enable_if template specialization custom traits

Question

I trying to understand how std::enable_if works inn template parameters.

#include <type_traits>
#include <iostream>
#include <memory>

using namespace std;


class Interface {};
class Value {};


class Stream
{
public:

    template<typename T,
            typename enable_if<
                is_base_of< Interface, T>{}
            >::type* = nullptr>
    void
    write(const shared_ptr<T>& v)
    {
        cerr << "Writing interface class" << endl;
    }

    template<typename T,
            typename enable_if<
                is_base_of< Value, T>{}
            >::type* = nullptr>
    void
    write(const shared_ptr<T>& v)
    {
        cerr << "Writing value class" << endl;
    }
};

class UserI : public Interface{};
class User : public Value{};

int main(int, char**)
{

    auto interface = make_shared<UserI>();
    auto value = make_shared<User>();

    Stream s;
    s.write(interface);
    s.write(value);

    return 0;
}

Them I tried to simplified the code by providing a custom traits to check if a object is an interface or value instance.

#include <type_traits>
#include <iostream>
#include <memory>

using namespace std;


class Interface {};
class Value {};

template<typename T>
struct is_interface
{
    const static bool value = is_base_of< Interface, T>::value;
};

template<typename T>
struct is_value
{
    const static bool value = is_base_of< Value, T>::value;
};

class Stream
{
public:

    template<typename T,
            typename enable_if<
                is_interface<T>{}
            >::type* = nullptr>
    void
    write(const shared_ptr<T>& v)
    {
        cerr << "Writing interface class" << endl;
    }

    template<typename T,
            typename enable_if<
                is_value<T>{}
            >::type* = nullptr>
    void
    write(const shared_ptr<T>& v)
    {
        cerr << "Writing value class" << endl;
    }
};

class UserI : public Interface{};
class User : public Value{};

int main(int, char**)
{

    auto interface = make_shared<UserI>();
    auto value = make_shared<User>();

    Stream s;
    s.write(interface);
    s.write(value);

    return 0;
}

But the second version fails to compile with the following error:

test_t2.cc: In function ‘int main(int, char**)’:
test_t2.cc:58:26: error: no matching function for call to ‘Stream::write(std::shared_ptr<UserI>&)’
        s.write(interface);
                        ^
test_t2.cc:58:26: note: candidates are:
test_t2.cc:32:9: note: template<class T, typename std::enable_if<is_interface<T>{}>::type* <anonymous> > void Stream::write(const std::shared_ptr<_Tp1>&)
        write(const shared_ptr<T>& v)
        ^
test_t2.cc:32:9: note:   template argument deduction/substitution failed:
test_t2.cc:30:28: error: could not convert template argument ‘is_interface<UserI>{}’ to ‘bool’
                >::type* = nullptr>

Can someone explain what is the issue with the second version?

Answer 1

typename enable_if<
    is_base_of< Interface, T>{}
>::type* = nullptr

That works because std::is_base_of provides operator bool to implicitly convert an instance of it to bool . Your traits do not provide that, so using is_interface<T>{} as a bool is not valid.

You could write operator bool for your traits, but the easy fix is to just use ::value instead:

template<typename T,
        typename enable_if<
            is_interface<T>::value //here
        >::type* = nullptr>
void
write(const shared_ptr<T>& v)

---

Personally, I think a tag-dispatch method would be cleaner and more maintainable in your case. Here's a possible implementation:

class Stream
{
private:
    struct interface_tag{};
    struct value_tag{};
    //if you ever need more tags, add them here
    struct invalid_tag{};

    template <typename T>
    struct get_tag {
        static interface_tag tagger (Interface*);
        static value_tag tagger (Value*);
        //add any mappings for other tags
        static invalid_tag tagger (...);

        using tag = decltype(tagger(std::declval<T*>()));
    };

    //convenience alias
    template <typename T> using tag_t = typename get_tag<T>::tag;

public:
    //clean public interface
    template <typename T>
    void write (const shared_ptr<T>& v) {
        write(v, tag_t<T>{});   
    }

private:
    //no more horrible std::enable_if
    template<typename T>
    void
    write(const shared_ptr<T>& v, interface_tag)
    {
        cerr << "Writing interface class" << endl;
    }

    template<typename T>
    void
    write(const shared_ptr<T>& v, value_tag)
    {
        cerr << "Writing value class" << endl;
    }
};

Live Demo

If you need any clarification on this method, just ask.