使用pthread实现多线程CPU+GPU计算

本次测试目的仅仅为了尝试能否使用pthread实现多线程对CPU和GPU分别进行控制:

  文件结构如下:

  main.cc:控制线程的主函数

  vector.cu:实现对核函数进行调用的任务函数

  vector.h:任务函数声明

  vector_kernel.cu:核函数

  vector_kernel.h:核函数声明

  生成文件如下:

  vector:最终生成的可执行文件

  main.o:生成的函数体obj

  vector.o:生成的任务函数obj

  vector_kernel.o:生成的核函数obj

  功能描述:

  线程1:执行GPU代码(向量加法)

  线程2:执行CPU代码(循环打印数据)

  线程3:执行CPU代码(循环打印数据)

  线程4:执行CPU代码(循环打印数据)

  文件代码:

  main.cc:

#include  < stdio.h >
#include 
< pthread.h >
#include 
< sys / time.h >
#include 
< stdlib.h >
#include 
< string .h >
#include 
" vector.h "


pthread_t thread[
4 ];


void   * thread1( void   * )
{
// do GPU task
float  time, start;
start 
=  clock();
printf(
" thread1:I'm thread1\n " );
int  i, n  =   100 ;
float   * a,  * b,  * c;
=  ( float   * )malloc(n  *   sizeof ( float ));
=  ( float   * )malloc(n  *   sizeof ( float ));
=  ( float   * )malloc(n  *   sizeof ( float ));
for (i  =   0 ; i  <  n; i ++ )
{
a[i] 
=   1.0f ;
b[i] 
=   1.0f ;
}
for (i  =   0 ; i  <   100 ; i ++ )
{
vectorAdd(a, b, c, n);
}
printf(
" thread1:c[%d] = %f\n " 0 , c[ 0 ]);
free(a);
free(b);
free(c);
time 
=  clock()  -  start;
printf(
" thread1: task was finished!\ncostTime1 : %f\n " , time  /  CLOCKS_PER_SEC);
pthread_exit(NULL);
}


void   * thread2( void   * )
{
// do CPU task
float  time, start;
start 
=  clock();
printf(
" thread2:I'm thread2\n " );
int  i, j, k  =   1
for (i  =   0 ; i  <   1000 ; i ++ )
{
for (j  =   0 ; j  <   1000 ; j ++ )
{
printf(
" thread2:k = %d\n " , k);
k
++ ;
}
}
time  
=  clock()  -  start;
printf(
" thread2: task was finished!\ncostTime2 : %f\n " , time  /  CLOCKS_PER_SEC);
pthread_exit(NULL);
}


void   * thread3( void   * )
{
// do CPU task
float  time, start;
start 
=  clock();
printf(
" thread3:I'm thread3\n " );
int  i, j, k  =   1
for (i  =   0 ; i  <   1000 ; i ++ )
{
for (j  =   0 ; j  <   1000 ; j ++ )
{
printf(
" thread3:k = %d\n " , k);
k
++ ;
}
}
time  
=  clock()  -  start;
printf(
" thread3: task was finished!\ncostTime3 : %f\n " , time  /  CLOCKS_PER_SEC);
pthread_exit(NULL);
}


void   * thread4( void   * )
{
// do CPU task
float  time, start;
start 
=  clock();
printf(
" thread4:I'm thread4\n " );
int  i, j, k  =   1
for (i  =   0 ; i  <   1000 ; i ++ )
{
for (j  =   0 ; j  <   1000 ; j ++ )
{
printf(
" thread4:k = %d\n " , k);
k
++ ;
}
}
time  
=  clock()  -  start;
printf(
" thread4: task was finished!\ncostTime4 : %f\n " , time  /  CLOCKS_PER_SEC);
pthread_exit(NULL);
}


void  thread_create()
{
int  temp;
memset(
& thread,  0 sizeof (thread));
if ((temp  =  pthread_create( & thread[ 0 ], NULL, thread1, NULL))  !=   0 )
printf(
" 线程1创建失败!\n " );
else
printf(
" 线程1被创建!\n " );
if ((temp  =  pthread_create( & thread[ 1 ], NULL, thread2, NULL))  !=   0 )
printf(
" 线程2创建失败!\n " );
else
printf(
" 线程2被创建!\n " );
if ((temp  =  pthread_create( & thread[ 2 ], NULL, thread3, NULL))  !=   0 )
printf(
" 线程3创建失败!\n " );
else
printf(
" 线程3被创建!\n " );
if ((temp  =  pthread_create( & thread[ 3 ], NULL, thread4, NULL))  !=   0 )
printf(
" 线程4创建失败!\n " );
else
printf(
" 线程4被创建!\n " );
}
void  thread_wait()
{
if (thread[ 0 !=   0 )
{
pthread_join(thread[
0 ], NULL);
printf(
" 线程1已经结束\n " );
}
if (thread[ 1 !=   0 )
{
pthread_join(thread[
1 ], NULL);
printf(
" 线程2已经结束\n " );
}
if (thread[ 2 !=   0 )
{
pthread_join(thread[
2 ], NULL);
printf(
" 线程3已经结束\n " );
}
if (thread[ 3 !=   0 )
{
pthread_join(thread[
3 ], NULL);
printf(
" 线程4已经结束\n " );
}
}


int  main()
{
float  time, start;
printf(
" 我是主函数,正在创建线程\n " );
start 
=  clock();
thread_create();
printf(
" 我是主函数,正在等待线程完成任务\n " );
thread_wait();
time 
=  clock()  -  start;
printf(
" costTime0 : %f\n " , time  /  CLOCKS_PER_SEC);
return   0 ;
}

  vector.cu:

#include  < stdio.h >
#include 
< stdlib.h >
#include 
< cuda.h >
#include 
" vector.h "
#include 
" vector_kernel.h "


void  vectorAdd( float   * a,  float   * b,  float   * c,  int  n)
{
float   * d_a,  * d_b,  * d_c;
cudaMalloc((
void   ** ) & d_a, n  *   sizeof ( float ));
cudaMemcpy(d_a, a, n 
*   sizeof ( float ), cudaMemcpyHostToDevice);
cudaMalloc((
void   ** ) & d_b, n  *   sizeof ( float ));
cudaMemcpy(d_b, b, n 
*   sizeof ( float ), cudaMemcpyHostToDevice);
cudaMalloc((
void   ** ) & d_c, n  *   sizeof ( float ));
cudaMemcpy(d_b, b, n 
*   sizeof ( float ), cudaMemcpyHostToDevice);
vectorAddKernel
<<< 1 , n >>> (d_a, d_b, d_c, n);
cudaMemcpy(c, d_c, n 
*   sizeof ( float ), cudaMemcpyDeviceToHost);
cudaFree(d_a);
cudaFree(d_b);
cudaFree(d_c);
}

  vector.h:

#include  < stdio.h >
#include 
< stdlib.h >

void  vectorAdd( float   * a,  float   * b,  float   * c,  int  n);

  vector_kernel.cu:

#include  < stdio.h >
#include 
< stdlib.h >
#include 
< cuda.h >
#include 
" vector_kernel.h "


__global__ 
void  vectorAddKernel( float   * a,  float   * b,  float   * c,  int  n)
{
int  tid;
tid 
=  threadIdx.x;
if (tid  <  n)
c[tid] 
=  a[tid]  +  b[tid];
}

  vector_kernel.h:

#include  < stdio.h >
#include 
< cuda.h >
#include 
< stdlib.h >


__global__ 
void  vectorAddKernel( float   * a,  float   * b,  float   * c,  int  n);

  而最主要的是makefile的内容,由于nvcc会将.cu生成obj默认为c++方式,所以,需要将主函数改为.cc文件,不然将会编译出错!

  makefile:

vector : main.o vector.o vector_kernel.o
nvcc 
- o vector main.o vector.o vector_kernel.o


vector_kernel.o : vector_kernel.cu vector_kernel.h
nvcc 
- c vector_kernel.cu


vector.o : vector.cu vector.h vector_kernel.h
nvcc 
- c vector.cu


main.o : main.cc vector.h
cc 
- lpthread  - c main.cc

  通过测试时间可以知道几个线程之间的执行关系,大家自己测试啦~

  以下是一个运行结果:

使用pthread实现多线程CPU+GPU计算_第1张图片

        更多内容请点击:

        CUDA专区:http://cuda.it168.com/

        CUDA论坛:http://cudabbs.it168.com/


原文链接:地址

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