template
void F(T&& x) {}
If we call it with argument of type int& everything is clear - reference collapsing takes pla
Forwarding references deduce lvalue reference for lvalues and rvalue reference for rvalues. For example, even for lvalue of type int&&
it will still deduce int&
(as it would for lvalue of type int
), likewise, for rvalues of type int
or int&&
it will deduce int&&
:
template<typename T>
class show;
template<typename T>
void F(T&& x)
{ show<decltype(x)>();}
int main() {
int&& x = 5;
F(x);
}
main.cpp:6:3: error: implicit instantiation of undefined template 'show<int &>'
int main() {
F(5);
}
main.cpp:6:3: error: implicit instantiation of undefined template 'show<int &&>'
int main() {
int x = 5;
F(x);
}
main.cpp:6:3: error: implicit instantiation of undefined template 'show<int &>'
int main() {
F([]()->int{return 5;}());
}
main.cpp:6:3: error: implicit instantiation of undefined template 'show<int &&>'
Since C++17, there is the concept of forwarding reference in the C++ standard. Normaly the template parameter is deduced as a non reference. But for the specific case of a forwarding reference if the corresponding argument is a lvalue, the parameter is deduced as a reference. C++ standard (N4700)[temp.over.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])). 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.[...]
For the concern of function call, it has the same meaning as this equivalent paragraph in the C++11(N337) and C++14(N414) [temp.over.deduct.call]/3:
[...]If P is an rvalue reference to a cv-unqualified template parameter and the argument is an lvalue, the type “lvalue reference to A” is used in place of A for type deduction[...]