过去几天我一直在调试，无法运行这个OpenCL矩阵乘法内核。每当我运行程序时，GPU的输出会产生类似于-198746573.0000的大负数。我想知道具有HPC经验的人是否可以在我的代码中指出错误或者是否是驱动程序错误。
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <string.h>
#define widthA 2
#define heightA 2
#define widthB heightA
#define heightB 2
#define widthC widthA
#define heightC heightB
#ifdef __APPLE__
#include < OpenCL/opencl.h >
#else
#include <opencl.h>
#endif
#define MEM_SIZE (128)
#define MAX_SOURCE_SIZE (0x100000)
int main()
{
float * A = (float *)malloc(sizeof(float)*widthA*heightA);
float * B = (float *)malloc(sizeof(float)*widthB*heightB);
float * C = (float *)malloc(sizeof(float)*widthC*heightC);
float * Res = (float *)malloc(sizeof(float)*widthC*heightC);
float * D= (float *)malloc(sizeof(float)*widthC*heightC);
float ref[widthC][heightC];
int i, j, k;
FILE * fp1 = fopen("matAdata.txt", "w");
if (!fp1) {
fprintf(stderr, "Failed to open matAdata.\n");
exit(1);
}
for(i = 0;i < widthA; i++)
{
for(j=0;j < heightA; j++) {
float p=(rand()%100)/7.0;
//*(A+i*heightA+j)=rand()%100 + p;
*(A+i*heightA+j)=4.0;
fprintf(fp1, "%f ",*(A+i*heightA+j));
}
fprintf(fp1, "\n");
}
fclose(fp1);
fp1 = fopen("matBdata.txt", "w");
if (!fp1) {
fprintf(stderr, "Failed to open matAdata.\n");
exit(1);
}
for(i = 0;i < widthB; i++)
{
for(j=0; j < heightB; j++) {
float p=(rand()%100)/7.0;
//*((B+i*heightB+j))=rand()%100 + p;
*((B+i*heightB+j))=4.0;
fprintf(fp1, "%f ",*(B+i*heightA+j));
}
fprintf(fp1, "\n");
}
fclose(fp1);
cl_device_id device_id = NULL;
cl_context context = NULL;
cl_command_queue command_queue = NULL;
cl_mem memobjA = NULL;
cl_mem memobjB = NULL;
cl_mem memobjC = NULL;
cl_mem rowA = NULL;
cl_mem colC = NULL;
cl_program program = NULL;
cl_kernel kernel = NULL;
cl_platform_id platform_id[10];
cl_platform_id platform = NULL;
cl_uint ret_num_devices;
cl_uint ret_num_platforms;
cl_int ret;
cl_event GPUDone[0];
//char string[MEM_SIZE];
FILE *fp;
char fileName[] = "matrixMultiplication.cl";
char *source_str;
size_t source_size;
int row = widthA;
int col = heightC;
/* Load the source code containing the kernel*/
fp = fopen(fileName, "r");
if (!fp) {
fprintf(stderr, "Failed to load kernel.\n");
exit(1);
}
source_str = (char*)malloc(MAX_SOURCE_SIZE);
source_size = fread( source_str, 1, MAX_SOURCE_SIZE, fp);
fclose( fp );
/* Get Platform and Device Info */
ret = clGetPlatformIDs(10, platform_id, &ret_num_platforms);
char cBuffer[1024];
cl_uint c;
for(c = 0; c < ret_num_platforms; c++)
{
clGetPlatformInfo(platform_id[c], CL_PLATFORM_NAME, 1024, &cBuffer, NULL);
if (strstr(cBuffer, "NVIDIA") != NULL)
{
platform = platform_id[c];
break;
}
}
printf("Found Platform %s\n", cBuffer);
ret = clGetDeviceIDs( platform, CL_DEVICE_TYPE_GPU, 1, &device_id, &ret_num_devices);
printf("Found %d devices.\n", ret_num_devices);
/* Create OpenCL context */
context = clCreateContext( NULL, 1, &device_id, NULL, NULL, &ret);
/* Create Command Queue */
command_queue = clCreateCommandQueue(context, device_id, 0, &ret);
/* Create Memory Buffer */
memobjA = clCreateBuffer(context, CL_MEM_READ_ONLY, widthA * heightA * sizeof(float), NULL, &ret);
memobjB = clCreateBuffer(context, CL_MEM_READ_ONLY, widthB * heightB * sizeof(float), NULL, &ret);
memobjC = clCreateBuffer(context, CL_MEM_READ_WRITE, widthC * heightC * sizeof(float), NULL, &ret);
rowA = clCreateBuffer(context, CL_MEM_READ_ONLY, sizeof(int), NULL, &ret);
colC = clCreateBuffer(context, CL_MEM_READ_ONLY, sizeof(int), NULL, &ret);
// Copy the lists A and B to their respective memory buffers
ret = clEnqueueWriteBuffer(command_queue,memobjA, CL_TRUE, 0,
widthA * heightA * sizeof(float), A, 0, NULL, NULL);
ret = clEnqueueWriteBuffer(command_queue, memobjB, CL_TRUE, 0,
widthB * heightB * sizeof(float), B, 0, NULL, NULL);
ret = clEnqueueWriteBuffer(command_queue, rowA, CL_TRUE, 0, sizeof(int), &row, 0, NULL, NULL);
ret = clEnqueueWriteBuffer(command_queue, colC, CL_TRUE, 0, sizeof(int), &col, 0, NULL, NULL);
/* Create Kernel Program from the source */
program = clCreateProgramWithSource(context, 1, (const char **)&source_str,
(const size_t *)&source_size, &ret);
/* Build Kernel Program */
ret = clBuildProgram(program, 1, &device_id, NULL, NULL, NULL);
/* Create OpenCL Kernel */
kernel = clCreateKernel(program, "matrixMultiplication", &ret);
/* Set OpenCL Kernel Arguments */
ret = clSetKernelArg(kernel, 0, sizeof(cl_mem), (void *)&memobjA);
ret = clSetKernelArg(kernel, 1, sizeof(cl_mem), (void *)&memobjB);
ret = clSetKernelArg(kernel, 2, sizeof(cl_mem), (void *)&memobjC);
ret = clSetKernelArg(kernel, 3, sizeof(int), (void *)&row);
ret = clSetKernelArg(kernel, 4, sizeof(int), (void *)&col);
/* Execute OpenCL Kernel */
//ret = clEnqueueTask(command_queue, kernel, 0, NULL,NULL);
size_t globalThreads[2] = {widthA, heightB};
size_t localThreads[2] = {16,16};
clEnqueueNDRangeKernel(command_queue, kernel, 2, NULL, globalThreads, localThreads, 0, NULL, NULL);
//clFlush(command_queue);
//clFinish(command_queue);
/* Copy results from the memory buffer */
ret = clEnqueueReadBuffer(command_queue, memobjC, CL_TRUE, 0,
widthA * heightC * sizeof(float), Res, 0, NULL, &GPUDone[0]);
printf("Buffer Read ended with %d.\n", ret);
clWaitForEvents(1, GPUDone);
fp1 = fopen("matGPURes.txt", "w");
if (!fp1) {
fprintf(stderr, "Failed to open matAdata.\n");
exit(1);
}
printf("\nResult\n");
for(i = 0;i < widthA; i++)
{
for(j=0;j < heightC; j++)
{
fprintf(fp1, "%f ",*(Res+i*heightC+j));
ref[i][j] = *(Res+i*heightC+j);
printf("GPU Output: %f\n", *(Res+i*heightC+j));
}
fprintf(fp1, "\n");
}
fclose(fp1);
ret = clFlush(command_queue);
ret = clFinish(command_queue);
ret = clReleaseKernel(kernel);
ret = clReleaseProgram(program);
ret = clReleaseMemObject(memobjA);
ret = clReleaseMemObject(memobjB);
ret = clReleaseMemObject(memobjC);
ret = clReleaseCommandQueue(command_queue);
ret = clReleaseContext(context);
ret = clReleaseEvent(GPUDone[0]);
free(source_str);
float sum=0.0;
for(i = 0;i < widthA; i++)
{
for(j = 0; j < heightC; j++)
{
sum = 0;
for(k = 0; k < widthB; k++)
{
sum += A[i*col+k] * B[k*row+j];
printf("Multiplying A: %f, B: %f\n", A[i*col+k], B[k*row+j]);
}
D[i*heightC+j] = sum;
}
}
fp1 = fopen("matNormalMultiplicationRes.txt", "w");
if (!fp1) {
fprintf(stderr, "Failed to open matNormalMultiplicationRes.txt\n");
exit(1);
}
for(i = 0; i<widthA; i++)
{
for(j = 0; j<heightA; j++)
{
if (ref[i][j] != D[i*heightA+j])
{
printf("Calculation error[ CPU: %f, GPU: %f ]\n", D[i*heightA+j], ref[i][j]);
}
}
}
printf("\nResult\n");
for(i = 0;i < widthA; i++)
{
for(j=0;j < heightC; j++)
{
fprintf(fp1, "%f ",*(D+i*heightC+j));
}
fprintf(fp1, "\n");
}
free(A);
free(B);
free(C);
free(D);
free(Res);
return 0;
}
这是内核
#define BLOCK_SIZE 16
__kernel
void matrixMultiplication(__global float* A, __global float* B, __global float* C, int wA, int wB )
{
//int i = get_global_id(0);
//int j = get_global_id(1);
float Csub = 0.0f;
int bx = get_group_id(0);
int by = get_group_id(1);
int tx = get_local_id(0);
int ty = get_local_id(1);
int aBegin = wA * BLOCK_SIZE * by;
int aEnd = aBegin + wA - 1;
int aStep = BLOCK_SIZE;
int bBegin = BLOCK_SIZE * bx;
int bStep = BLOCK_SIZE * wB;
for (int a = aBegin, b=bBegin;
a <= aEnd;
a += aStep, b+=bStep)
{
__local float As[BLOCK_SIZE][BLOCK_SIZE];
__local float Bs[BLOCK_SIZE][BLOCK_SIZE];
As[ty][tx] = A[a + wA * ty + tx];
Bs[ty][tx] = B[b + wB * ty + tx];
barrier(CLK_LOCAL_MEM_FENCE);
for( int k = 0; k < BLOCK_SIZE; ++k)
Csub += As[ty][k] * Bs[k][tx];
barrier(CLK_LOCAL_MEM_FENCE);
}
int c = wB * BLOCK_SIZE * by + BLOCK_SIZE * bx;
C[c + wB * ty + tx] = Csub;
/*
float value=0;
for ( int k = 0; k < widthA; k++)
{
value = value + A[k + j * widthA] * B[k*widthB + i];
}
C[i + widthA * j] = value;
*/
}
我一遍又一遍地检查，但根本找不到任何错误。在结束驱动程序问题之前，我想确保它不是代码错误。
谢谢！
答案 0 :(得分：3)
你真的需要这样的复杂内核吗？如果你真的想做简单的矩阵乘法 你可以写一个像这样的简单内核，这很容易调试。
__kernel void matrixMultiplication (__global float* A,
__global float* B,
__global float* C,
int widthA, int widthB )
{
//y direction
int row = get_global_id(1);
int col = get_global_id(0);
float cSum = 0.0f;
//calculate the result
for (int i=0; i<widthA; i++)
{
cSum += A[row*widthA+ i] * B[i*widthB+col];
}
C[row*widthB+col] = cSum;
}
答案 1 :(得分：1)
案件可能已经关闭，但为了谷歌的目的： 不应该在主机上显式声明共享内存并作为内核参数传递给源？在这种情况下，__ local关键字不是您要查找的关键字。
有关详细说明，请参阅How to declare local memory in OpenCL?上的帖子。
答案 2 :(得分：0)
检查主机的功能。这里有一些让你入门的事情......
1）您不需要创建缓冲区并将其排入标量常量Int，如row和col。只需将其设置为内核arg。
2）使用事件等待clEnqueueNDRangeKernel。你想确保计算已经完成。
3）在内核中添加printf语句以打印所选值，以查看输入和输出值是否符合预期。
尝试
if（get_local_id（0）％8 == 0） {
printf some useful value of a,b,c
}
3）尝试使用哑内核的主机代码，该内核将输入数组复制到输出数组。这将证实你已经处理缓冲区创建和enqeue读/写代码正确！