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Why is my code printing rvalue 2 times instead of rvalue & lvalue?

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7

So I wanted to practice the usage of std::forward and created a Test class with 2 constructors. 1 with T& and the other with T&& as overload. T& prints lvalue, and T&& prints rvalue so I know which one of the constructors is being used. I create 2 instances of class on stack and to my surprise both of which are using the T&& overload.

#include <iostream>
#include <type_traits>
#include <utility>

template <class T> auto forward(T &&t) 
 if constexpr (std::is_lvalue_reference<T>::value) 
 return t;
 
 return std::move(t);


template <class T> class Test 
public:
 Test(T &) std::cout << "lvalue" << std::endl; ;
 Test(T &&) std::cout << "rvalue" << std::endl; ;
;

int main() 
 int x = 5;
 Test<int> a(forward(3));
 Test<int> b(forward(x));
 return 0;

I tried using the original std::forward function and implementing it but both times it printed rvalue x2. What am I doing wrong?

c++ forward rvalue lvalue stdmove

share|improve this question

edited 9 hours ago

Evg

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asked 9 hours ago

3l4x3l4x

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3l4x is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  auto is never a reference.
  
  – Evg
  9 hours ago
  

  
  
  
  

add a comment | 

7

So I wanted to practice the usage of std::forward and created a Test class with 2 constructors. 1 with T& and the other with T&& as overload. T& prints lvalue, and T&& prints rvalue so I know which one of the constructors is being used. I create 2 instances of class on stack and to my surprise both of which are using the T&& overload.

#include <iostream>
#include <type_traits>
#include <utility>

template <class T> auto forward(T &&t) 
 if constexpr (std::is_lvalue_reference<T>::value) 
 return t;
 
 return std::move(t);


template <class T> class Test 
public:
 Test(T &) std::cout << "lvalue" << std::endl; ;
 Test(T &&) std::cout << "rvalue" << std::endl; ;
;

int main() 
 int x = 5;
 Test<int> a(forward(3));
 Test<int> b(forward(x));
 return 0;

I tried using the original std::forward function and implementing it but both times it printed rvalue x2. What am I doing wrong?

c++ forward rvalue lvalue stdmove

share|improve this question

edited 9 hours ago

Evg

5,68833 gold badges1919 silver badges4242 bronze badges

asked 9 hours ago

3l4x3l4x

3833 bronze badges

New contributor

3l4x is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  auto is never a reference.
  
  – Evg
  9 hours ago
  

  
  
  
  

add a comment | 

So I wanted to practice the usage of std::forward and created a Test class with 2 constructors. 1 with T& and the other with T&& as overload. T& prints lvalue, and T&& prints rvalue so I know which one of the constructors is being used. I create 2 instances of class on stack and to my surprise both of which are using the T&& overload.

#include <iostream>
#include <type_traits>
#include <utility>

template <class T> auto forward(T &&t) 
 if constexpr (std::is_lvalue_reference<T>::value) 
 return t;
 
 return std::move(t);


template <class T> class Test 
public:
 Test(T &) std::cout << "lvalue" << std::endl; ;
 Test(T &&) std::cout << "rvalue" << std::endl; ;
;

int main() 
 int x = 5;
 Test<int> a(forward(3));
 Test<int> b(forward(x));
 return 0;

I tried using the original std::forward function and implementing it but both times it printed rvalue x2. What am I doing wrong?

c++ forward rvalue lvalue stdmove

share|improve this question

edited 9 hours ago

Evg

5,68833 gold badges1919 silver badges4242 bronze badges

asked 9 hours ago

3l4x3l4x

3833 bronze badges

New contributor

3l4x is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

#include <iostream>
#include <type_traits>
#include <utility>

template <class T> auto forward(T &&t) 
 if constexpr (std::is_lvalue_reference<T>::value) 
 return t;
 
 return std::move(t);


template <class T> class Test 
public:
 Test(T &) std::cout << "lvalue" << std::endl; ;
 Test(T &&) std::cout << "rvalue" << std::endl; ;
;

int main() 
 int x = 5;
 Test<int> a(forward(3));
 Test<int> b(forward(x));
 return 0;

share|improve this question

edited 9 hours ago

Evg

5,68833 gold badges1919 silver badges4242 bronze badges

asked 9 hours ago

3l4x3l4x

3833 bronze badges

New contributor

3l4x is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 9 hours ago

Evg

5,68833 gold badges1919 silver badges4242 bronze badges

asked 9 hours ago

3l4x3l4x

3833 bronze badges

New contributor

3l4x is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 9 hours ago

Evg

5,68833 gold badges1919 silver badges4242 bronze badges

edited 9 hours ago

Evg

5,68833 gold badges1919 silver badges4242 bronze badges

asked 9 hours ago

3l4x3l4x

3833 bronze badges

New contributor

3l4x is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 9 hours ago

3l4x3l4x

3833 bronze badges

1 Answer
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12

Your problem stems from the return type of forward. You use auto as the return type which will not deduce a reference for you. That means when you do return, no matter which branch it returns from, you return by value which means you have a prvalue.

What you need is decltype(auto) so you return an rvalue or lvalue reference, depending on the return statement. Using

template <class T> decltype(auto) forward(T &&t) 
 if constexpr (std::is_lvalue_reference<T>::value)
 return t;
 else
 return std::move(t);

gives you the output:

rvalue
lvalue

share|improve this answer

edited 9 hours ago

answered 9 hours ago

NathanOliverNathanOliver

106k1919 gold badges157157 silver badges233233 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Isn't there such a thing as an auto&, or am I making up silliness?
  
  – Chipster
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Chipster There is, but that will always return an lvalue reference, which would not be what you want in a forwarding function.
  
  – NathanOliver
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you, problem solved
  
  – 3l4x
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @3l4x You'e welcome. Good first effort BTW. Generic code and forwarding can be pretty tough to get right.
  
  – NathanOliver
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @NathanOliver, even for compilers. ;)
  
  – Evg
  8 hours ago
  
  

  
  
  
  

add a comment | 

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12

Your problem stems from the return type of forward. You use auto as the return type which will not deduce a reference for you. That means when you do return, no matter which branch it returns from, you return by value which means you have a prvalue.

What you need is decltype(auto) so you return an rvalue or lvalue reference, depending on the return statement. Using

template <class T> decltype(auto) forward(T &&t) 
 if constexpr (std::is_lvalue_reference<T>::value)
 return t;
 else
 return std::move(t);

gives you the output:

rvalue
lvalue

share|improve this answer

edited 9 hours ago

answered 9 hours ago

NathanOliverNathanOliver

106k1919 gold badges157157 silver badges233233 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Isn't there such a thing as an auto&, or am I making up silliness?
  
  – Chipster
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Chipster There is, but that will always return an lvalue reference, which would not be what you want in a forwarding function.
  
  – NathanOliver
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you, problem solved
  
  – 3l4x
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @3l4x You'e welcome. Good first effort BTW. Generic code and forwarding can be pretty tough to get right.
  
  – NathanOliver
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @NathanOliver, even for compilers. ;)
  
  – Evg
  8 hours ago
  
  

  
  
  
  

add a comment | 

12

Your problem stems from the return type of forward. You use auto as the return type which will not deduce a reference for you. That means when you do return, no matter which branch it returns from, you return by value which means you have a prvalue.

What you need is decltype(auto) so you return an rvalue or lvalue reference, depending on the return statement. Using

template <class T> decltype(auto) forward(T &&t) 
 if constexpr (std::is_lvalue_reference<T>::value)
 return t;
 else
 return std::move(t);

gives you the output:

rvalue
lvalue

share|improve this answer

edited 9 hours ago

answered 9 hours ago

NathanOliverNathanOliver

106k1919 gold badges157157 silver badges233233 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Isn't there such a thing as an auto&, or am I making up silliness?
  
  – Chipster
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Chipster There is, but that will always return an lvalue reference, which would not be what you want in a forwarding function.
  
  – NathanOliver
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you, problem solved
  
  – 3l4x
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @3l4x You'e welcome. Good first effort BTW. Generic code and forwarding can be pretty tough to get right.
  
  – NathanOliver
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @NathanOliver, even for compilers. ;)
  
  – Evg
  8 hours ago
  
  

  
  
  
  

add a comment | 

Your problem stems from the return type of forward. You use auto as the return type which will not deduce a reference for you. That means when you do return, no matter which branch it returns from, you return by value which means you have a prvalue.

What you need is decltype(auto) so you return an rvalue or lvalue reference, depending on the return statement. Using

template <class T> decltype(auto) forward(T &&t) 
 if constexpr (std::is_lvalue_reference<T>::value)
 return t;
 else
 return std::move(t);

gives you the output:

rvalue
lvalue

share|improve this answer

edited 9 hours ago

answered 9 hours ago

NathanOliverNathanOliver

106k1919 gold badges157157 silver badges233233 bronze badges

template <class T> decltype(auto) forward(T &&t) 
 if constexpr (std::is_lvalue_reference<T>::value)
 return t;
 else
 return std::move(t);

rvalue
lvalue

share|improve this answer

edited 9 hours ago

answered 9 hours ago

NathanOliverNathanOliver

106k1919 gold badges157157 silver badges233233 bronze badges

answered 9 hours ago

NathanOliverNathanOliver

106k1919 gold badges157157 silver badges233233 bronze badges

answered 9 hours ago

NathanOliverNathanOliver

106k1919 gold badges157157 silver badges233233 bronze badges