This is an unofficial snapshot of the ISO/IEC JTC1 SC22 WG21 Core Issues List revision 114a. See http://www.open-std.org/jtc1/sc22/wg21/ for the official list.
2024-04-18
template <class T> class Foo {
public:
typedef int Bar;
Bar f();
};
template <class T> typename Foo<T>::Bar Foo<T>::f() { return 1;}
--------------------
In the class template definition, the declaration of the member function
is interpreted as:
int Foo<T>::f();In the definition of the member function that appears outside of the class template, the return type is not known until the member function is instantiated. Must the return type of the member function be known when this out-of-line definition is seen (in which case the definition above is ill-formed)? Or is it OK to wait until the member function is instantiated to see if the type of the return type matches the return type in the class template definition (in which case the definition above is well-formed)?
Suggested resolution: (John Spicer)
My opinion (which I think matches several posted on the reflector recently) is that the out-of-class definition must match the declaration in the template. In your example they do match, so it is well formed.
I've added some additional cases that illustrate cases that I think either are allowed or should be allowed, and some cases that I don't think are allowed.
template <class T> class A { typedef int X; };
template <class T> class Foo {
public:
typedef int Bar;
typedef typename A<T>::X X;
Bar f();
Bar g1();
int g2();
X h();
X i();
int j();
};
// Declarations that are okay
template <class T> typename Foo<T>::Bar Foo<T>::f()
{ return 1;}
template <class T> typename Foo<T>::Bar Foo<T>::g1()
{ return 1;}
template <class T> int Foo<T>::g2() { return 1;}
template <class T> typename Foo<T>::X Foo<T>::h() { return 1;}
// Declarations that are not okay
template <class T> int Foo<T>::i() { return 1;}
template <class T> typename Foo<T>::X Foo<T>::j() { return 1;}
In general, if you can match the declarations up using only information
from the template, then the declaration is valid.
Declarations like Foo::i and Foo::j are invalid because for a given instance of A<T>, A<T>::X may not actually be int if the class is specialized.
This is not a problem for Foo::g1 and Foo::g2 because for any instance of Foo<T> that is generated from the template you know that Bar will always be int. If an instance of Foo is specialized, the template member definitions are not used so it doesn't matter whether a specialization defines Bar as int or not.