上一節(jié)我們寫了個(gè)一維向量相加的程序。這節(jié)我們來(lái)看一個(gè)4×4矩陣轉(zhuǎn)置程序。

4X4矩陣我們采用二維數(shù)組進(jìn)行存儲(chǔ)，在程序設(shè)計(jì)上，我們讓轉(zhuǎn)置過(guò)程分4次轉(zhuǎn)置完成，就是一次轉(zhuǎn)一行。注意這里的OpenCL的工作維數(shù)是二維。（當(dāng)然用一維的方式也可以，只是在CL代碼中要用到循環(huán)，效率不高）

程序分兩部份：

（1）transposition.cl代碼

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__kernel void transposition(__global int* A, ????????????????????__global int* B) { ????//獲取索引號(hào)，這里是二維的，所以可以取兩個(gè) ????//否則另一個(gè)永遠(yuǎn)是0 ????int col = get_global_id(0); ????int row = get_global_id(1); ????B[col*4+row] = A[row*4+col]; }

（2）main.cpp代碼

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#include <iostream> #include <stdio.h> #include <string.h> #include <string> #include <CL/cl.h>//包含CL的頭文件 using namespace std; //4x4數(shù)組 #define dim_x 4 #define dim_y 4 //從外部文件獲取cl內(nèi)核代碼 bool GetFileData(const char* fname,string& str) { ????FILE* fp = fopen(fname,"r"); ????if(fp==NULL) ????{ ????????printf("no found file\n"); ????????return false; ????} ????int n=0; ????while(feof(fp)==0) ????{ ????????str += fgetc(fp); ????} ????return true; } int main() { ????//先讀外部CL核心代碼，如果失敗則退出。 ????//代碼存buf_code里面 ????string code_file; ????if(false == GetFileData("transposition.cl",code_file)) ????????return 0; ????char* buf_code = new char[code_file.size()]; ????strcpy(buf_code,code_file.c_str()); ????buf_code[code_file.size()-1] = NULL; ????//聲明CL所需變量。 ????cl_device_id device; ????cl_platform_id platform_id = NULL; ????cl_context context; ????cl_command_queue cmdQueue; ????cl_mem bufferA,bufferB,bufferC; ????cl_program program; ????cl_kernel kernel = NULL; ????//我們使用的是二維向量 ????//設(shè)定向量大小（維數(shù)） ????size_t globalWorkSize[2]; ????globalWorkSize[0] = dim_x ; ????globalWorkSize[1] = dim_y; ????cl_int err; ????/* ????????定義輸入變量和輸出變量，并設(shè)定初值 ????*/ ????int buf_A[dim_x][dim_y]; ????int buf_B[dim_x][dim_y]; ????size_t datasize = sizeof(int) * dim_x * dim_y; ????int n=0; ????int m=0; ????for(n=0;n<dim_x;n++) ????{ ????????for(m=0;m<dim_y;m++) ????????{ ????????????buf_A[m][n] = m + n*dim_x; ????????} ????} ????//step 1:初始化OpenCL ????err = clGetPlatformIDs(1,&platform_id,NULL); ????if(err!=CL_SUCCESS) ????{ ????????cout<<"clGetPlatformIDs error"<<endl; ????????return 0; ????} ????//這次我們只用CPU來(lái)進(jìn)行并行運(yùn)算，當(dāng)然你也可以該成GPU ????clGetDeviceIDs(platform_id,CL_DEVICE_TYPE_GPU,1,&device,NULL); ????//step 2:創(chuàng)建上下文 ????context = clCreateContext(NULL,1,&device,NULL,NULL,NULL); ????//step 3:創(chuàng)建命令隊(duì)列 ????cmdQueue = clCreateCommandQueue(context,device,0,NULL); ????//step 4:創(chuàng)建數(shù)據(jù)緩沖區(qū) ????bufferA = clCreateBuffer(context, ?????????????????????????????CL_MEM_READ_ONLY, ?????????????????????????????datasize,NULL,NULL); ????bufferB = clCreateBuffer(context, ?????????????????????????????CL_MEM_WRITE_ONLY, ?????????????????????????????datasize,NULL,NULL); ????//step 5:將數(shù)據(jù)上傳到緩沖區(qū) ????clEnqueueWriteBuffer(cmdQueue, ?????????????????????????bufferA,CL_FALSE, ?????????????????????????0,datasize, ?????????????????????????buf_A,0, ?????????????????????????NULL,NULL); ????//step 6:加載編譯代碼,創(chuàng)建內(nèi)核調(diào)用函數(shù) ????program = clCreateProgramWithSource(context,1, ????????????????????????????????????????(const char**)&buf_code, ????????????????????????????????????????NULL,NULL); ????clBuildProgram(program,1,&device,NULL,NULL,NULL); ????kernel = clCreateKernel(program,"transposition",NULL); ????//step 7:設(shè)置參數(shù)，執(zhí)行內(nèi)核 ????clSetKernelArg(kernel,0,sizeof(cl_mem),&bufferA); ????clSetKernelArg(kernel,1,sizeof(cl_mem),&bufferB); ????//<span style="color: #ff0000;"><strong>注意這里第三個(gè)參數(shù)已經(jīng)改成2，表示二維數(shù)據(jù)。</strong></span> ????clEnqueueNDRangeKernel(cmdQueue,kernel, ???????????????????????????2,NULL, ???????????????????????????globalWorkSize, ???????????????????????????NULL,0,NULL,NULL); ????//step 8:取回計(jì)算結(jié)果 ????clEnqueueReadBuffer(cmdQueue,bufferB,CL_TRUE,0, ????????????????????????datasize,buf_B,0,NULL,NULL); ????//輸出計(jì)算結(jié)果 ????for(n=0;n<dim_x;n++) ????{ ????????for(m=0;m<dim_y;m++) ????????{ ????????????cout<< buf_A[m][n] <<","; ????????} ????????cout<<endl; ????} ????cout<<endl<<"====transposition===="<<endl<<endl; ????for(n=0;n<dim_x;n++) ????{ ????????for(m=0;m<dim_y;m++) ????????{ ????????????cout<< buf_B[m][n] <<","; ????????} ????????cout<<endl; ????} ????//釋放所有調(diào)用和內(nèi)存 ????clReleaseKernel(kernel); ????clReleaseProgram(program); ????clReleaseCommandQueue(cmdQueue); ????clReleaseMemObject(bufferA); ????clReleaseMemObject(bufferB); ????clReleaseContext(context); ????delete buf_code; ????return 0; }

運(yùn)算結(jié)果：

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總結(jié)

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