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目前，我有如下就像穿越在OpenCL內核。如果有人對這個相當大的內核進行優化，我會很高興。的OpenCL內核遍歷 - 進一步優化
問題是，我正在SAH BVH上運行此代碼，並且我希望在他的論文（瞭解GPU上的光線遍歷效率）中使用遍歷的方式獲得與Timo Aila類似的性能，當然他的代碼使用SplitBVH（我可能會考慮使用它代替SAH BVH，但在我看來它的構建時間確實很慢）。但我問的是遍歷，而不是BVH（我迄今爲止只適用於場景，SplitBVH不會給你帶來比SAH BVH更多的優勢）。
首先，這裏是我迄今爲止（標準的同時，同時遍歷內核）。
__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;
}
}
當天的第一個問題是，怎麼能寫在他的OpenCL而持久，而和投機性，同時，而內核？
廣告持續而-而，我得到它的權利，我其實剛開始與全球等同於當地的工作尺寸工作尺寸的內核，而這兩個數字應該等於以翹曲的GPU /波陣面尺寸？ 我得到與CUDA持久線程實現看起來是這樣的：
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);
怎麼可能相當於OpenCL的樣子，我知道我必須做在那裏的一些障礙，我也知道，一個後應我原子地將WARP_SIZE添加到warpCounter的分數。
廣告投機穿越 - 好，我可能沒有任何想法如何這應該在OpenCL的實施，所以任何提示的歡迎。我也不知道在哪裏放置障礙物（因爲將它們放在模擬__中會導致駕駛員碰撞）。
如果你在這裏做到了，感謝閱讀&任何提示，答案等，歡迎！