Function template does not recognize lvalue value

Question

I have a problem in my code

Here is a simplified version:

#include <iostream> class A { public : template <class T> void func(T&&)//accept rvalue { std::cout<<"in rvalue\n"; } template <class T> void func(const T&)//accept lvalue { std::cout<<"in lvalue\n"; } }; int main() { A a; double n=3; a.func(n); a.func(5); } 

I expect the output to be:

 in lvalue in rvalue 

but this

 in rvalue in rvalue 

why?!

Answer 1

template <class T> void func(T&&) is a reference link for a universal link .

To check what you want try: ( Live example )

 template <typename T> class A { public: void func(T&&)//accept rvalue { std::cout<<"in rvalue\n"; } void func(T&)//accept lvalue { std::cout<<"in lvalue\n"; } }; int main() { A<double> a; double n = 3; a.func(n); a.func(5.); } 

Answer 2

To build the perfect answer on Jarod42, if you want to preserve the design of the main function template, you can decide based on the type of universal reference parameter deduced:

 #include <iostream> #include <type_traits> struct A { template <typename T> // T is either U or U & void func(T && x) { func_impl<T>(std::forward<T>(x)); } template <typename U> void func_impl(typename std::remove_reference<U>::type & u) { std::cout << "lvalue\n"; } template <typename U> void func_impl(typename std::remove_reference<U>::type && u) { std::cout << "rvalue\n"; } }; 

Answer 3

I think the surprise comes from how template arguments are derived. You will get what you expect if you write:

 a.func<double>(n);