C++中接口的多重继承

Question

我有一个对象接口和派生对象可能希望支持的开放式接口集合。

// An object 
class IObject 
{ 
    getAttribute() = 0 
} 

// A mutable object 
class IMutable 
{ 
    setAttribute() = 0 
} 

// A lockable object 
class ILockable 
{ 
    lock() = 0 
} 

// A certifiable object 
class ICertifiable 
{ 
    setCertification() = 0 
    getCertification() = 0 
} 

一些派生的对象可能是这样的：

class Object1 : public IObject, public IMutable, public ILockable {} 
class Object2 : public IObject, public ILockable, public ICertifiable {} 
class Object3 : public IObject {} 

这里是我的问题：有没有写功能，将只需要这些接口的某些组合的方式？例如：

void doSomething(magic_interface_combiner<IObject, IMutable, ILockable> object); 

doSomething(Object1()) // OK, all interfaces are available. 
doSomething(Object2()) // Compilation Failure, missing IMutable. 
doSomething(Object3()) // Compilation Failure, missing IMutable and ILockable. 

我发现的最接近的东西是boost :: mpl :: inherit。我有一些有限的成功，但它并不完全符合我的需要。

例如：

class Object1 : public boost::mpl::inherit<IObject, IMutable, ILockable>::type 
class Object2 : public boost::mpl::inherit<IObject, ILockable, ICertifiable>::type 
class Object3 : public IObject 

void doSomething(boost::mpl::inherit<IObject, ILockable>::type object); 

doSomething(Object1()) // Fails even though Object1 derives from IObject and ILockable. 
doSomething(Object2()) // Fails even though Object2 derives from IObject and ILockable. 

我认为类似的boost :: MPL ::继承的东西，但会产生与提供的类型的所有可能的排列组合可能会奏效继承树。

我也很好奇其他解决这个问题的方法。理想的情况是编译时间检查，而不是运行时（即没有dynamic_cast）。

Answer 1

也许这不是最优雅的方式，因为它与C++语法03

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

这是给你的伎俩的精神来完成。 现在你的情况：

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); 
} 

我想它可以更明智的方式使用C++ 11个型特征来完成。

Answer 2

您应该使用static_assert在函数内检查类型：

#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"); 

    // ... 
} 

，这将给很不错的错误信息，告诉你哪些接口不满意你。如果你需要重载函数，有一个版本是只适用于某个接口的组合，你也可以使用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) 
{ 
    // ... 
} 

这将使得功能消失从过载时设置的接口要求没有得到满足。

Answer 3

可以使用递归可变参数继承写一个接口检查类：

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有每个接口检查一个指针的足迹;这是因为它对实际参数的声明类型执行类型擦除。

Answer 4

只给你的C++ 11小情趣：

单一类型的无递归：

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; } 
}; 

Answer 5

一个解决方案使用std::enable_if和std::is_base_of：

#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 
}