function template does not recognize lvalue

问题

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 ?!

回答1:

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.);
}

回答2:

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";
    }
};

回答3:

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);

来源：https://stackoverflow.com/questions/22166767/function-template-does-not-recognize-lvalue