为什么右值引用类型的模板参数可以绑定到左值类型？

Question

As I know, a rvalue reference cannot be bound to a lvalue. eg,

void func(Foo &&f) {}
int main() {
 Foo f;
 func(f);
}

compiler complains: error: cannot bind rvalue reference of type 'Foo&&' to lvalue of type 'Foo

But, why a template argument of rvalue reference type can be bound to a lvalue? eg,

template <typename T> void funcTemp(T &&arg) {}
int main() {
 Foo f;
 funcTemp(f);
}

The compiler won't complain the error. Why?

Answer 1

You can read this article Universal References in C++11 to understand. Here some part of it:

If a variable or parameter is declared to have type T&& for some deduced type T, that variable or parameter is a universal reference.

 Widget&& var1 = someWidget; // here, “&&” means rvalue reference auto&& var2 = var1; // here, “&&” does not mean rvalue reference template<typename T> void f(std::vector<T>&& param); // here, “&&” means rvalue reference template<typename T> void f(T&& param); // here, “&&”does not mean rvalue reference 

Here a relevant for your case excerpt from the Standard :

... function template parameter type (call it P) ... If P is a forwarding reference and the argument is an lvalue, the type “lvalue reference to A” is used in place of A for type deduction.