Template reference argument deduction failure in C++

Question

Why following code doesn't compile in C++14 compiler? If I use

const int i = 10;
int n = fun(i);

The compiler gives an error.

But, If I use

int n = fun(10);

instead of above statements, it's working fine.

Example:

template<typename T>
int fun(const T&&)
{
    cout<<"fun"<<endl;
}

int main()
{
 // int i = 10;         // Not work
    const int i = 10;   // Not work
    int n = fun(i);  
 // int n = fun(10);    // Working fine
}

Answer 1

It fails because adding the const prevents it from being a forwarding reference. It becomes a regular reference to a const rvalue:

[temp.deduct.call/3]

... A forwarding reference is an rvalue reference to a cv-unqualified template parameter that does not represent a template parameter of a class template (during class template argument deduction ([over.match.class.deduct])). ...

And you pass it an lvalue. It fails the match.