Question

在我的光线追踪器中，我正在构建一个边界体积层次结构。我用了几天试图解决这个错误，但我似乎做了一些根本错误的事情。

第一种方法是使用typedef std :: vector Objects数组调用的构造。

void BVH::build(Objects* objs)
{
    // construct the bounding volume hierarchy
    int size = objs->size();

    // Calculate the bounding box for this node
    BBox bb = (*objs)[0]->boundingBox();
    for ( int p = 1; p < size; p++ )
        bb.expandToInclude( (*objs)[p]->boundingBox());
    Vector3 pivot = (bb.max + bb.min) * 0.5f;
    tree.bbox = bb;

    int split = qsplit(objs, size, pivot.x, 0);

    tree.left = subdivision(objs, split, 1);
    tree.right = subdivision(&objs[split], size - split, 1);
}

下面使用此方法在我的二叉树中构造叶节点。

Node* BVH::makeLeaf(Objects* objs, int num)
{
    Node* node = new Node;
    if ( num == 1 ) { node->objs = &objs[0]; }
    else if ( num == 2 ) { node->objs = ((&objs)[0],(&objs)[1]); }
    node->isLeaf = true;
    return node;
}

这是一个递归方法，使用Objects数组来拆分和构造节点和叶节点。

Node* BVH::subdivision(Objects* objs, int size, int axis)
{
    if ( size == 1 ) { return makeLeaf(objs, 1); }
    if ( size == 2 ) { return makeLeaf(objs, 2); }

    Node* node = new Node;
    node->isLeaf = false;

    BBox bb = (*objs)[0]->boundingBox();
    for ( int p = 1; p < size; p++ )
        bb.expandToInclude((*objs)[p]->boundingBox());
    node->bbox = bb;
    Vector3 pivot = (bb.max + bb.min) * 0.5f;

    int split = qsplit(objs, size, pivot[axis], axis);

    node->left = subdivision(objs, split, (axis + 1) % 3);
    node->right = subdivision(&objs[split], size - split, (axis + 1) % 3);

    return node;
}

运行此代码会在细分内部给出分段错误。

这是来自gdb的空指针的帖子：

debug: Loading "teapot.obj"...
debug: Loaded "teapot.obj" with 576 triangles

Program received signal SIGSEGV, Segmentation fault.
0x000000000040bb9e in BVH::subdivision (this=0x61f0c0, objs=0x61f100, size=5, axis=1) at BVH.cpp:44
44      BBox bb = (*objs)[0]->boundingBox();
(gdb) bt
0  0x000000000040bb9e in BVH::subdivision (this=0x61f0c0, objs=0x61f100, size=5, axis=1) at BVH.cpp:44
1  0x000000000040bdef in BVH::subdivision (this=0x61f0c0, objs=0x61f0a0, size=9, axis=0) at BVH.cpp:56
2  0x000000000040bd62 in BVH::subdivision (this=0x61f0c0, objs=0x61f0a0, size=18, axis=2) at BVH.cpp:53
3  0x000000000040bd62 in BVH::subdivision (this=0x61f0c0, objs=0x61f0a0, size=36, axis=1) at BVH.cpp:53
4  0x000000000040bd62 in BVH::subdivision (this=0x61f0c0, objs=0x61f0a0, size=72, axis=0) at BVH.cpp:53
5  0x000000000040bd62 in BVH::subdivision (this=0x61f0c0, objs=0x61f0a0, size=144, axis=2) at BVH.cpp:53
6  0x000000000040bd62 in BVH::subdivision (this=0x61f0c0, objs=0x61f0a0, size=288, axis=1) at BVH.cpp:53
7  0x000000000040b9ca in BVH::build (this=0x61f0c0, objs=0x61f0a0) at BVH.cpp:21
8  0x00000000004121ac in Scene::preCalc (this=0x61f0a0) at Scene.cpp:42
9  0x0000000000402fde in makeTeapotScene () at assignment1.cpp:113
10 0x000000000040f973 in main (argc=1, argv=0x7fffffffe798) at main.cpp:65
(gdb) p (*objs)[0]
$1 = (Object *&) @0x0: <error reading variable>

Answer 1

(&objs)[0]

与

相同

*(&objs)

与

相同

objs

以同样的方式，

(&objs)[1]

与

相同

objs + 1

最有可能不有效对象的地址。

该行

node->objs = ((&objs)[0],(&objs)[1]);

因为它使用逗号运算符，会抛弃(&objs)[0]的值并将node->objs设置为objs + 1，这是无效的。

目前还不清楚你希望完成任务的目的是什么，所以我不知道你应该用它替换它。

Answer 2

在这个有趣的typedef上使用了很多小时之后，我得到了朋友的帮助，我们写了一个转换器方法来转

vector<Object*>* into vector<Object*>&

使每种方法都更加清洁和美观。

void BVH::build(Objects* objs_orig)
{
    m_objects = objs_orig;
    std::vector<Object*>* objs = objs_orig;
    return build(*objs);
}

void BVH::build(const std::vector<Object*>& objs)
{
    tree = makeTree(objs, 1);
}

与向量关联的NULL指针

2 个答案: