> The template "funcT1" is not a previous declaration of the specialization
> B::funcT1<int>. But to figure out that the specialization B::funcT1<int> is
> static we either need the specialization itself to be marked static or we
> need to walk through getTemplateSpecializationInfo. The member specialization
> info is for something like:
What is going on is that if a template has static storage class, we
say that the instantiated function has it too. So for example, if I
comment out the call to getTemplateSpecializationInfo, we fail to
compile
struct B { template <class T> static void funcT1(); };
template <class T> void B::funcT1() {}
void test() { B::funcT1<int>(); }
but can still compile
struct B { template <class T> static void funcT1(); };
void test() { B::funcT1<int>(); }
because there is a static to be copied in this case. Similarly, the
instantiated function in
template<typename T>
struct C { static void funcT2(); };
template<typename T> void C<T>::funcT2(){}
void test() { C<int>::funcT2(); }
gets marked as static.
Looks like "as written" is a somewhat fuzzy concept for
instantiations. Maybe it would be more regular to say that a
instantiated function gets a static storage class if the template
isStatic()? That should allow us to remove the
getTemplateSpecializationInfo in isStatic().
> - Doug
Cheers,
Rafael
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