我的以下极简主义Cuda代码返回不正确的结果(所有多边形在末尾都有0个顶点),而在C ++中以串行方式运行的相同代码运行良好。问题是令人尴尬的并行:没有通信,没有同步传输等,而且Cuda内存分配是成功的。即使是我的虚拟变量,存储用于调试目的的输入数组的内容,对于Cuda版本也是0。由于我的阵列足够大,所以没有超出范围的访问权限。用Cuda中的循环替换memcpy不会改变任何东西 我真的不明白会发生什么......有什么想法吗?谢谢!
Cuda代码:
#include <stdio.h>
#include <iostream>
#include <stdlib.h>
#include <cuda.h>
class Point2D {
public:
__device__ Point2D(double xx=0, double yy=0):x(xx),y(yy){};
double x, y;
};
__device__ double dot(const Point2D &A, const Point2D &B) {
return A.x*B.x + A.y*B.y;
}
__device__ Point2D operator*(double a, const Point2D &P) {
return Point2D(a*P.x, a*P.y);
}
__device__ Point2D operator+(Point2D A, const Point2D &B) {
return Point2D(A.x + B.x, A.y + B.y);
}
__device__ Point2D operator-(Point2D A, const Point2D &B) {
return Point2D(A.x - B.x, A.y - B.y);
}
__device__ Point2D inter(const Point2D &A, const Point2D &B, const Point2D &C, const Point2D &D) { //intersects AB by *the mediator* of CD
Point2D M = 0.5*(C+D);
return A - (dot(A-M, D-C)/dot(B-A, D-C)) * (B-A);
}
class Polygon {
public:
__device__ Polygon():nbpts(0){};
__device__ void addPts(Point2D pt) {
pts[nbpts] = pt;
nbpts++;
};
__device__ Polygon& operator=(const Polygon& rhs) {
nbpts = rhs.nbpts;
dummy = rhs.dummy;
memcpy(pts, rhs.pts, nbpts*sizeof(Point2D));
return *this;
}
__device__ void cut(const Point2D &inside_pt, const Point2D &outside_pt) {
int new_nbpts = 0;
Point2D newpts[128];
Point2D AB(outside_pt-inside_pt);
Point2D M(0.5*(outside_pt+inside_pt));
double ABM = dot(AB, M);
Point2D S = pts[nbpts-1];
for (int i=0; i<nbpts; i++) {
Point2D E = pts[i];
double ddot = -ABM + dot(AB, E);
if (ddot<0) { // E inside clip edge
double ddot2 = -ABM + dot(AB, S);
if (ddot2>0) {
newpts[new_nbpts] = inter(S,E, inside_pt, outside_pt);
new_nbpts++;
}
newpts[new_nbpts] = E;
new_nbpts++;
} else {
double ddot2 = -ABM + dot(AB, S);
if (ddot2<0) {
newpts[new_nbpts] = inter(S,E, inside_pt, outside_pt);
new_nbpts++;
}
}
S = E;
}
memcpy(pts, newpts, min(128, new_nbpts)*sizeof(Point2D));
nbpts = new_nbpts;
}
//private:
Point2D pts[128];
int nbpts;
float dummy;
};
__global__ void cut_poly(float *a, Polygon* polygons, int N)
{
int idx = blockIdx.x * blockDim.x + threadIdx.x;
if (idx>=N/2) return;
Polygon pol;
pol.addPts(Point2D(0.,0.));
pol.addPts(Point2D(1.,0.));
pol.addPts(Point2D(1.,1.));
pol.addPts(Point2D(0.,1.));
Point2D curPt(a[2*idx], a[2*idx+1]);
for (int i=0; i<N/2; i++) {
Point2D other_pt(a[2*i], a[2*i+1]);
pol.cut(curPt, other_pt);
}
pol.dummy = a[idx];
polygons[idx] = pol;
}
int main(int argc, unsigned char* argv[])
{
const int N = 100;
float a_h[N], *a_d;
Polygon p_h[N/2], *p_d;
size_t size = N * sizeof(float);
size_t size_pol = N/2 * sizeof(Polygon);
cudaError_t err = cudaMalloc((void **) &a_d, size);
cudaError_t err2 = cudaMalloc((void **) &p_d, size_pol);
for (int i=0; i<N; i++) a_h[i] = (float)(rand()%1000)*0.001;
cudaMemcpy(a_d, a_h, size, cudaMemcpyHostToDevice);
int block_size = 4;
int n_blocks = N/block_size + (N%block_size == 0 ? 0:1);
cut_poly <<< n_blocks, block_size >>> (a_d, p_d, N);
cudaMemcpy(a_h, a_d, sizeof(float)*N, cudaMemcpyDeviceToHost);
cudaMemcpy(p_h, p_d, sizeof(Polygon)*N/2, cudaMemcpyDeviceToHost);
for (int i=0; i<N/2; i++)
printf("%f \t %f \t %u\n", a_h[i], p_h[i].dummy, p_h[i].nbpts);
cudaFree(a_d);
cudaFree(p_d);
return 0;
}
C ++中的相同代码可以正常工作:
#include <stdio.h>
#include <iostream>
#include <stdlib.h>
class Point2D {
public:
Point2D(double xx=0, double yy=0):x(xx),y(yy){};
double x, y;
};
double dot(const Point2D &A, const Point2D &B) {
return A.x*B.x + A.y*B.y;
}
Point2D operator*(double a, const Point2D &P) {
return Point2D(a*P.x, a*P.y);
}
Point2D operator+(Point2D A, const Point2D &B) {
return Point2D(A.x + B.x, A.y + B.y);
}
Point2D operator-(Point2D A, const Point2D &B) {
return Point2D(A.x - B.x, A.y - B.y);
}
Point2D inter(const Point2D &A, const Point2D &B, const Point2D &C, const Point2D &D) { //intersects AB by *the mediator* of CD
Point2D M = 0.5*(C+D);
return A - (dot(A-M, D-C)/dot(B-A, D-C)) * (B-A);
}
class Polygon {
public:
Polygon():nbpts(0){};
void addPts(Point2D pt) {
pts[nbpts] = pt;
nbpts++;
};
Polygon& operator=(const Polygon& rhs) {
nbpts = rhs.nbpts;
dummy = rhs.dummy;
memcpy(pts, rhs.pts, nbpts*sizeof(Point2D));
return *this;
}
void cut(const Point2D &inside_pt, const Point2D &outside_pt) {
int new_nbpts = 0;
Point2D newpts[128];
Point2D AB(outside_pt-inside_pt);
Point2D M(0.5*(outside_pt+inside_pt));
double ABM = dot(AB, M);
Point2D S = pts[nbpts-1];
for (int i=0; i<nbpts; i++) {
Point2D E = pts[i];
double ddot = -ABM + dot(AB, E);
if (ddot<0) { // E inside clip edge
double ddot2 = -ABM + dot(AB, S);
if (ddot2>0) {
newpts[new_nbpts] = inter(S,E, inside_pt, outside_pt);
new_nbpts++;
}
newpts[new_nbpts] = E;
new_nbpts++;
} else {
double ddot2 = -ABM + dot(AB, S);
if (ddot2<0) {
newpts[new_nbpts] = inter(S,E, inside_pt, outside_pt);
new_nbpts++;
}
}
S = E;
}
memcpy(pts, newpts, std::min(128, new_nbpts)*sizeof(Point2D));
/*for (int i=0; i<128; i++) {
pts[i] = newpts[i];
}*/
nbpts = new_nbpts;
}
//private:
Point2D pts[128];
int nbpts;
float dummy;
};
void cut_poly(int idx, float *a, Polygon* polygons, int N)
{
if (idx>=N/2) return;
Polygon pol;
pol.addPts(Point2D(0.,0.));
pol.addPts(Point2D(1.,0.));
pol.addPts(Point2D(1.,1.));
pol.addPts(Point2D(0.,1.));
Point2D curPt(a[2*idx], a[2*idx+1]);
for (int i=0; i<N/2; i++) {
if (idx==i) continue;
Point2D other_pt(a[2*i], a[2*i+1]);
pol.cut(curPt, other_pt);
}
pol.dummy = a[idx];
polygons[idx] = pol;
}
int main(int argc, unsigned char* argv[])
{
const int N = 100; // Number of elements in arrays
float a_h[N], *a_d; // Pointer to host & device arrays
Polygon p_h[N/2], *p_d;
for (int i=0; i<N; i++) a_h[i] = (float)(rand()%1000)*0.001;
for (int idx=0; idx<N; idx++)
cut_poly(idx, a_h, p_h, N);
for (int i=0; i<N/2; i++)
printf("%f \t %f \t %u\n", a_h[i], p_h[i].dummy, p_h[i].nbpts);
return 0;
}
答案 0 :(得分:3)
嗯,我想你可以忽略我的大部分评论。我错误地使用了我用CUDA 3.2设置的机器,并且它在内核启动失败的情况下表现不同。当我切换到CUDA 4.1和CUDA 5.0时,事情开始变得有意义了。为我的困惑道歉。
无论如何,我很快就注意到你的CPU和GPU实现之间存在差异。具体在这里(查看CPU代码):
void cut_poly(int idx, float *a, Polygon* polygons, int N)
{
if (idx>=N/2) return;
Polygon pol;
pol.addPts(Point2D(0.,0.));
pol.addPts(Point2D(1.,0.));
pol.addPts(Point2D(1.,1.));
pol.addPts(Point2D(0.,1.));
Point2D curPt(a[2*idx], a[2*idx+1]);
for (int i=0; i<N/2; i++) {
if (idx==i) continue; /* NOTE THIS LINE MISSING FROM YOUR GPU CODE */
Point2D other_pt(a[2*i], a[2*i+1]);
pol.cut(curPt, other_pt);
}
pol.dummy = a[idx];
polygons[idx] = pol;
}
参考我在上面添加注释的行,如果你将那段精确的代码行添加到cut_poly
内核的GPU代码中的相应位置,那么对我来说无论如何GPU代码都会生成相同的代码。打印结果作为CPU代码。
我要做的另一个观察是你不必要地运行只有4个线程的块。当你解决代码中的问题时没有错,但是一旦你为了“生产”目的而运行它,你很可能想要定位一个更高的数字,如256,并确保选择一个数字为32的整数倍,以获得最佳性能。
在回复评论中发布的问题时,我认为数据正在被正确复制,但很可能您没有在主机上正确访问它。 (我不知道你是怎么确定“我的阵列没有正确地返回给主机”)。您的大部分课程定义仅为__device__
。因此,很难访问主机上的类内的结构(例如Point2D pts
类中的Polygon
类)。我在这里插入修改后的代码,我认为这些代码表明数据正在传输回主机:
#include <stdio.h>
#include <iostream>
#include <stdlib.h>
// #include <cuda.h>
#define cudaCheckErrors(msg) \
do { \
cudaError_t __err = cudaGetLastError(); \
if (__err != cudaSuccess) { \
fprintf(stderr, "Fatal error: %s (%s at %s:%d)\n", \
msg, cudaGetErrorString(__err), \
__FILE__, __LINE__); \
fprintf(stderr, "*** FAILED - ABORTING\n"); \
exit(1); \
} \
} while (0)
class Point2D {
public:
__host__ __device__ Point2D(double xx=0, double yy=0):x(xx),y(yy){};
double x, y;
};
__host__ __device__ double dot(const Point2D &A, const Point2D &B) {
return A.x*B.x + A.y*B.y;
}
__host__ __device__ Point2D operator*(double a, const Point2D &P) {
return Point2D(a*P.x, a*P.y);
}
__host__ __device__ Point2D operator+(Point2D A, const Point2D &B) {
return Point2D(A.x + B.x, A.y + B.y);
}
__host__ __device__ Point2D operator-(Point2D A, const Point2D &B) {
return Point2D(A.x - B.x, A.y - B.y);
}
__host__ __device__ Point2D inter(const Point2D &A, const Point2D &B, const Point2D &C, const Point2D &D) { //intersects AB by *the mediator* of CD
Point2D M = 0.5*(C+D);
return A - (dot(A-M, D-C)/dot(B-A, D-C)) * (B-A);
}
class Polygon {
public:
__host__ __device__ Polygon():nbpts(0){};
__host__ __device__ void addPts(Point2D pt) {
pts[nbpts] = pt;
nbpts++;
};
__host__ __device__ Polygon& operator=(const Polygon& rhs) {
nbpts = rhs.nbpts;
dummy = rhs.dummy;
memcpy(pts, rhs.pts, nbpts*sizeof(Point2D));
return *this;
}
__host__ __device__ Point2D getpoint(unsigned i){
if (i<128) return pts[i];
else return pts[0];
}
__host__ __device__ void cut(const Point2D &inside_pt, const Point2D &outside_pt) {
int new_nbpts = 0;
Point2D newpts[128];
Point2D AB(outside_pt-inside_pt);
Point2D M(0.5*(outside_pt+inside_pt));
double ABM = dot(AB, M);
Point2D S = pts[nbpts-1];
for (int i=0; i<nbpts; i++) {
Point2D E = pts[i];
double ddot = -ABM + dot(AB, E);
if (ddot<0) { // E inside clip edge
double ddot2 = -ABM + dot(AB, S);
if (ddot2>0) {
newpts[new_nbpts] = inter(S,E, inside_pt, outside_pt);
new_nbpts++;
}
newpts[new_nbpts] = E;
new_nbpts++;
} else {
double ddot2 = -ABM + dot(AB, S);
if (ddot2<0) {
newpts[new_nbpts] = inter(S,E, inside_pt, outside_pt);
new_nbpts++;
}
}
S = E;
}
memcpy(pts, newpts, min(128, new_nbpts)*sizeof(Point2D));
nbpts = new_nbpts;
}
//private:
Point2D pts[128];
int nbpts;
float dummy;
};
__global__ void cut_poly(float *a, Polygon* polygons, int N)
{
int idx = blockIdx.x * blockDim.x + threadIdx.x;
if (idx>=N/2) return;
Polygon pol;
pol.addPts(Point2D(0.,0.));
pol.addPts(Point2D(1.,0.));
pol.addPts(Point2D(1.,1.));
pol.addPts(Point2D(0.,1.));
Point2D curPt(a[2*idx], a[2*idx+1]);
for (int i=0; i<N/2; i++) {
if (idx==i) continue;
Point2D other_pt(a[2*i], a[2*i+1]);
pol.cut(curPt, other_pt);
}
pol.dummy = pol.getpoint(0).x;
polygons[idx] = pol;
}
int main(int argc, unsigned char* argv[])
{
const int N = 100;
float a_h[N], *a_d;
Polygon p_h[N/2], *p_d;
size_t size = N * sizeof(float);
size_t size_pol = N/2 * sizeof(Polygon);
cudaMalloc((void **) &a_d, size);
cudaCheckErrors("cm1");
cudaMalloc((void **) &p_d, size_pol);
cudaCheckErrors("cm2");
for (int i=0; i<N; i++) a_h[i] = (float)(rand()%1000)*0.001;
cudaMemcpy(a_d, a_h, size, cudaMemcpyHostToDevice);
cudaCheckErrors("cmcp1");
int block_size = 128;
int n_blocks = N/block_size + (N%block_size == 0 ? 0:1);
cut_poly <<< n_blocks, block_size >>> (a_d, p_d, N);
cudaCheckErrors("kernel");
cudaMemcpy(a_h, a_d, sizeof(float)*N, cudaMemcpyDeviceToHost);
cudaCheckErrors("cmcp2");
cudaMemcpy(p_h, p_d, sizeof(Polygon)*N/2, cudaMemcpyDeviceToHost);
cudaCheckErrors("cmcp3");
for (int i=0; i<N/2; i++)
printf("%f \t %f \t %f \t %u\n", a_h[i], p_h[i].dummy, p_h[i].getpoint(0).x, p_h[i].nbpts);
cudaFree(a_d);
cudaFree(p_d);
return 0;
}
我建议使用针对这些内容发布新问题。