C++ partial template specialization in combination with std::is_base_of and std::enable_if

问题

Let's say I have a two classes: Serializable and Printable.

So a simple template function which accepts all derived classes of Printable could look like:

template <class T, class B = Printable, class = typename std::enable_if<std::is_base_of<B,     T>::value>::type>
void print(T value) {
    cout << value << endl;
}

However, if I want it to accept also all derived classes of Serializable while I still have control over the function body, this would obviously not work:

template <class T, class B = Printable, class = typename std::enable_if<std::is_base_of<B,     T>::value>::type>
void print(T value) {
    cout << value << endl;
}

template <class T, class B = Serializable, class = typename std::enable_if<std::is_base_of<B,     T>::value>::type>
void print(T value) {
    cout << value << endl;
}

// Error: Redefinition of ...

So I figured the remaining solutions for this problem are template specializations.

But I just can't figure out, how I can specialize a template in combination with std::is_base_of and std::enable_if.

I hope someone is willing to help me!

回答1:

Try a logical operator:

std::enable_if<std::is_base_of<Serializable, T>::value ||
               std::is_base_of<Printable, T>::value>::type

You can easily write a variadic template like:

is_base_of_any<T, Printable, Serialiable, Googlable, Foobarable>::value

For example:

template <typename T, typename ...> struct is_base_of_any : std::true_type {};

template <typename T, typename Head, typename ...Rest>
struct is_base_of_any<T, Head, Rest...>
: std::integral_constant<bool, std::is_base_of<T, Head>::value ||
                               is_base_of_any<T, Rest...>::value>
{ };

If you want different implementations:

template <bool...> struct tag_type {};

template <typename T>
void foo(T, tag_type<true, false>) { }   // for Printable

template <typename T>
void foo(T, tag_type<false, true>) { }   // for Serializable

template <typename T>
void foo(T x)
{
    foo(x, tag_type<std::is_base_of<Printable, T>::value,
                    std::is_base_of<Serializable, T>::value>());
}

The last overload (the "user-facing" one) should probably be endowed with the above enable_if to not create overly many overload candidates.

You can probably also make a variadic template <typename ...Bases> with a tag like:

tag_type<std::is_base_of<Bases, T>::value...>

回答2:

A little less machinery than Kerrek's answer, but I'm afraid no more readable:

template <class T, typename std::enable_if<std::is_base_of<Printable, T>::value>::type* = nullptr>
void print(const T& value) {
    std::cout << "printable(" << &value << ")\n";
}

template <class T, typename std::enable_if<std::is_base_of<Serializable, T>::value>::type* = nullptr>
void print(const T& value) {
    std::cout << "serializable(" << &value << ")\n";
}

See it live at ideone.

回答3:

Consider this:

void print(const Printable& value) {
    cout << value << endl;
}

void print(const Serializable& value) {
    cout << value << endl;
}

Naturally you will have the appropriate operator<< calling a virtual function in the right hand side operand, which would do the actual printing.

来源：https://stackoverflow.com/questions/17200755/c-partial-template-specialization-in-combination-with-stdis-base-of-and-std

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templates

c++11

specialization