最近在百度知道上回答了一道問(wèn)題，并為他封裝了一個(gè)類來(lái)解決。大意是不開(kāi)辟新的數(shù)組同時(shí)排序兩個(gè)數(shù)組。
http://zhidao.baidu.com/question/112777967.html

之后想要給這個(gè)類添加iterator好直接使用sort之類的函數(shù)，經(jīng)過(guò)1個(gè)多小時(shí)的忙碌（能力有限，可以鄙視一下），終于實(shí)現(xiàn)了random_access_iterator。
上源碼：

template<typename T>
class Obj;

template<typename T>
class Obj_iterator : public std::iterator <std::random_access_iterator_tag , T>
{
public:
        Obj_iterator() : m_Obj(NULL) , m_index(0) {}
        Obj_iterator( Obj<T> *p , size_t index ) : m_Obj(p) , m_index(index){}
        Obj_iterator& operator = ( const  Obj_iterator &t )
        {
                m_index = t.m_index;
                m_Obj = t.m_Obj;
                return *this;
        }
        bool operator ==( const Obj_iterator &rh ) const
        {
                return rh.m_Obj == m_Obj && rh.m_index == m_index;
        }
        bool operator !=( const Obj_iterator &rh ) const
        {
                return !( *this == rh );
        }
        Obj_iterator operator ++()
        {
                ++m_index;
                return Obj_iterator(m_Obj,m_index);
        }
        Obj_iterator operator ++( int )
        {
                Obj_iterator x = *this;
                ++(*this);
                return x;
        }
        Obj_iterator operator -- ()
        {
                --m_index;
                return Obj_iterator(m_Obj , m_index);
        }
        //一開(kāi)始這里總是過(guò)不去，后來(lái)回想起以前遇到的問(wèn)題，才知道需要在后面加上const來(lái)修飾，原因如下
        /*
        C++不鼓勵(lì)把非常量引用綁定到臨時(shí)對(duì)象
        非const的operator*有可能改變對(duì)象值，從而對(duì)象整體的const性質(zhì)得不到保證
        */
        T& operator *() const
        {
                return (*m_Obj)[m_index];
        }
        T* operator ->() const
        {
                return &(*m_Obj[m_index]);
        }
        const Obj_iterator operator +(size_t n)
        {
                return Obj_iterator(m_Obj , m_index + n);
        }
        const Obj_iterator operator += (size_t n)
        {
                m_index += n;
                return *this;
        }
        difference_type operator -(const Obj_iterator &n)
        {
                return m_index - n.m_index;
        }
        Obj_iterator operator -(size_t n)
        {
                return Obj_iterator(m_Obj , m_index-n);
        }
        bool  operator < (const Obj_iterator n) const
        {
                return m_index<n.m_index;
        }
protected:
private:
        Obj<T> *m_Obj;
        size_t m_index;
};

template <typename T>
class Obj
{
public:
        Obj(T *p1,size_t sizeofp1,T *p2,size_t sizeofp2):
              m_p1(p1),
              m_p2(p2),
              m_sizeofp1(sizeofp1),
              m_sizeofp2(sizeofp2)
      {
              m_sumsize = sizeofp1+sizeofp2;
      }

      typedef Obj_iterator<T>  iterator;

      T& operator [](size_t index)
      {
              if (index > m_sizeofp1 + m_sizeofp2)
              {
                      throw std::out_of_range("out_of_range");
              }
              if (index<m_sizeofp1)
              {
                      return m_p1[index];
              }
              else
              {
                      return m_p2[index - m_sizeofp1];
              }
      }
      size_t size() const
      {
              return m_sumsize;
      }
      iterator begin()
      {
              return iterator(this,0);
      }
      iterator end()
      {
              return iterator(this,m_sumsize);
      }
protected:
private:
        T *m_p1;
        T *m_p2;
        size_t m_sizeofp1;
        size_t m_sizeofp2;
        size_t m_sumsize;
};

int _tmain(int argc, _TCHAR* argv[])
{
        int p1[] = {1,6,3,123456,2,67,2,8};
        int p2[] = {4,2,6,234,9,4,0,5};
        Obj<int> h(p1 , sizeof(p1)/sizeof(*p1) , p2 , sizeof(p2)/sizeof(*p2));
        std::copy(h.begin() , h.end() , std::ostream_iterator<int>(std::cout," "));
        std::cout<<std::endl;
        std::sort(h.begin() , h.end());
        std::copy(h.begin() , h.end(),std::ostream_iterator<int>(std::cout,"\n"));
        return EXIT_SUCCESS;
}

這段程序在我的vs2008上正確編譯執(zhí)行。