Forward declare C++ [duplicate] - c++

Question

The following compiles on GCC 4.8.1 (with --std=c++11):
struct non_default_constructible { non_default_constructible() = delete; };
template<class T>
struct dummy {
T new_t() { return T(); }
};
int main(int argc, char** argv) {
dummy<non_default_constructible> d;
return 0;
}
The tricky part is that dummy<non_default_constructible>::new_t() is obviously ill-formed, but that does not prevent the compiler from instantiating dummy<non_default_constructible>.
Is this the behaviour specified by the standard? And what would be the relevant sections/keywords?

Answer 1

The member functions of a class template are instantiated only when required by a context, which means you will not see any error until you try to use new_t(). The related section from the C++ standard is:
§ 14.7.1 Implicit instantiation [temp.inst]
Unless a function template specialization has been explicitly instantiated or explicitly specialized, the function template specialization is implicitly instantiated when the specialization is referenced in a context that requires a function definition to exist. Unless a call is to a function template explicit specialization or to a member function of an explicitly specialized class template, a default argument for a function template or a member function of a class template is implicitly instantiated when the function is called in a context that requires the value of the default argument.
[ Example:
template<class T> struct Z {
void f();
void g();
};
void h() {
Z<int> a; // instantiation of class Z<int> required
Z<char>* p; // instantiation of class Z<char> not required
Z<double>* q; // instantiation of class Z<double> not required
a.f(); // instantiation of Z<int>::f() required
p->g(); // instantiation of class Z<char> required, and
// instantiation of Z<char>::g() required
}
Nothing in this example requires class Z<double>, Z<int>::g(), or Z<char>::f() to be implicitly
instantiated. — end example ]