CUDA共享内存的使用示例

CUDA共享内存使用示例如下：参考教材《GPU高性能编程CUDA实战》。P54-P65

教材下载地址：http://download.csdn.net/download/yizhaoyanbo/10150300。如果没有下载分可以评论区留下邮箱，我发你。

 #include <cuda.h>
 #include <cuda_runtime.h>
 #include <device_launch_parameters.h>
 #include <device_functions.h>
 #include <iostream>
 #include <string>

 using namespace std;

 #define imin(a,b) (a<b? a:b)
 const int N =  * ;
 const int threadsPerBlock = ;
 const int blocksPerGrid = imin(, (N + threadsPerBlock - ) / threadsPerBlock);

 __global__  void dot(float *a, float *b, float *c)
 {
     __shared__ float cache[threadsPerBlock];
     int tid = threadIdx.x + blockDim.x*blockIdx.x;
     int cacheIndex = threadIdx.x;

     float temp = ;
     //每个线程负责计算的点乘，再加和
     while (tid<N)
     {
         temp += a[tid] * b[tid];
         tid += blockDim.x*gridDim.x;
     }

     //每个线程块中线程计算的加和保存到缓冲区cache，一共有blocksPerGrid个缓冲区副本
     cache[cacheIndex] = temp;
     //对线程块中的线程进行同步
     __syncthreads();

     //归约运算，将每个缓冲区中的值加和，存放到缓冲区第一个元素位置
     int i = blockDim.x / ;
     while (i != )
     {
         if (cacheIndex < i)
         {
             cache[cacheIndex] += cache[cacheIndex + i];
         }
         __syncthreads();
         i /= ;
     }
     //使用第一个线程取出每个缓冲区第一个元素赋值到C数组
     if (cacheIndex == )
     {
         c[blockIdx.x] = cache[];
     }
 }

 void main()
 {
     float *a, *b, c, *partial_c;
     float *dev_a, *dev_b, *dev_partial_c;

     //分配CPU内存
     a = (float*)malloc(N * sizeof(float));
     b = (float*)malloc(N * sizeof(float));
     partial_c = (float*)malloc(blocksPerGrid * sizeof(float));

     //分配GPU内存
     cudaMalloc(&dev_a, N * sizeof(float));
     cudaMalloc(&dev_b, N * sizeof(float));
     cudaMalloc(&dev_partial_c, blocksPerGrid * sizeof(float));

     float sum = ;
     for (int i = ; i < N; i++)
     {
         a[i] = i;
         b[i] = i * ;
     }

     //将数组上传到GPU
     cudaMemcpy(dev_a, a, N * sizeof(float), cudaMemcpyHostToDevice);
     cudaMemcpy(dev_b, b, N * sizeof(float), cudaMemcpyHostToDevice);

     dot << <blocksPerGrid, threadsPerBlock >> > (dev_a, dev_b, dev_partial_c);

     cudaMemcpy(partial_c, dev_partial_c, blocksPerGrid * sizeof(float), cudaMemcpyDeviceToHost);

     //CPU 完成最终求和
     c = ;
     for (int i = ; i < blocksPerGrid; i++)
     {
         c += partial_c[i];
     }

 #define sum_squares(x) (x*(x+1)*(2*x+1)/6)
     printf("does GPU value %.6g = %.6g?\n", c,  * sum_squares((float)(N - )));

     cudaFree(dev_a);
     cudaFree(dev_b);
     cudaFree(dev_partial_c);

     free(a);
     free(b);
     free(partial_c);
 }

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