SFINAE: Compiler doesn't pick the specialized

问题:

I have an SFINAE problem:

In the following code, I want the C++ compiler to pick the specialized functor and print "special", but it's printing "general" instead.

#include <iostream>
#include <vector>

template<class T, class V = void>
struct Functor {
  void operator()() const {
    std::cerr << "general" << std::endl;
  }
};

template<class T>
struct Functor<T, typename T::Vec> {
  void operator()() const {
    std::cerr << "special" << std::endl;
  }
};

struct Foo {
  typedef std::vector<int> Vec;
};

int main() {
  Functor<Foo> ac;
  ac();
}

How can I fix it so that the specialized struct is used automatically? Note I don't want to directly specialize the Functor struct on Foo, but I want to specialize it on all types that have a Vec type.

P.S.: I am using g++ 4.4.4

回答1:

Sorry for misleading you in the last answer, I thought for a moment that it would be simpler. So I will try to provide a complete solution here. The general approach to solve this type of problems is to write a traits helper template and use it together with enable_if (either C++11, boost or manual implementation) to decide a class specialization:

Trait

A simple approach, not necessarily the best, but simple to write would be:

template <typename T>
struct has_nested_Vec {
    typedef char yes;
    typedef char (&no)[2];
    template <typename U>
    static yes test( typename U::Vec* p );
    template <typename U>
    static no test( ... );

    static const bool value = sizeof( test<T>(0) ) == sizeof(yes);
};

The approach is simple, provide two template functions, that return types of different sizes. One of which takes the nested Vec type and the other takes ellipsis. For all those types that have a nested Vec the first overload is a better match (ellipsis is the worst match for any type). For those types that don't have a nested Vec SFINAE will discard that overload and the only option left will be the ellipsis. So now we have a trait to ask whether any type has a nested Vec type.

Enable if

You can use this from any library, or you can roll your own, it is quite simple:

template <bool state, typename T = void>
struct enable_if {};

template <typename T>
struct enable_if<true,T> {
    typedef T type;
};

When the first argument is false, the base template is the only option, and that does not have a nested type, if the condition is true, then enable_if has a nested type that we can use with SFINAE.

Implementation

Now we need to provide the template and the specialization that will use SFINAE for only those types with a nested Vec:

template<class T, class V = void>
struct Functor {
    void operator()() const {
        std::cerr << "general" << std::endl;
    }
};
template<class T>
struct Functor<T, typename enable_if<has_nested_Vec<T>::value>::type > {
    void operator()() const {
        std::cerr << "special" << std::endl;
    }
};

Whenever we instantiate Functor with a type, the compiler will try to use the specialization, which will in turn instantiate has_nested_Vec and obtain a truth value, passed to enable_if. For those types for which the value is false, enable_if does not have a nested type type, so the specialization will be discarded in SFINAE and the base template will be used.

Your particular case

In your particular case, where it seems that you don't really need to specialize the whole type but just the operator, you can mix the three elements into a single one: a Functor that dispatches to one of two internal templated functions based on the presence of Vec, removing the need for enable_if and the traits class:

template <typename T>
class Functor {
   template <typename U>
   void op_impl( typename U::Vec* p ) const {
      std::cout << "specialized";
   }
   template <typename U>
   void op_impl( ... ) const {
      std::cout << "general";
   }
public:
   void operator()() const {
      op_impl<T>(0);
   }
};

回答2:

Even though this is an old question, I think it's still worth providing a couple more alternatives for quickly fixing the original code.

Basically, the problem is not with the use of SFINAE (that part is fine, actually), but with the matching of the default parameter in the primary template (void) to the argument supplied in the partial specialization(typename T::Vec). Because of the default parameter in the primary template, Functor<Foo> actually means Functor<Foo, void>. When the compiler tries to instantiate that using the specialization, it tries to match the two arguments with the ones in the specialization and fails, as void cannot be substituted for std::vector<int>. It then falls back to instantiating using the primary template.

So, the quickest fix, which assumes all your Vecs are std::vector<int>s, is to replace the line

template<class T, class V = void>

with this

template<class T, class E = std::vector<int>>

The specialization will now be used, because the arguments will match. Simple, but too limiting. Clearly, we need to better control the type of the argument in the specialization, in order to make it match something that we can specify as the default parameter in the primary template. One quick solution that doesn't require defining new traits is this:

#include <iostream>
#include <vector>
#include <type_traits>

template<class T, class E = std::true_type>
struct Functor {
  void operator()() const {
    std::cerr << "general" << std::endl;
  }
};

template<class T>
struct Functor<T, typename std::is_reference<typename T::Vec&>::type> {
  void operator()() const {
    std::cerr << "special" << std::endl;
  }
};

struct Foo {
  typedef std::vector<int> Vec;
};

int main() {
  Functor<Foo> ac;
  ac();
}

This will work for any Vec type that could make sense here, including fundamental types and arrays, for example, and references or pointers to them.

回答3:

Another alternative for detecting the existence of a member type is to use void_t. As valid partial specialisations are preferable to the general implementation as long as they match the default parameter(s), we want a type that evaluates to void when valid, and is only valid when the specified member exists; this type is commonly (and, as of C++17, canonically) known as void_t.

template<class...>
using void_t = void;

If your compiler doesn't properly support it (in early C++14 compilers, unused parameters in alias templates weren't guaranteed to ensure SFINAE, breaking the above void_t), a workaround is available.

template<typename... Ts> struct make_void { typedef void type; };
template<typename... Ts> using void_t = typename make_void<Ts...>::type;

As of C++17, void_t is available in the utilities library, in type_traits.

#include <iostream>
#include <vector>
#include <type_traits> // For void_t.

template<class T, class V = void>
struct Functor {
  void operator()() const {
    std::cerr << "general" << std::endl;
  }
};

// Use void_t here.
template<class T>
struct Functor<T, std::void_t<typename T::Vec>> {
  void operator()() const {
    std::cerr << "special" << std::endl;
  }
};

struct Foo {
  typedef std::vector<int> Vec;
};

int main() {
  Functor<Foo> ac;
  ac();
}

With this, the output is special, as intended.

---

In this case, since we're checking for the existence of a member type, the process is very simple; it can be done without expression SFINAE or the type_traits library, allowing us to rewrite the check to use C++03 facilities if necessary.

// void_t:
// Place above Functor's definition.
template<typename T> struct void_t { typedef void type; };

// ...

template<class T>
struct Functor<T, typename void_t<typename T::Vec>::type> {
  void operator()() const {
    std::cerr << "special" << std::endl;
  }
};

To my knowledge, this should work on most, if not all, SFINAE-capable C++03-, C++11-, C++14-, or C++1z-compliant compilers. This can be useful when dealing with compilers that lag behind the standard a bit, or when compiling for platforms that don't have C++11-compatible compilers yet.

---

For more information on void_t, see cppreference.