Class template argument deduction with partial specialization - c++

Question

I'm having some trouble understanding all the limitations of the new C++17 feature that allows template deduction on constructors.
In particular, this example compiles correctly:
struct B {};
template <typename T, typename = T>
struct A {
A(T) {}
};
int main() {
B b;
A a(b); // ok
}
While this one does not:
struct B {};
template <typename T, typename = T>
struct A;
template <typename T>
struct A<T> {
A(T) {}
};
int main() {
B b;
A a(b); // error
}
The error in this second case is:
main.cpp: In function ‘int main()’:
main.cpp:17:14: error: class template argument deduction failed:
A a(b);
^
main.cpp:17:14: error: no matching function for call to ‘A(B&)’
main.cpp:4:12: note: candidate: template<class T, class> A(A<T, <template-parameter-1-2> >)-> A<T, <template-parameter-1-2> >
struct A;
^
main.cpp:4:12: note: template argument deduction/substitution failed:
main.cpp:17:14: note: ‘B’ is not derived from ‘A<T, <template-parameter-1-2> >’
A a(b);
^
Why is this happening?

Answer 1

Class template argument deduction only considers constructors from the primary class template in order to do deduction. In the first example, we have one constructor that we synthesize a function template for:
template <class T> A<T> __f(T );
The result of __f(b) is A<B>, and we're done.
But in the second example, the primary class template is just:
template <typename T, typename = T>
struct A;
It has no constructors, so we have no function templates to synthesize from them. All we have is a hypothetical default constructor and the copy deduction guide, which together give us this overload set:
template <class T> A<T> __f();
template <class T> A<T> __f(A<T> );
Neither of which are viable for __f(b) (the compile error you get is about trying to match the copy deduction guide), so deduction fails.
If you want this to succeed, you'll have to write a deduction guide:
template <class T>
A(T ) -> A<T>;
Which would allow A a(b) to work.