in class?
How to stop compiler from generating special member functions?
Compiler Generated Destructor is always non-virtual
How to make a class object un-copyable?
Why virtual functions should not be called in constructor & destructor ?
Explaining C++ casts
How pointer to class members are different ?
How std::forward( ) works?
What is reference collapsing?
How std::move() function works?
How delete keyword can be used to filter polymorphism
Rule of Three

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