问题描述
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我目前正在寻找扩展任意标准符合分配器类型的最佳方法。需要明确的是:我不想编写自定义分配器。我只想向现有的扩展名“添加”特定的扩展名或行为。我已经创建了一个样例。请注意,以下代码仅用于说明目的。
#ifndef HPP_SMART_ALLOCATOR_INCLUDED
#define HPP_SMART_ALLOCATOR_INCLUDED
#include <memory>
#include <map>
template<typename T>
struct allocator_traits;
template<typename T,class allocator_type = std::allocator<T>>
class smart_allocator;
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 smart_allocator<void,allocator_type>
: public allocator_traits<void>
{
public:
template<typename U> struct rebind { typedef smart_allocator<U,typename allocator_type::rebind<U>::other> other; };
};
template<typename T,class allocator_type>
class smart_allocator
: public allocator_traits<T>,private allocator_type
{
public:
using typename allocator_traits<T>::const_pointer;
using typename allocator_traits<T>::const_reference;
using typename allocator_traits<T>::difference_type;
using typename allocator_traits<T>::pointer;
using typename allocator_traits<T>::reference;
using typename allocator_traits<T>::size_type;
using typename allocator_traits<T>::value_type;
template<typename U> struct rebind { typedef smart_allocator<U,typename allocator_type::rebind<U>::other> other; };
smart_allocator() throw() /*noexcept*/;
smart_allocator(allocator_type const&) throw() /*noexcept*/;
virtual ~smart_allocator() throw();
virtual ~smart_allocator()
{
std::map<pointer,size_type>::iterator i = this->m_map.begin();
while (i != this->m_map.end())
{
this->allocator_type::deallocate(i->first,i->second);
++i;
}
}
pointer allocate(size_type n,allocator_traits<void>::const_pointer hint = 0)
{
pointer p = this->allocator_type::allocate(n,hint);
this->m_map.insert(std::pair<pointer,size_type>(p,n));
return p;
}
void deallocate(pointer p,size_type n) /*noexcept*/
{
std::map<pointer,size_type>::iterator iter = this->m_map.find(p);
if (iter != this->m_map.end())
this->allocator_type::deallocate(iter->first,iter->second);
}
using allocator_type::address;
using allocator_type::construct;
using allocator_type::destroy;
using allocator_type::max_size;
private:
smart_allocator(smart_allocator const&) throw();
smart_allocator& operator=(smart_allocator const&);
std::map<pointer,size_type> m_map;
};
#endif /* HPP_SMART_ALLOCATOR_INCLUDED */
请考虑以下注意事项:
模板参数allocator_type可以是任何标准符合类型。它不限于std :: allocator。这是所有STL实现都使用的相同技术。
从allocator_type派生时,我们需要使用私有继承,因为std :: allocator成员函数中的任何一个都不是虚拟的。但是,std :: allocator&alloc = smart_allocator()不会做您可能期望的事情。
您认为这适用吗?
解决方法
您肯定需要实现一个复制构造函数和一个复制赋值运算符,否则当容器按值传递分配器时,您的映射可能会被打乱(特别是您可能会重复删除)。我可能没有其他注意事项。
, 立刻想到的是装饰器。如参考说明所述,“装饰器可用于使对象适应新情况,而无需重写原始对象的代码。”如果我理解您的问题,这听起来像您在追求什么。