class template partial specialization parametrized on member function return type

Question

The following code, which attempts to specialize class template 'special', based on the return type of member function pointer types, results in a compile error with VC9:

template<class F> struct special {};
template<class C> struct special<void(C::*)()> {};
template<class R, class C> struct special<R(C::*)()> {};

struct s {};

int main()
{
  special<void(s::*)()> instance;
  return 0;
}

error C2752: 'special' : more than one partial specialization matches the template argument list

The same code is accepted by GCC-4.3.4, as shown by: http://ideone.com/ekWGg
Is this a bug in VC9 and if so, has this bug persisted into VC10?

I have however come up with a horrendously intrusive workaround (for this specific use case, at least. More general solutions welcome):

#include <boost/function_types/result_type.hpp>
#include <boost/type_traits/is_same.hpp>

template<typename F, typename R>
struct is_result_same :
  boost::is_same<
    typename boost::function_types::result_type<F>::type,
    R
  >
{};

template<class F, bool = is_result_same<F, void>::value>
struct special {};

template<class R, class C> struct special<R(C::*)(), true>  {};
template<class R, class C> struct special<R(C::*)(), false> {};

I imagine this is because when resolving the case of R being `void` the signatures of both specializations are in fact the same ... some clever use of the `typename` keyword may help but I can't think of anything clever at the moment. — AJG85, Feb 24 '11 at 22:02

score 3 · Accepted Answer · answered Feb 24 '11 at 22:36

This is a bug.

template <class C> struct special<void(C::*)()>;        // specialization 1
template <class R, class C> struct special<R(C::*)()>;  // specialization 2

According to 14.5.4.2, the partial ordering of these two class template specializations are the same as the partial ordering of these imaginary function templates:

template <class C> void f(special<void(C::*)()>);       // func-template 3
template <class R, class C> void f(special<R(C::*)()>); // func-template 4

According to 14.5.5.2, the partial ordering of these two function templates is determined by substituting invented types for each type template parameter in the argument list of one and attempting template argument deduction using that argument list in the other function template.

// Rewrite the function templates with different names -
// template argument deduction does not involve overload resolution.
template <class C> void f3(special<void(C::*)()>);
template <class R, class C> void f4(special<R(C::*)()>);

struct ty5 {}; struct ty6 {}; struct ty7 {};
typedef special<void(ty5::*)()> arg3;
typedef special<ty6 (ty7::*)()> arg4;

  // compiler internally tests whether these are well-formed and
  // the resulting parameter conversion sequences are "exact":
  f3(arg4());
  f4(arg3());

The details of template argument deduction are in 14.8.2. Among the valid deductions are from template_name<dependent_type> and dependent_type1 (dependent_type2::*)(arg_list). So the f4(arg3()) deduction succeeds, deducing f4<void,ty5>(arg3());. The f3(arg4()) deduction can obviously never succeed, since void and ty6 do not unify.

Therefore function template 3 is more specialized than function template 4. And class template specialization 1 is more specialized than class template specialization 2. So although special<void(s::*)()> matches both specializations, it unambiguously instantiates specialization 1.

class template partial specialization parametrized on member function return type

1 Answers1