Make C++ fail compilation on specific instantiation of template function

Question

I'm working on a project which has an template function as so:

template <class T> 
T foo<T>(T val) { return someFunc(val); }

template <>
bool foo<bool>(bool val) { return otherFunc(val); };

Now, I have a class Bar, which I don't want to accept as input. In fact, I want it to generate an easy to spot compile error. The problem is that if I do this:

template <>
Bar foo<Bar>(Bar val) { static_assert(false,"uh oh..."); }

It fails on every compile. I found https://stackoverflow.com/a/3926854/7673414, which says that I need to make reference to the template type, otherwise the static assert always takes place. The problem is I don't have a template type here. If I do:

template< typename T >
struct always_false { 
    enum { value = false };  
};

template <>
Bar foo<Bar>(Bar val) { static_assert(always_false<Bar>::value,"uh oh..."); }

then it also always fails compiling. Is there a way to ensure that an instantiation of the template with type Bar always causes a compile error?

Answer 1

Since foo is a complete specialization, it will always get compiled, and the static assert will always get called.

However, there’s an easier way:

template <>
Bar foo<Bar>(Bar val) = delete;

This will say that this specific version is deleted, and cannot be called.

Answer 2

Another way is enable the template (not specialized version) only if the type is different from Bar

template <class T> 
typename std::enable_if< ! std::is_same<T, Bar>::value, T>::type foo<T>(T val)
 { return someFunc(val); }

If you can use C++14, is can be simplified using std::enable_if_t

template <class T> 
std::enable_if_t< ! std::is_same<T, Bar>::value, T> foo<T>(T val)
 { return someFunc(val); }

Answer 3

You can use std::is_same to help with your requirement.

template <class T> 
T foo<T>(T val)
{
   // Make sure T is not Bar
   static_assert(std::is_same<T, Bar>::value == false, "uh oh...");
   return someFunc(val);
}