function template parameter loses const when template argument is explicity passed?

问题

template <typename T>
void test(const T& x) {}

int a {};
int& ref = a;
const int& c_ref = a;

test(c_ref)  // T = int, x = const int&
test<int&>(ref); // T = int& , x = int&

Why does the function template parameter x loses the const qualifier?

回答1:

In the explicit (non-deduced) instantiation

test<int&>(ref);

this is the (theoretical) signature you get

void test<int&>(const (int&)& x)

which shows that the const-qualification applies to the whole (int&), and not only the int. const applies to what is left, and if there's nothing, it applies to what is right: int&, but as a whole - there, it applies to &, again because const applies to what's on its left. But there are no const references (they aren't changeable at all, i.e., they can't rebind), the const is dropped, and reference collapsing rules contract the two & into one.

来源：https://stackoverflow.com/questions/64893722/function-template-parameter-loses-const-when-template-argument-is-explicity-pass