本文整理汇总了C++中BBox::inside方法的典型用法代码示例。如果您正苦于以下问题：C++ BBox::inside方法的具体用法？C++ BBox::inside怎么用？C++ BBox::inside使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类BBox的用法示例。
在下文中一共展示了BBox::inside方法的2个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: intersectKD
bool Mesh::intersectKD(KDTree& node, const BBox& box, const FastRay& ray, IntersectionInfo& info) const
{
if (node.axis == AXIS_NONE) {
vector<int>& trios = *node.data.triangles;
double foundlambda2 = 0, foundlambda3 = 0, dist = INF;
int foundIdx = -1;
for (int i = 0; i < (int) trios.size(); i++) {
const Triangle& t = triangles[trios[i]];
double l2, l3;
if (intersectTriangleFast(ray, vertices[t.v[0]], t, l2, l3, dist)) {
foundIdx = trios[i];
foundlambda2 = l2;
foundlambda3 = l3;
}
}
if (foundIdx != -1) {
Vector ip = ray.start + ray.dir * dist;
if (box.inside(ip)) {
fillInfo(ray, info, foundIdx, foundlambda2, foundlambda3, dist);
return true;
}
}
return false;
} else {
BBox boxChild[2];
box.split(node.axis, node.sp, boxChild[0], boxChild[1]);
int order[2] = { 0, 1 };
if (ray.start[node.axis] > node.sp) swap(order[0], order[1]);
// if the ray intersects the common wall between the two sub-boxes, then it invariably
// intersects both boxes (we can skip the testIntersect() checks):
// (see http://raytracing-bg.net/?q=node/68 )
if (box.intersectWall(node.axis, node.sp, ray)) {
if (intersectKD(node.data.children[order[0]], boxChild[order[0]], ray, info)) return true;
return intersectKD(node.data.children[order[1]], boxChild[order[1]], ray, info);
} else {
// if the wall isn't hit, then we intersect exclusively one of the sub-boxes;
// test one, if the test fails, then it's in the other:
if (boxChild[order[0]].testIntersect(ray))
return intersectKD(node.data.children[order[0]], boxChild[order[0]], ray, info);
else
return intersectKD(node.data.children[order[1]], boxChild[order[1]], ray, info);
}
}
return false;
}示例2: intersectKD
bool Mesh::intersectKD(KDTreeNode& node, const BBox& bbox, const Ray& ray, IntersectionData& data)
{
if (node.axis == AXIS_NONE) {
// leaf node; try intersecting with the triangle list:
bool found = false;
for (size_t i = 0; i < node.triangles->size(); i++) {
int triIdx = (*node.triangles)[i];
if (intersectTriangle(ray, data, triangles[triIdx])) {
found = true;
}
}
// the found intersection has to be inside "our" BBox, otherwise we can miss a triangle,
// as explained in the presentations:
if (found && bbox.inside(data.p)) return true;
return false;
} else {
// a in-node; intersect with the two children, starting with the closer one first:
int childOrder[2] = { 0, 1 };
if (ray.start[node.axis] > node.splitPos)
swap(childOrder[0], childOrder[1]);
//
BBox childBB[2];
bbox.split(node.axis, node.splitPos, childBB[0], childBB[1]);
// name the children bboxen:
BBox& firstBB = childBB[childOrder[0]];
BBox& secondBB = childBB[childOrder[1]];
KDTreeNode& firstChild = node.children[childOrder[0]];
KDTreeNode& secondChild = node.children[childOrder[1]];
// if the ray intersects the common wall between the two sub-boxes, then it invariably
// intersects both boxes (we can skip the testIntersect() checks):
// (see http://raytracing-bg.net/?q=node/68 )
if (bbox.intersectWall(node.axis, node.splitPos, ray)) {
if (intersectKD(firstChild, firstBB, ray, data)) return true;
return intersectKD(secondChild, secondBB, ray, data);
} else {
// if the wall isn't hit, then we intersect exclusively one of the sub-boxes;
// test one, if the test fails, then it's in the other:
if (firstBB.testIntersect(ray))
return intersectKD(firstChild, firstBB, ray, data);
else
return intersectKD(secondChild, secondBB, ray, data);
}
}
}