向量迭代器不兼容

发布于 2024-11-10 12:19:54 字数 2545 浏览 1 评论 0原文

我目前正在开发 C++ 图形库,现在陷入了运行时在调试模式下出现断言错误的问题。我还在这里查看了其他一些问题,但没有一个问题和答案能引导我找到解决方案。在阅读一些论坛后,我的印象是,发生此错误是因为一旦向量内容发生更改,迭代器就会变得无效。 (例如,当使用 erase() 时)但正如您在我的代码中看到的,我没有修改向量,只是进行迭代。

错误出现在我用 //ASSERTION 标记的行中。奇怪的是,neighbor_it 并不指向 (*vertex_it)->neighbors() 中的第一个对象,而是指向 0xfeeefeee。通过代码进行调试时,我可以清楚地看到邻居向量至少包含一项。 neighbor_it 不应该指向这个向量中的第一个对象吗?

有关更多信息:m_vertices 是图中所有顶点的向量,vertex::neighbors() 返回边向量(具有指向邻居/目的地的指针)顶点)。在这个方法中,我想删除指向某个顶点的所有边。如果已找到并删除相应的边,则返回 true;如果没有指向 p_vertex 的边,则返回 false。

bool graph::remove_edges_pointing_to( vertex* p_vertex )
{
    bool res = false;

    std::vector<vertex*>::iterator vertex_it = m_vertices.begin();

    // iterate through all vertices
    while( vertex_it != m_vertices.end() )
    {
        // iterator on first element of neighbors of vertex
        std::vector<edge*>::iterator neighbor_it = (*vertex_it)->neighbors().begin();

        // iterate through all successors of each vertex
        while( neighbor_it != (*vertex_it)->neighbors().end() ) //ASSERTION
        {
            if( (*neighbor_it)->dest() == p_vertex )
            {
                if( (*vertex_it)->remove_edge( *neighbor_it ) )
                {
                    res = true;
                }
            }

            neighbor_it++;
        }

        vertex_it++;
    }

    return res;
}

编辑:(解决方案)

好吧,这是我的新代码,可以正常工作。 remove_edge() 现在返回一个迭代器,指向从中删除边缘的向量中的下一个对象。此外,neighbors() 现在返回相应向量的引用

bool graph::remove_edges_pointing_to( vertex* p_vertex )
{
    bool res = false;

    std::vector<vertex*>::iterator vertex_it = m_vertices.begin();

    // iterate through all vertices
    while( vertex_it != m_vertices.end() )
    {
        // iterator on first element of neighbors of vertex
        std::vector<edge*>::iterator neighbor_it = (*vertex_it)->neighbors().begin();

        // iterate through all successors of each vertex
        while( neighbor_it != (*vertex_it)->neighbors().end() )
        {
            if( (*neighbor_it)->dest() == p_vertex )
            {
                neighbor_it = (*vertex_it)->remove_edge( *neighbor_it );
                res = true;
            }
            else
            {
                neighbor_it++;
            }
        }

        vertex_it++;
    }

    return res;
}

再次感谢您的回答! :)

I'm currently working on a graph library for C++ and now got stuck at a point where I get an assertion error in debug mode during runtime. I also had a look an some other question here on SO but none of the questions and answers lead me to a solution. After reading in some forums I have the impression that this error happens because iterators become invalid as soon as the vector content is changed. (for example when using erase()) But as you can see in my code, I'm not modifying the vector, just iterating.

The error is in the line I marked with //ASSERTION. The strange thing is that neighbor_it doesn't point to the first object in (*vertex_it)->neighbors() but to 0xfeeefeee. When debugging through the code I can clearly see that the neighbors-vector contains at least one item. Shouldn't neighbor_it point to the first object in this vector?

For further information: m_vertices is a vector of all vertices in a graph and vertex::neighbors() returns a vector of edges (which have a pointer to the neighbor/destination vertex). In this method I want to remove all edges pointing to a certain vertex. Returns true if an according edge has been found and removed, false if there is no edge pointing to p_vertex.

bool graph::remove_edges_pointing_to( vertex* p_vertex )
{
    bool res = false;

    std::vector<vertex*>::iterator vertex_it = m_vertices.begin();

    // iterate through all vertices
    while( vertex_it != m_vertices.end() )
    {
        // iterator on first element of neighbors of vertex
        std::vector<edge*>::iterator neighbor_it = (*vertex_it)->neighbors().begin();

        // iterate through all successors of each vertex
        while( neighbor_it != (*vertex_it)->neighbors().end() ) //ASSERTION
        {
            if( (*neighbor_it)->dest() == p_vertex )
            {
                if( (*vertex_it)->remove_edge( *neighbor_it ) )
                {
                    res = true;
                }
            }

            neighbor_it++;
        }

        vertex_it++;
    }

    return res;
}

EDIT: (Solution)

Alright, here is my new code which works properly. remove_edge() now returns an iterator to the next object in the vector it removed the edge from. In addition neighbors() now returns a reference to the according vector.

bool graph::remove_edges_pointing_to( vertex* p_vertex )
{
    bool res = false;

    std::vector<vertex*>::iterator vertex_it = m_vertices.begin();

    // iterate through all vertices
    while( vertex_it != m_vertices.end() )
    {
        // iterator on first element of neighbors of vertex
        std::vector<edge*>::iterator neighbor_it = (*vertex_it)->neighbors().begin();

        // iterate through all successors of each vertex
        while( neighbor_it != (*vertex_it)->neighbors().end() )
        {
            if( (*neighbor_it)->dest() == p_vertex )
            {
                neighbor_it = (*vertex_it)->remove_edge( *neighbor_it );
                res = true;
            }
            else
            {
                neighbor_it++;
            }
        }

        vertex_it++;
    }

    return res;
}

Thanks again for your answers! :)

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评论(2

蓝眼睛不忧郁 2024-11-17 12:19:54

我的猜测是,鉴于您提供的有限上下文,neighbours() 返回 std::vector 的副本,而不是引用,即 std ::向量<边缘*>&。因此,在 begin() 调用之后,临时对象被释放,并且获得的迭代器指向垃圾。

My guess, given limited context you provided, is that neighbours() returns copy of std::vector<edge*>, instead of reference, i.e. std::vector<edge*>&. So after begin() call temporary object is disposed, and obtained iterator points to rubbish.

别挽留 2024-11-17 12:19:54

我猜想 remove_edge 修改了 neighbor_it 的底层容器,从而使其无效,但如果没有看到更多代码,我无法确定。

如果是这种情况,可能的解决方案是将迭代器返回到已删除元素之后的下一个元素,例如通过 std::vector::erase 完成的操作。

I would guess that remove_edge modifies the underlying container of neighbor_it, thereby invalidating it, but I cannot be certain without seeing more of your code.

If this is a case, a possible solution would be to return an iterator to the next element after the deleted one, as done for instance by std::vector::erase.

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