C++ template specialization without default function

Question

I have the following code that compiles and works well:

template<typename T>
T GetGlobal(const char *name);

template<>
int GetGlobal<int>(const char *name);

template<>
double GetGlobal<double>(const char *name);

However I want to remove the "default" function. That is, I want to make all calls to GetGlobal<t> where 't' is not an int or a double an error.

For example, GetGlobal<char>() should be a compile time error.

I tried to just delete the default function, but, as I imagined, I received a lot of errors.. So is there a way to "disable" it and allow calls only to the specialized versions of the function?

Thanks!

Answer 1

To get a compile-time error implement it as:

template<typename T>
T GetGlobal(const char *name) { T::unimplemented_function; }
// `unimplemented_function` identifier should be undefined

If you use Boost you could make it more elegant:

template<typename T>
T GetGlobal(const char *name) { BOOST_STATIC_ASSERT(sizeof(T) == 0); }

C++ Standard guarantees that there is no such type which has sizeof equal to 0, so you'll get a compile-time error.

As sbi suggested in his comments the last could be reduced to:

template<typename T>
T GetGlobal(const char *name) { char X[!sizeof(T)]; }

I prefer the first solution, because it gives more clear error message (at least in Visual C++) than the others.

Answer 2

Though it is an old and outdated question, it may worth noting that C++11 had solved this issue using deleted functions:

template<typename T>
T GetGlobal(const char *name) = delete;

template<>
int GetGlobal<int>(const char *name);

---

UPDATE

This will not compile under MacOS llvm 8. It is due to a still hanging 4 years old defect (see this bug report).

The following workaround will fit the issue (using a static_assert construct).

template<typename T>
T GetGlobal(const char *name) {
    static_assert(sizeof(T) == 0, "Only specializations of GetGlobal can be used");
}

template<>
int GetGlobal<int>(const char *name);

---

UPDATE

Visual studio 15.9 has the same bug. Use the previous workaround for it.

Answer 3

If you don't implement it, you'll at least get a linker error. If you want a compile-time error, you could do this with class templates:

template<typename T>
struct GlobalGetter;

template<>
struct GlobalGetter<int> {
  static int GetGlobal(const char *name);
};

template<>
struct GlobalGetter<double> {
  static double GetGlobal(const char *name);
};

template<typename T>
T GetGlobal(const char *name)
{
  return GlobalGetter<T>::GetGlobal(name);
}

Answer 4

I would suggest not to actually provide an implementation, just a bare declaration of the method.

The other option would be to use a compile-time assert. Boost has a number of such beasts.

namespace mpl = boost::mpl;
BOOST_MPL_ASSERT((mpl::or_< boost::same_type<T, double>,
                            boost::same_type<T, int> >));

There is also its message version counterpart, which would help.

Answer 5

The following are alternative techniques to using boost:

Declare a typedef to a dependent name

This works because name lookup for DONT only occurs when 'T' has been replaced. This is a similar (but legal) version of the example given by Kirill

template <typename T>
T GetGlobal (const char * name) {
    typedef typename T::DONT CALL_THIS_FUNCTION;
}

Use an incomplete return type

This technique doesn't work for specializations, but it will work for overloads. The idea is that its legal to declare a function which returns an incomplete type, but not to call it:

template <typename T>
class DONT_CALL_THIS_FUNCTION GetGlobal (const char * name);