Question

我写了一个程序来添加两个2D数组来检查CPU和GPU的性能。我使用clock（）函数来测量CPU执行，使用cudaEvent来测量GPU中的内核执行时间。由于我在Udacity下学习CUDA，我试图在他们的服务器上执行该程序并找到结果，

 Output:
 GPU: 0.001984 ms
 CPU : 30.000000 ms

现在问我真正的问题，我发现这些结果在GPU上速度非常快，现在我对这些结果是否准确或我的程序中是否有任何错误持怀疑态度？

这是我的计划：

 #include "stdio.h"
 #include<time.h>
 #define COLUMNS 900
 #define ROWS 900
 long a[ROWS][COLUMNS], b[ROWS][COLUMNS], c[ROWS][COLUMNS],d[ROWS][COLUMNS];
__global__ void add(long *a, long *b, long *c,long *d)
{
 int x = blockIdx.x;
 int y = blockIdx.y;
 int i = (COLUMNS*y) + x;
 c[i] = a[i] + b[i];
 a[i]=d[i];
}

int main()
{
  long *dev_a, *dev_b, *dev_c,*dev_d;
  float ms;
  clock_t startc, end;
  double cpu_time_used;
  cudaEvent_t start,stop;


 cudaMalloc((void **) &dev_a, ROWS*COLUMNS*sizeof(int));
 cudaMalloc((void **) &dev_b, ROWS*COLUMNS*sizeof(int));
 cudaMalloc((void **) &dev_c, ROWS*COLUMNS*sizeof(int));
 cudaMalloc((void **) &dev_d, ROWS*COLUMNS*sizeof(int));

 startc = clock();
 for (long y = 0; y < ROWS; y++) // Fill Arrays
 for (long x = 0; x < COLUMNS; x++)
 {
     a[y][x] = x;
     b[y][x] = y;
     d[y][x]=rand()%4;
     c[y][x]=a[y][x]+b[y][x];
 }
 end = clock();

cpu_time_used = ((double) (end - startc)) / CLOCKS_PER_SEC;
cpu_time_used*=1000;


cudaMemcpy(dev_a, a, ROWS*COLUMNS*sizeof(int),
cudaMemcpyHostToDevice);
cudaMemcpy(dev_b, b, ROWS*COLUMNS*sizeof(int),
cudaMemcpyHostToDevice);
cudaMemcpy(dev_d, d, ROWS*COLUMNS*sizeof(int),
cudaMemcpyHostToDevice);


cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaEventRecord(start, 0); 
cudaEventRecord(stop, 0);


add<<<dim3(1024,1024),dim3(128,128)>>>(dev_a, dev_b, dev_c,dev_d);

cudaEventSynchronize(stop);
cudaEventElapsedTime(&ms, start, stop);
cudaMemcpy(c, dev_c, ROWS*COLUMNS*sizeof(int),cudaMemcpyDeviceToHost);
cudaEventDestroy(start);
cudaEventDestroy(stop);




printf("GPU: %f ms",ms);
printf("\n CPU : %f ms",cpu_time_used);

 return 0;
}

感谢大家提供给我的查询的答案，以下是我对代码所做的更改以及更新后的结果，

更新代码：

#include "stdio.h"
#include <time.h>
#include <sys/time.h>
#include <unistd.h>
#define COLUMNS 500
#define ROWS 500
long a[ROWS][COLUMNS], b[ROWS][COLUMNS], c[ROWS][COLUMNS],d[ROWS][COLUMNS];



__global__ void add(long *a, long *b, long *c,long *d)
{
 int x = blockIdx.x;
 int y = blockIdx.y;
 int i = (COLUMNS*y) + x;
 c[i] = a[i] + b[i];
 a[i]=d[i];
}
int main()
{
 long *dev_a, *dev_b, *dev_c,*dev_d;
 struct timeval startc, end;
 float ms;
 long mtime, seconds, useconds;
 //   clock_t startc, end;
 //  double cpu_time_used;
 long ns;
 cudaEvent_t start,stop;


 cudaMalloc((void **) &dev_a, ROWS*COLUMNS*sizeof(int));
 cudaMalloc((void **) &dev_b, ROWS*COLUMNS*sizeof(int));
 cudaMalloc((void **) &dev_c, ROWS*COLUMNS*sizeof(int));
 cudaMalloc((void **) &dev_d, ROWS*COLUMNS*sizeof(int));

 gettimeofday(&startc, NULL);
 for (long y = 0; y < ROWS; y++) // Fill Arrays
 for (long x = 0; x < COLUMNS; x++)
 {
  a[y][x] = x;
  b[y][x] = y;
  d[y][x]=rand()%4;
  c[y][x]=a[y][x]+b[y][x];
 }
  gettimeofday(&end, NULL);

 seconds  = end.tv_sec  - startc.tv_sec;
 useconds = end.tv_usec - startc.tv_usec;
 mtime = ((seconds) * 1000 + useconds/1000.0) + 0.5;


for (long y = ROWS-1; y < ROWS; y++) // Output Arrays
 {
 for (long x = COLUMNS-1; x < COLUMNS; x++)
 {
    // printf("\n[%ld][%ld]=%ld ",y,x,c[y][x]);
   //   printf("[%d][%d]=%d ",y,x,d[y][x]);
 }
 printf("\n");
 }



cudaMemcpy(dev_a, a, ROWS*COLUMNS*sizeof(int),
cudaMemcpyHostToDevice);
cudaMemcpy(dev_b, b, ROWS*COLUMNS*sizeof(int),
cudaMemcpyHostToDevice);
cudaMemcpy(dev_d, d, ROWS*COLUMNS*sizeof(int),
cudaMemcpyHostToDevice);


cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaEventRecord(start, 0); 



add<<<dim3(1024,1024),dim3(128,128)>>>(dev_a, dev_b, dev_c,dev_d);

cudaThreadSynchronize();
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
cudaEventElapsedTime(&ms, start, stop);

cudaMemcpy(c, dev_c, ROWS*COLUMNS*sizeof(int),cudaMemcpyDeviceToHost);
cudaEventDestroy(start);
cudaEventDestroy(stop);



//cpu_time_used = ((double) (end - start)) / CLOCKS_PER_SEC;
printf("GPU: %f ms",ms);
printf("\n CPU : %ld ms",mtime);
 for (long y = ROWS-1; y < ROWS; y++) // Output Arrays
 {
     for (long x = COLUMNS-1; x < COLUMNS; x++)
     {
      //   printf("\n[%ld][%ld]=%ld ",y,x,c[y][x]);
      //   printf("[%d][%d]=%d ",y,x,d[y][x]);
     }
     printf("\n");
 }
 return 0;
}

输出：

GPU: 0.011040 ms
CPU : 9 ms

现在我可以安全地判断它是否正确吗？

Answer 1

你认为加速太快是正确的，CPU的时间太长。使用此方法为CPU C++ obtaining milliseconds time on Linux -- clock() doesn't seem to work properly计时也可能需要将cudaEventRecord(stop, 0);移到内核之后。

我在你的内核中看到了5个读写。以5*4Bytes*500*500/(1024^3*0.009)为例，你的记忆力大约为0.517 GB/s，这只是可用的一小部分。我会说你的CPU版本需要一些工作。相比之下，您的GPU位于5*4Bytes*500*500/(1024^3*0.01104e-3)大约是421GB/s。我会说你不在那里。

所以，这么多错误......

#include "stdio.h"
#include <time.h>
#include <sys/time.h>
#include <unistd.h>
#include <cuda.h>
#include <cuda_runtime.h>

#define COLUMNS 500
#define ROWS 500
long a[ROWS*COLUMNS], b[ROWS*COLUMNS], c[ROWS*COLUMNS],d[ROWS*COLUMNS];



__global__ void add(long *a, long *b, long *c,long *d)
{
 int x = blockIdx.x;
 int y = blockIdx.y;
 int i = (COLUMNS*y) + x;
 c[i] = a[i] + b[i];
 a[i]=d[i];
}
int main()
{
 long *dev_a, *dev_b, *dev_c,*dev_d;
 struct timeval startc, end;
 float ms;
 long seconds, useconds;
 double mtime;
 cudaEvent_t start,stop;


 for(int i=0; i<ROWS*COLUMNS; i++)
     d[i]=rand()%4;

 for(int i=0; i<ROWS; i++){
     for(int j=0; j<COLUMNS; j++){
         a[i*COLUMNS+j]=j;
         b[i*COLUMNS+j]=i;
     }
 }

 cudaMalloc((void **) &dev_a, ROWS*COLUMNS*sizeof(int));
 cudaMalloc((void **) &dev_b, ROWS*COLUMNS*sizeof(int));
 cudaMalloc((void **) &dev_c, ROWS*COLUMNS*sizeof(int));
 cudaMalloc((void **) &dev_d, ROWS*COLUMNS*sizeof(int));



 gettimeofday(&startc, NULL);
 for (long i = 0; i < ROWS*COLUMNS; i++){ // Fill Arrays
     c[i]=a[i]+b[i];
     a[i]=d[i];
 }
  gettimeofday(&end, NULL);

 seconds  = end.tv_sec  - startc.tv_sec;
 useconds = end.tv_usec - startc.tv_usec;
 mtime = useconds;
 mtime/=1000;
 mtime+=seconds*1000;

for (long y = ROWS-1; y < ROWS; y++) // Output Arrays
 {
 for (long x = COLUMNS-1; x < COLUMNS; x++)
 {
    // printf("\n[%ld][%ld]=%ld ",y,x,c[y][x]);
   //   printf("[%d][%d]=%d ",y,x,d[y][x]);
 }
 printf("\n");
 }



cudaMemcpy(dev_a, a, ROWS*COLUMNS*sizeof(int),
cudaMemcpyHostToDevice);
cudaMemcpy(dev_b, b, ROWS*COLUMNS*sizeof(int),
cudaMemcpyHostToDevice);
cudaMemcpy(dev_d, d, ROWS*COLUMNS*sizeof(int),
cudaMemcpyHostToDevice);


cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaEventRecord(start, 0);



add<<<dim3(1024,1024),dim3(128,128)>>>(dev_a, dev_b, dev_c,dev_d);



cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
cudaEventElapsedTime(&ms, start, stop);

cudaMemcpy(c, dev_c, ROWS*COLUMNS*sizeof(int),cudaMemcpyDeviceToHost);
cudaEventDestroy(start);
cudaEventDestroy(stop);

printf("GPUassert: %s\n", cudaGetErrorString(cudaGetLastError()));

//cpu_time_used = ((double) (end - start)) / CLOCKS_PER_SEC;
double memXFers=5*4*COLUMNS*ROWS;
memXFers/=1024*1024*1024;


printf("GPU: %f ms bandwidth %g GB/s",ms, memXFers/(ms/1000.0));
printf("\n CPU : %g ms bandwidth %g GB/s",mtime, memXFers/(mtime/1000.0));
 for (long y = ROWS-1; y < ROWS; y++) // Output Arrays
 {
     for (long x = COLUMNS-1; x < COLUMNS; x++)
     {
      //   printf("\n[%ld][%ld]=%ld ",y,x,c[y][x]);
      //   printf("[%d][%d]=%d ",y,x,d[y][x]);
     }
     printf("\n");
 }

 return 0;
}

我当前的结果（显然不正确）......

GPU: 0.001792 ms bandwidth 2598.56 GB/s
CPU : 0.567 ms bandwidth 8.21272 GB/s

CUDA C - 使用clock（）和cudaEvent的CPU和GPU执行时间，是否正确？

1 个答案: