4
目前，我有如下就像穿越在OpenCL内核。如果有人对这个相当大的内核进行优化，我会很高兴。的OpenCL内核遍历 - 进一步优化
问题是，我正在SAH BVH上运行此代码，并且我希望在他的论文（了解GPU上的光线遍历效率）中使用遍历的方式获得与Timo Aila类似的性能，当然他的代码使用SplitBVH（我可能会考虑使用它代替SAH BVH，但在我看来它的构建时间确实很慢）。但我问的是遍历，而不是BVH（我迄今为止只适用于场景，SplitBVH不会给你带来比SAH BVH更多的优势）。
首先，这里是我迄今为止（标准的同时，同时遍历内核）。
__constant sampler_t sampler = CLK_FILTER_NEAREST;
// Inline definition of horizontal max
inline float max4(float a, float b, float c, float d)
{
return max(max(max(a, b), c), d);
}
// Inline definition of horizontal min
inline float min4(float a, float b, float c, float d)
{
return min(min(min(a, b), c), d);
}
// Traversal kernel
__kernel void traverse(__read_only image2d_t nodes,
__global const float4* triangles,
__global const float4* rays,
__global float4* result,
const int num,
const int w,
const int h)
{
// Ray index
int idx = get_global_id(0);
if(idx < num)
{
// Stack
int todo[32];
int todoOffset = 0;
// Current node
int nodeNum = 0;
float tmin = 0.0f;
float depth = 2e30f;
// Fetch ray origin, direction and compute invdirection
float4 origin = rays[2 * idx + 0];
float4 direction = rays[2 * idx + 1];
float4 invdir = native_recip(direction);
float4 temp = (float4)(0.0f, 0.0f, 0.0f, 1.0f);
// Traversal loop
while(true)
{
// Fetch node information
int2 nodeCoord = (int2)((nodeNum << 2) % w, (nodeNum << 2)/w);
int4 specs = read_imagei(nodes, sampler, nodeCoord + (int2)(3, 0));
// While node isn't leaf
while(specs.z == 0)
{
// Fetch child bounding boxes
float4 n0xy = read_imagef(nodes, sampler, nodeCoord);
float4 n1xy = read_imagef(nodes, sampler, nodeCoord + (int2)(1, 0));
float4 nz = read_imagef(nodes, sampler, nodeCoord + (int2)(2, 0));
// Test ray against child bounding boxes
float oodx = origin.x * invdir.x;
float oody = origin.y * invdir.y;
float oodz = origin.z * invdir.z;
float c0lox = n0xy.x * invdir.x - oodx;
float c0hix = n0xy.y * invdir.x - oodx;
float c0loy = n0xy.z * invdir.y - oody;
float c0hiy = n0xy.w * invdir.y - oody;
float c0loz = nz.x * invdir.z - oodz;
float c0hiz = nz.y * invdir.z - oodz;
float c1loz = nz.z * invdir.z - oodz;
float c1hiz = nz.w * invdir.z - oodz;
float c0min = max4(min(c0lox, c0hix), min(c0loy, c0hiy), min(c0loz, c0hiz), tmin);
float c0max = min4(max(c0lox, c0hix), max(c0loy, c0hiy), max(c0loz, c0hiz), depth);
float c1lox = n1xy.x * invdir.x - oodx;
float c1hix = n1xy.y * invdir.x - oodx;
float c1loy = n1xy.z * invdir.y - oody;
float c1hiy = n1xy.w * invdir.y - oody;
float c1min = max4(min(c1lox, c1hix), min(c1loy, c1hiy), min(c1loz, c1hiz), tmin);
float c1max = min4(max(c1lox, c1hix), max(c1loy, c1hiy), max(c1loz, c1hiz), depth);
bool traverseChild0 = (c0max >= c0min);
bool traverseChild1 = (c1max >= c1min);
nodeNum = specs.x;
int nodeAbove = specs.y;
// We hit just one out of 2 childs
if(traverseChild0 != traverseChild1)
{
if(traverseChild1)
{
nodeNum = nodeAbove;
}
}
// We hit either both or none
else
{
// If we hit none, pop node from stack (or exit traversal, if stack is empty)
if (!traverseChild0)
{
if(todoOffset == 0)
{
break;
}
nodeNum = todo[--todoOffset];
}
// If we hit both
else
{
// Sort them (so nearest goes 1st, further 2nd)
if(c1min < c0min)
{
unsigned int tmp = nodeNum;
nodeNum = nodeAbove;
nodeAbove = tmp;
}
// Push further on stack
todo[todoOffset++] = nodeAbove;
}
}
// Fetch next node information
nodeCoord = (int2)((nodeNum << 2) % w, (nodeNum << 2)/w);
specs = read_imagei(nodes, sampler, nodeCoord + (int2)(3, 0));
}
// If node is leaf & has some primitives
if(specs.z > 0)
{
// Loop through primitives & perform intersection with them (Woop triangles)
for(int i = specs.x; i < specs.y; i++)
{
// Fetch first point from global memory
float4 v0 = triangles[i * 4 + 0];
float o_z = v0.w - origin.x * v0.x - origin.y * v0.y - origin.z * v0.z;
float i_z = 1.0f/(direction.x * v0.x + direction.y * v0.y + direction.z * v0.z);
float t = o_z * i_z;
if(t > 0.0f && t < depth)
{
// Fetch second point from global memory
float4 v1 = triangles[i * 4 + 1];
float o_x = v1.w + origin.x * v1.x + origin.y * v1.y + origin.z * v1.z;
float d_x = direction.x * v1.x + direction.y * v1.y + direction.z * v1.z;
float u = o_x + t * d_x;
if(u >= 0.0f && u <= 1.0f)
{
// Fetch third point from global memory
float4 v2 = triangles[i * 4 + 2];
float o_y = v2.w + origin.x * v2.x + origin.y * v2.y + origin.z * v2.z;
float d_y = direction.x * v2.x + direction.y * v2.y + direction.z * v2.z;
float v = o_y + t * d_y;
if(v >= 0.0f && u + v <= 1.0f)
{
// We got successful hit, store the information
depth = t;
temp.x = u;
temp.y = v;
temp.z = t;
temp.w = as_float(i);
}
}
}
}
}
// Pop node from stack (if empty, finish traversal)
if(todoOffset == 0)
{
break;
}
nodeNum = todo[--todoOffset];
}
// Store the ray traversal result in global memory
result[idx] = temp;
}
}
当天的第一个问题是，怎么能写在他的OpenCL而持久，而和投机性，同时，而内核？
广告持续而-而，我得到它的权利，我其实刚开始与全球等同于当地的工作尺寸工作尺寸的内核，而这两个数字应该等于以翘曲的GPU /波阵面尺寸？ 我得到与CUDA持久线程实现看起来是这样的：
do
{
volatile int& jobIndexBase = nextJobArray[threadIndex.y];
if(threadIndex.x == 0)
{
jobIndexBase = atomicAdd(&warpCounter, WARP_SIZE);
}
index = jobIndexBase + threadIndex.x;
if(index >= totalJobs)
return;
/* Perform work for task numbered 'index' */
}
while(true);
怎么可能相当于OpenCL的样子，我知道我必须做在那里的一些障碍，我也知道，一个后应我原子地将WARP_SIZE添加到warpCounter的分数。
广告投机穿越 - 好，我可能没有任何想法如何这应该在OpenCL的实施，所以任何提示的欢迎。我也不知道在哪里放置障碍物（因为将它们放在模拟__中会导致驾驶员碰撞）。
如果你在这里做到了，感谢阅读&任何提示，答案等，欢迎！