我对 OpenCL 很陌生，但我已经做了一年多的并行编程了。我正在制作我的第一个 openCL 代码（矩阵乘法）。我写了以下代码，
//#include<stdio.h>
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <string.h>
#include <SDKCommon.hpp>
#include <SDKApplication.hpp>
#include <SDKCommandArgs.hpp>
#include <SDKFile.hpp>
#include <CL/cl.h>
#define MAX_SOURCE_SIZE (0x100000)
#define MATSIZE 16
void initmat(float *Aa,float *Bb,float *Cc,int row,int colrow,int col);
void initmat(float *Aa,float *Bb,float *Cc,int row,int colrow,int col)
{
unsigned int i;
for(i=0;i<row*colrow;i++){
Aa[i]=1;
}
for(i=0;i<colrow*col;i++){
Bb[i]=2;
}
for(i=0;i<row*col;i++){
Cc[i]=0;
}
}
int main(void)
{
printf(" Here: 1 \n");
// Load the kernel source code into the array source_str
FILE *fp;
char *source_str;
size_t source_size;
fp = fopen("matmul.cl", "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 );
printf(" Here: 2 \n");
// matrix declaration
float *A;
float *B;
float *C;
// set dimesions
int Arow,AcolBrow,Bcol;
Arow=AcolBrow=Bcol=MATSIZE;
// no. of elements in matrix
int sizea, sizeb, sizec;
// Error code from opencl
int err;
// Setting up matrices
sizea= Arow*AcolBrow;
sizeb= AcolBrow*Bcol;
sizec= Arow*Bcol;
A = (float *) malloc(sizeof(float)*sizea);
B = (float *) malloc(sizeof(float)*sizeb);
C = (float *) malloc(sizeof(float)*sizec);
printf(" Here: 3 \n");
initmat(A,B,C,Arow,AcolBrow,Bcol);
// Displaying inputs
unsigned long int ii;
printf("Input A: \n");
for(ii=0;ii<sizea;ii++)
printf("%f ",A[ii]);
printf("\n \n \n \n");
printf("Input B: \n");
for(ii=0;ii<sizeb;ii++)
printf("%f ",B[ii]);
printf("\n \n \n");
// get platform id & device id
cl_uint numplatform;
cl_platform_id platformid=NULL;
cl_device_id deviceid=NULL;
err= clGetPlatformIDs(1,&platformid,&numplatform);
err=clGetDeviceIDs(platformid,CL_DEVICE_TYPE_GPU,1,&deviceid,NULL);
cl_context_properties properties[]=
{
CL_CONTEXT_PLATFORM, (cl_context_properties)platformid,0
};
// create context
cl_context context= clCreateContext(properties,1,&deviceid,NULL,NULL,&err);
/* when more than one gpu is installed on the system than we make use of the approach as we stated in the notes !! */
printf(" Here: 4 \n");
// create command queue
cl_command_queue queue = clCreateCommandQueue(context,deviceid,0,&err); // I have disabled profiling option
// Allocate buffer object for Ad,Bd,Cd
cl_mem Ad = clCreateBuffer(context,CL_MEM_READ_ONLY,sizeof(cl_float)*sizea,NULL,NULL);
cl_mem Bd = clCreateBuffer(context,CL_MEM_READ_ONLY,sizeof(cl_float)*sizeb,NULL,NULL);
cl_mem Cd = clCreateBuffer(context,CL_MEM_WRITE_ONLY,sizeof(cl_float)*sizec,NULL,NULL);
printf(" Here: 5 \n");
// We are not explicitely making kernel. We are putting the kernel code here itself (see notes)
cl_program program= clCreateProgramWithSource(context,1,(const char **)&source_str, (const size_t *)&source_size,&err);
printf(" Here: 6 \n");
// Build program using program object just created
err = clBuildProgram(program,0,NULL,NULL,NULL,NULL);
if(err !=CL_SUCCESS)
{
size_t len;
char buffer[2048];
printf("ERROR: Failed to build executable \n ");
clGetProgramBuildInfo(program,deviceid,CL_PROGRAM_BUILD_LOG, sizeof(buffer), buffer , &len);
printf("%s \n",buffer);
//return FAILURE;
}
printf(" Here: 7 \n");
// Create kernel object
cl_kernel kernel = clCreateKernel(program,"matmul",NULL);
printf(" Here: 8 \n");
// set kernel argument values
err=0;
err= clSetKernelArg(kernel,0,sizeof(int),&Arow);
err|= clSetKernelArg(kernel,1,sizeof(int),&AcolBrow);
err|= clSetKernelArg(kernel,2,sizeof(int),&Bcol);
err|= clSetKernelArg(kernel,3,sizeof(cl_mem),&Ad);
err|= clSetKernelArg(kernel,4,sizeof(cl_mem),&Bd);
err|= clSetKernelArg(kernel,5,sizeof(cl_mem),&Cd);
printf(" Here: 9 \n");
// Write to device buffers. Ad=A and Bd=B : Equivalent to CUDAmemcpy
err=clEnqueueWriteBuffer(queue,Ad,CL_TRUE,0,sizeof(cl_float)*sizea,A,0,NULL,NULL);
err=clEnqueueWriteBuffer(queue,Bd,CL_TRUE,0,sizeof(cl_float)*sizeb,B,0,NULL,NULL);
printf(" Here: 10 \n");
// since we have set the copy as synchronous we will be creating event
cl_event event;
// Execute the kernel over entire range of C matrix
size_t global[2];
size_t local[2];
cl_uint * ndim; // no. of dimension in ND range. 3rd parameter in kernel call signifies the dimension.
global[0]=(size_t)Arow;
global[1]=(size_t)Bcol;
* ndim=2; // because we want 2-D multiplication. Gives n
/* no local size declaration cause we are not making work groups ie blocks.We are just make making oneblock where everythread takes one element of A,B and computes C */
printf(" Here: 11 \n");
err = clEnqueueNDRangeKernel(queue,kernel,*ndim,NULL,global,NULL,0,NULL,&event); // the NULL position after global is for passing local dimension. In this case we don't have one.
clFinish(queue); // wait for kernel to finish before we begin copying the result back on host
printf(" Here: 12 \n");
//read back the result
err=clEnqueueReadBuffer(queue,Cd,CL_TRUE,0,sizeof(cl_float)*sizec,C,0,NULL,NULL);
// Checking whether the computations done are on CPU or GPU
cl_device_type dev_type;
clGetDeviceInfo(deviceid, CL_DEVICE_TYPE, sizeof(dev_type), &dev_type, NULL);
if (dev_type == CL_DEVICE_TYPE_GPU) {
printf("Following code was complied on GPU ! \n \n \n \n");
}
else
printf("Following code was complied on CPU ! \n \n \n \n");
printf(" Here: 13 \n \n \n");
// Displaying results
printf("Result is: \n");
for(ii=0;ii<sizec;ii++)
printf("%f ",C[ii]);
printf("\n \n \n");
// free all memory
printf(" Here: 14 \n");
clReleaseProgram(program);
clReleaseKernel(kernel);
clReleaseMemObject(Ad);
clReleaseMemObject(Bd);
clReleaseMemObject(Cd);
clReleaseCommandQueue(queue);
clReleaseContext(context);
printf(" Here: 15 \n");
return 0;
printf(" Here: 16 \n");
}
我的内核代码如下，
__kernel void matmul(const int Mdim, const int Ndim,const int Pdim,__global float* A,__global float* B,__global float* C)
{
float tmp;
int i = get_global_id(0);
int j = get_global_id(1);
if((i<Ndim)&(j<Mdim))
C[i*Ndim+j] = 3;
}
我可以成功编译它，它在 ../bin/x86/ 文件夹中创建了一个二进制文件。当我尝试使用 ./matmul 运行它时，它会引发以下错误，
93 > Sun Mar 17 : 04:22 PM : samkit@samkit:~/AMD/AMD-APP-SDK-v2.8-RC-lnx32/samples/opencl/bin/x86$ ./matmul 分段错误（核心转储）
这是我的示例输出：
135 > Sun Mar 17 : 07:49 PM : samkit@samkit:~/AMD/AMD-APP-SDK-v2.8-RC-lnx32/samples/opencl/bin/x86$ ./matmul
Here: 1
Here: 2
Here: 3
Here: 4
Here: 5
Here: 6
Here: 7
Here: 8
Here: 9
Here: 10
Here: 11
Here: 12
Following code was complied on GPU !
Here: 13
Result is:
Segmentation fault (core dumped)
我知道内核的功能与矩阵乘法内核的功能相差甚远，但我这样做只是为了检查这些是否是我的内核中的一些错误。请提供可以帮助我运行代码的建议或建议。
提前致谢。