C++中接口的多重继承

您可以使用递归可变继承编写接口检查类：

template<typename... Interfaces>
struct check_interfaces;
template<>
struct check_interfaces<> {
   template<typename T> check_interfaces(T *) {}
};
template<typename Interface, typename... Interfaces>
struct check_interfaces<Interface, Interfaces...>:
public check_interfaces<Interfaces...> {
   template<typename T> check_interfaces(T *t):
      check_interfaces<Interfaces...>(t), i(t) {}
   Interface *i;
   operator Interface *() const { return i; }
};

例如：

struct IObject { virtual int getAttribute() = 0; };
struct IMutable { virtual void setAttribute(int) = 0; };
struct ILockable { virtual void lock() = 0; };
void f(check_interfaces<IObject, IMutable> o) {
   static_cast<IObject *>(o)->getAttribute();
   static_cast<IMutable *>(o)->setAttribute(99);
}
struct MutableObject: IObject, IMutable {
   int getAttribute() { return 0; }
   void setAttribute(int) {}
};
struct LockableObject: IObject, ILockable {
   int getAttribute() { return 0; }
   void lock() {}
};
int main() {
   f(new MutableObject);
   f(new LockableObject);  // fails
}

注意，check_interfaces具有每个检查接口一个指针的占用空间；这是因为它对实际参数的声明类型执行类型擦除。

您应该使用static_assert来检查函数中的类型：

#include <type_traits>
template< typename T >
void doSomething( const T& t )
{
   static_assert( std::is_base_of<IObject,T>::value, "T does not satisfy IObject" );
   static_assert( std::is_base_of<IMutable,T>::value, "T does not satisfy IMutable" );
   // ...
}

这将给您非常好的错误消息，告诉您哪些接口不满意。如果您需要重载该功能，并且有一个仅适用于特定接口组合的版本，您也可以使用enable_if:

#include <type_traits>
template< typename T, typename... Is >
struct HasInterfaces;
template< typename T >
struct HasInterfaces< T > : std::true_type {};
template< typename T, typename I, typename... Is >
struct HasInterfaces< T, I, Is... >
  : std::integral_constant< bool,
      std::is_base_of< I, T >::value && HasInterfaces< T, Is... >::value > {};
template< typename T >
typename std::enable_if< HasInterfaces< T, IObject, IMutable >::value >::type
doSomething( const T& t )
{
  // ...
}

当不满足接口要求时，这将使函数从过载集中消失。

使用std::enable_if和std::is_base_of:的解决方案

#include <type_traits>
// An object
struct IObject
{
    virtual void getAttribute() = 0;
};
// A mutable object
struct IMutable
{
    virtual void setAttribute() = 0;
};
// A lockable object 
struct ILockable
{
    virtual void lock() = 0;
};
// A certifiable object 
struct ICertifiable
{
    virtual void setCertification() = 0;
    virtual void getCertification() = 0;
};
struct Object1 : public IObject, public IMutable, public ILockable
{
    void getAttribute() {}
    void setAttribute() {}
    void lock() {}
};
struct Object2 : public IObject, public ILockable, public ICertifiable
{
    void getAttribute() {}
    void lock() {}
    void setCertification() {}
    void getCertification() {}
};
struct Object3 : public IObject
{
    void getAttribute() {}
};
template<typename T>
void doSomething(
    typename std::enable_if<
        std::is_base_of<IObject, T>::value &&
        std::is_base_of<IMutable, T>::value &&
        std::is_base_of<ILockable, T>::value,
        T>::type& obj)
{
}
int main()
{
    Object1 object1;
    Object2 object2;
    Object3 object3;
    doSomething<Object1>(object1);  // Works
    doSomething<Object2>(object2);  // Compilation error
    doSomething<Object3>(object3);  // Compilation error
}

也许这不是最优雅的方式，因为它是用C++03语法完成的

template <typename T, typename TInterface>
void interface_checker(T& t)
{
    TInterface& tIfClassImplementsInterface = static_cast<TInterface&>(t);
}

这是给你的精神的把戏。现在以您为例：

template <typename T, typename TInterface1, typename TInterface2, typename TInterface3 >
void magic_interface_combiner(T& t)
{
    TInterface1& tIfClassImplementsInterface = static_cast<TInterface1&>(t);
    TInterface2& tIfClassImplementsInterface = static_cast<TInterface2&>(t);
    TInterface3& tIfClassImplementsInterface = static_cast<TInterface3&>(t);
}

我想使用C++11类型的特性可以做得更聪明。

只想让您尝一尝C++11:

无递归的单一类型：

template <typename... Ts>
class magic_interface_combiner {
  typedef std::tuple<Ts*...> Tpl;
  Tpl tpl;
  template <typename T, int I>
  T *as_(std::false_type)
  {
    static_assert(I < std::tuple_size<Tpl>::value, "T not found");
    return as_<T, I+1>(std::is_same<T, typename std::tuple_element<I+1, Tpl>::type>{});
  }
  template <typename T, int I>
  T *as_(std::true_type) { return std::get<I>(tpl); }
public:
  template <typename T>
  magic_interface_combiner(T * t) : tpl(static_cast<Ts*>(t)...) {}
  template <typename T> T * as() { return as_<T, 0>(std::false_type{}); }
};
// no template    
void doSomething(magic_interface_combiner<IObject, IMutable, ILockable> object)
{
}

两种类型但没有递归：

template <typename T>
class single_interface_combiner {
  T *p;
public:
  single_interface_combiner(T *t) : p(t) {}
  operator T* () { return p; }
};
template <typename... Ts>
struct magic_interface_combiner : single_interface_combiner<Ts>... {
  template <typename T>
  magic_interface_combiner(T* t) : single_interface_combiner<Ts>(t)... {}
  template <typename T>
  T * as() { return *this; }
};