分配器类型参数的C++设计模式

Question

当将类型传递给要作为分配器的类时,C++ 03标准库使用简单的模板类型参数.这是可能的,因为模板在C++中的工作方式.但是,它并不是非常简单,您可能不知道类型定义应该是什么样子 - 特别是在非标准类型的情况下.

我认为使用适配器类instread可能是个好主意.我已经创建了一个示例来向您展示我的意思:

#ifndef HPP_ALLOCATOR_ADAPTOR_INCLUDED
#define HPP_ALLOCATOR_ADAPTOR_INCLUDED


#include <memory>


template<typename T>
struct allocator_traits;

template<typename T, class allocator_type = std::allocator<T>>
class allocator_adaptor;


template<>
struct allocator_traits<void>
{
    typedef std::allocator<void>::const_pointer const_pointer;
    typedef std::allocator<void>::pointer       pointer;
    typedef std::allocator<void>::value_type    value_type;
};

template<typename T>
struct allocator_traits
{
    typedef typename std::allocator<T>::const_pointer   const_pointer;
    typedef typename std::allocator<T>::const_reference const_reference;
    typedef typename std::allocator<T>::difference_type difference_type;
    typedef typename std::allocator<T>::pointer         pointer;
    typedef typename std::allocator<T>::reference       reference;
    typedef typename std::allocator<T>::size_type       size_type;
    typedef typename std::allocator<T>::value_type      value_type;
};


template<class allocator_type>
class allocator_adaptor<void, allocator_type>
    : public allocator_traits<void>
{
public:
    template<typename U> struct rebind { typedef allocator_adaptor<U, allocator_type> other; };
};

template<typename T, class allocator_type>
class allocator_adaptor
    : public allocator_traits<T>
{
private:
    allocator_type m_impl;

public:
    template<typename U> struct rebind { typedef allocator_adaptor<U, allocator_type> other; };

    allocator_adaptor() throw() /*noexcept*/;
    allocator_adaptor(allocator_adaptor const&) throw() /*noexcept*/;
    allocator_adaptor(allocator_type const&) throw() /*noexcept*/;
    template<typename U> allocator_adaptor(allocator_adaptor<U, allocator_type> const&) throw() /*noexcept*/;
    ~allocator_adaptor() throw();

    pointer       address(reference x) const /*noexcept*/;
    const_pointer address(const_reference x) const /*noexcept*/;

    pointer   allocate  (size_type, allocator_traits<void>::const_pointer hint = 0);
    void      deallocate(pointer p, size_type n) /*noexcept*/;
    size_type max_size  () const throw() /*noexcept*/;

    template<class U, typename... argument_types> void construct(U* p, argument_types&&... args);
    template<class U> void destroy(U* p);
};


#endif /* HPP_ALLOCATOR_ADAPTOR_INCLUDED */

实施应该是显而易见的.这是一些用法示例.

template<class allocator_type>
int max_size(allocator_type const& alloc)
{
    // we don't know what kind of max_szie function will be called.
    return alloc.max_size();
}

template<typename T>
int max_size(allocator_adaptor<T> const& alloc)
{
    // we know what kind of max_size function will be called.
    return alloc.max_size();
}

与通常的方式相比,这是一种改进吗？

Answer 1

对我来说看起来不错......如果能解释一下你的设计何时会比通常的方式更有优势，那就太好了。一个例子对我很有帮助

“但是，这并不是很简单，您可能不知道类型定义到底应该是什么样子”

与 std::allocator 相比，您的实现如何改进类型定义部分，请解释