cpu和gpu(CUDA)的性能比较以及优化使用的探讨

各位大侠，我的笔记本电脑显卡是独立GF8400MG,128MB的显存。我做了试验，GPU计算块中最大线程数为512，支持的最大块数为21056，我用GPU计算512*21056个数加1，得到的运行时间为0.375秒，其中包括前100个数的打印。而我用CPU for循环对512*21066个数逐个加1，得到的运行时间为0.171秒，而且我做了很多试验，GPU运行的时间都比CPU逐个计算的运行时间多出了1倍，我就纳闷了，是这个技术提倡不成功，还是其他原因，请各位大侠探讨一下。。。

能把两个代码写出来吗？

下面这个是用GPU进行的并行运算的代码：（运行时间为：0.375秒）
#include <stdio.h>
#include <assert.h>
#include <time.h>

// Simple utility function to check for CUDA runtime errors
void checkCUDAError(const char* msg);

// Part 2 of 2: implement the kernel
__global__ void reverseArrayBlock( int*d_a)
{  int dx=blockDim.x*blockIdx.x+threadIdx.x;
  
  d_a[dx]=d_a[dx]+1;

}

////////////////////////////////////////////////////////////////////////////////
// Program main
////////////////////////////////////////////////////////////////////////////////
int main( int argc, char** argv)
{
clock_t start, finish;
double duration;
start = clock();
    // pointer for host memory and size
    int *h_a,transfer;
    int dimA = 512*21056; // 256K elements (1MB total)

    // pointer for device memory
    int *d_a;

    // define grid and block size
    int numThreadsPerBlock =512;

    // Part 1 of 2: compute number of blocks needed based on array size and desired block size
    int numBlocks = dimA/numThreadsPerBlock;  
    printf("%d\n",numBlocks);

    // allocate host and device memory
    size_t memSize = numBlocks * numThreadsPerBlock * sizeof(int);
    h_a = (int *) malloc(memSize);
    cudaMalloc( (void **) &d_a, memSize );

    // Initialize input array on host
    for (int i = 0; i < dimA; ++i)
    {
        h_a = i;
        //printf("%d ",h_a);
    }
  printf("\n");
   
    // Copy host array to device array
    cudaMemcpy( d_a, h_a, memSize, cudaMemcpyHostToDevice );

    // launch kernel
    dim3 dimGrid(numBlocks);
    dim3 dimBlock(numThreadsPerBlock);
    reverseArrayBlock < < < dimGrid, dimBlock >>>( d_a );

    // block until the device has completed
    cudaThreadSynchronize();

    // check if kernel execution generated an error
    // Check for any CUDA errors
    checkCUDAError("kernel invocation");

    // device to host copy
    cudaMemcpy( h_a, d_a, memSize, cudaMemcpyDeviceToHost );

    // Check for any CUDA errors
    checkCUDAError("memcpy");

    // verify the data returned to the host is correct
    for (int i = 0; i < 100; i++)
    {
        //assert(h_a == dimA - 1 - i );
        printf("%d ",h_a);
    }
   

    // free device memory
    cudaFree(d_a);


    // free host memory
    free(h_a);

    // If the program makes it this far, then the results are correct and
    // there are no run-time errors.  Good work!
    printf("Correct!\n");
finish = clock();
  duration = (double)(finish - start) / CLOCKS_PER_SEC;
printf( "the time is %f seconds\n", duration );
    return 0;
  
}
************************************************

下面是CPU进行的循环运算的代码：（运行时间为：0.171）
#include <stdio.h>
#include <assert.h>
#include <time.h>


////////////////////////////////////////////////////////////////////////////////
// Program main
////////////////////////////////////////////////////////////////////////////////
int main( int argc, char** argv)
{
clock_t start, finish;
double duration;
start = clock();
    // pointer for host memory and size
    int *h_a,transfer;
    int dimA = 512*21056; // 256K elements (1MB total)

   
   

  
    // allocate host memory
    size_t memSize = 512*21056* sizeof(int);
    h_a = (int *) malloc(memSize);
   

    // Initialize input array on host
    for (int i = 0; i < dimA; ++i)
    {
        h_a = i;
        //printf("%d ",h_a);
    }
  printf("\n");
    for( int j=0; j < (dimA/2) ; ++j )
    {  
   
        h_a[j]=h_a[j]+1;  
   
    }
  
  
  
  
    // verify the data returned to the host is correct
    for (int i = 0; i < 100; i++)
    {
        //assert(h_a == dimA - 1 - i );
        printf("%d ",h_a);
    }
   

  
    // free host memory
    free(h_a);

    // If the program makes it this far, then the results are correct and
    // there are no run-time errors.  Good work!
    printf("Correct!\n");
finish = clock();
  duration = (double)(finish - start) / CLOCKS_PER_SEC;
printf( "the time is %f seconds\n", duration );
    return 0;
  
}

CPU代码有点问题，以前只计算了一半，今天我改了，全部计算用的时间是0.281秒，也比GPU并行运算要快啊！
代码如下：
#include <stdio.h>
#include <assert.h>
#include <time.h>


////////////////////////////////////////////////////////////////////////////////
// Program main
////////////////////////////////////////////////////////////////////////////////
int main( int argc, char** argv)
{
clock_t start, finish;
double duration;
start = clock();
    // pointer for host memory and size
    int *h_a,transfer;
    int dimA = 512*21056; // 256K elements (1MB total)

   
   

  
    // allocate host memory
    size_t memSize = 512*21056* sizeof(int);
    h_a = (int *) malloc(memSize);
   

    // Initialize input array on host
    for (int i = 0; i < dimA; ++i)
    {
        h_a = i;
        //printf("%d ",h_a);
    }
  printf("\n");
    for( int j=0; j < dimA ; ++j )
    {  
   
        h_a[j]=h_a[j]+1;  
   
    }
  
  
  
  
    // verify the data returned to the host is correct
    for (int i = 0; i < 100; i++)
    {
        printf("%d ",h_a);
    }
   

  
    // free host memory
    free(h_a);

    // If the program makes it this far, then the results are correct and
    // there are no run-time errors.  Good work!
    printf("Correct!\n");
finish = clock();
  duration = (double)(finish - start) / CLOCKS_PER_SEC;
printf( "the time is %f seconds\n", duration );
    return 0;
  
}