如何正确链接boost :: mpl :: inherit_linearly和boost :: mpl :: inherit这样可以解析占位符？

Question

说我有这些类型:

template
<
    class T,
    template <class> class Storage
>
struct AbstractFactoryUnit
{
    virtual ~AbstractFactoryUnit() {}
    virtual typename Storage< T >::StoredType doCreate(Storage< T >) = 0;
};

和

template
<
    class TypeSequence,
    template <class> class ProductStorage,
    template <class, template <class> class> class Unit = AbstractFactoryUnit
>
struct AbstractFactory
    : boost::mpl::inherit_linearly
        <
            TypeSequence,
            boost::mpl::inherit
            <
                boost::mpl::_1,
                Unit< boost::mpl::_2, ProductStorage >
            >
        >::type
{
    typedef TypeSequence Products;

    template <class T>
    auto create() -> typename ProductStorage< T >::StoredType
    {
        Unit< T, ProductStorage >& unit = *this;
        unit.doCreate(ProductStorage< T >());
    }
};

现在我要实现le AbstractFactory...

一些lol类型:

struct Foo {};
struct Bar {};
struct Baz {};

一个lol 存储:

template <class T>
struct RawPointerStorage
{
    typedef T* StoredType;
};

最后实施:

struct FooBarBaz
    : AbstractFactory< boost::mpl::set< Foo, Bar, Baz >, RawPointerStorage >
{
    A* doCreate(RawPointerStorage< Foo >) override
    {
         return new A;
    }

    B* doCreate(RawPointerStorage< Bar >) override
    {
         return new B;
    }

    C* doCreate(RawPointerStorage< Baz >) override
    {
         return new C;
    }
};

不幸的是,编译器抱怨:

1>C:\Libs\boost\boost_1_51_0\boost/mpl/aux_/preprocessed/plain/inherit.hpp(20): error C2500: 'boost::mpl::inherit2<T1,T2>' : 'AbstractFactoryUnit<T,ProductStorage>' is already a direct base class
1>          with
1>          [
1>              T1=AbstractFactoryUnit<boost::mpl::_2,RawPointerStorage>,
1>              T2=AbstractFactoryUnit<boost::mpl::_2,RawPointerStorage>
1>          ]
1>          and
1>          [
1>              T=boost::mpl::_2,
1>              ProductStorage=RawPointerStorage
1>          ]

我有点困惑,因为它只AbstractFactoryUnit接受一个模板参数时编译得很好.我的猜测是编译器不能"解析"第二个占位符,但我应该承认我不知道为什么 - 因为我不知道boost如何调用apply占位符.

我使用VS2012与vc100或vc110.

任何的想法？(是的,我正在玩AbstractFactory现代C++设计中所描述的)

编辑:我最终决定AbstractFactory在我的问题和答案中不加掩饰地提供我的整个代码.

Answer 1

我不知道为什么——在这种情况下——第二个占位符不能“扩展”，但我发现包装表达式boost::mpl::inherit解决了我的问题。

所以你在这里，AbstractFactory简而言之：

我们将实现封装在命名空间中Impl：

namespace Impl
{

    template
    <
        class TypeSequence,
        template <class> class ProductStorage,
        template <class, template <class> class> class Unit
    >
    struct AbstractFactory
    {
    private:
        template <class T, class U>
        struct Inherit : boost::mpl::inherit< T, Unit< U, ProductStorage > >
        {};

    public:
        typedef typename boost::mpl::inherit_linearly
                            <
                                TypeSequence,
                                // the trick is on the following line
                                Inherit< boost::mpl::_1, boost::mpl::_2 >
                            >
                            ::type Type;
    };

} // namespace Impl

我们从中得出这样的结论：

template
<
    class TypeSequence,
    template <class> class ProductStorage = RawPointerStorage,
    template <class, template <class> class> class Unit = AbstractFactoryUnit
>
struct AbstractFactory
    : Impl::AbstractFactory< TypeSequence, ProductStorage, Unit >::Type
{
    typedef TypeSequence Products;

    template <class T>
    auto create() -> typename ProductStorage< T >::StoredType
    {
        Unit< T, ProductStorage >& unit = *this;
        return unit.doCreate(ProductStorage< T >());
    }
};