STL迭代器:"解除引用"迭代器到临时.可能吗？

Question

我正在为我的科学软件编写一个3D网格,我需要遍历网格的节点来获取它们的坐标.而不是将每个节点对象保存在容器中,我宁愿只是在迭代时动态计算坐标.问题是stl :: iterator需要返回对值的引用operator*()或者指针的引用operator->().

下面的一些代码:

class spGridIterator {
public:
    typedef forward_iterator_tag iterator_category;
    typedef spVector3D value_type;
    typedef int difference_type;
    typedef spVector3D* pointer;
    typedef spVector3D& reference;

    spGridIterator(spGrid* gr, int index);

    spGridIterator& operator++();
    spGridIterator& operator++(int);

    reference operator*() const;
    pointer operator->() const;

private:
    spGrid* m_grid;
    int m_idx;
};

spGridIterator::reference spGridIterator::operator*() const {
    // return m_grid->GetPoint(m_idx);
}

spGridIterator::pointer spGridIterator::operator->() const {
    // return m_grid->GetPoint(m_idx);
}

此方法通过提供的索引查询节点坐标

spVector3D spGrid::GetPoint(int idx) const {
    // spVector3D vec = ... calculate the coordinates here ...
    return vec;
}

有什么输入吗？

提前谢谢,伊利亚

Answer 1

您可以使用成员变量来保存当前指向的网格点:

class spGridIterator {
public:
    typedef forward_iterator_tag iterator_category;
    typedef spVector3D value_type;
    typedef int difference_type;
    typedef spVector3D* pointer;
    typedef const spVector3D* const_pointer;
    typedef const spVector3D& const_reference;
    typedef spVector3D& reference;

    spGridIterator(spGrid* gr, int index);

    spGridIterator& operator++();
    spGridIterator& operator++(int);

    reference operator*();
    const_reference operator*() const;

    pointer operator->();
    const_pointer operator->() const;

private:
    spGrid* m_grid;
    int m_idx;
    mutable spVector3D CurrentPoint;
};

然后解除引用运算符可能如下所示:

spGridIterator::const_reference spGridIterator::operator*() const {
    CurrentPoint = m_grid->GetPoint(m_idx);
    return CurrentPoint;
}

感谢@greg指出CurrentPoint需要mutable为此工作.这将是一个惰性实现(仅在迭代器实际解除引用时检索点).一个急切的实现将更新CurrentPoint迭代器的mutator方法中的成员(operator++本例中的变体),这使得mutable多余.