Hi all,
This is regarding 14.5.6.2 Partial ordering of function template.
Test case - 1
struct A { };
template struct B {
template int operator*(R&); // #1
};
template<class T, class R> int operator*(T&, R&); // #2
// The declaration of B::operator* is transformed into the equivalent of
// template int operator*(B&, R&); // #1a
int main() {
A a;
b * a; // calls #1a
}
The above test case works fine as per standards as #1a is more specialized than #2.
However when we tweak the test case in this way:
struct A{};
template struct B
{
template int operator*(R&); // #1
};
template int operator*(T&, A&); // #2
// The declaration of B::operator* is transformed into the equivalent of
// template int operator*(B&, R&); // #1a
clang throws an error saying:
error: use of overloaded operator ‘*’ is ambiguous (with operand types ‘B’ and ‘A’)
How is partial ordering done in this case? How is the specialization decided for the below mentioned
operator overloading declarations?
template int operator*(B&, R&); // #1a
template int operator*(T&, A&); // #2
It would be great if someone could explain whether this behavior is correct and more importantly
how clang does partial ordering and judge specialization between two function template overloads?
Thanks,
Rahul
