I have a problem in my code
Here is simplified version of it :
#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 it is
in rvalue in rvalue why ?!
template <class T> void func(T&&) is universal reference forwarding reference.
To test 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.); } 
To build on Jarod42's fine answer, if you want to keep the design of having a primary function template, you can decide based on the deduced type of the universal reference parameter:
#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"; } }; I think the surprise comes from the way the template argument are deduced. You'll get what you expect if you write:
a.func<double>(n); 
&?