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。

 1 #include <cuda.h>
 2 #include <cuda_runtime.h>
 3 #include <device_launch_parameters.h>
 4 #include <device_functions.h>
 5 #include <iostream>
 6 #include <string>
 7 
 8 using namespace std;
 9 
10 #define imin(a,b) (a<b? a:b)
11 const int N = 33 * 1024;
12 const int threadsPerBlock = 256;
13 const int blocksPerGrid = imin(32, (N + threadsPerBlock - 1) / threadsPerBlock);
14 
15 __global__  void dot(float *a, float *b, float *c)
16 {
17     __shared__ float cache[threadsPerBlock];
18     int tid = threadIdx.x + blockDim.x*blockIdx.x;
19     int cacheIndex = threadIdx.x;
20 
21     float temp = 0;
22     //每个线程负责计算的点乘，再加和
23     while (tid<N)
24     {
25         temp += a[tid] * b[tid];
26         tid += blockDim.x*gridDim.x;
27     }
28     
29     //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
30     cache[cacheIndex] = temp;
31     //对线程块中的线程进行同步
32     __syncthreads();
33 
34     //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
35     int i = blockDim.x / 2;
36     while (i != 0)
37     {
38         if (cacheIndex < i)
39         {
40             cache[cacheIndex] += cache[cacheIndex + i];
41         }
42         __syncthreads();
43         i /= 2;
44     }
45     //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
46     if (cacheIndex == 0)
47     {
48         c[blockIdx.x] = cache[0];
49     }
50 }
51 
52 void main()
53 {
54     float *a, *b, c, *partial_c;
55     float *dev_a, *dev_b, *dev_partial_c;
56 
57     //分配CPU内存
58     a = (float*)malloc(N * sizeof(float));
59     b = (float*)malloc(N * sizeof(float));
60     partial_c = (float*)malloc(blocksPerGrid * sizeof(float));
61 
62     //分配GPU内存
63     cudaMalloc(&dev_a, N * sizeof(float));
64     cudaMalloc(&dev_b, N * sizeof(float));
65     cudaMalloc(&dev_partial_c, blocksPerGrid * sizeof(float));
66 
67     float sum = 0;
68     for (int i = 0; i < N; i++)
69     {
70         a[i] = i;
71         b[i] = i * 2;
72     }
73 
74     //将数组上传到GPU
75     cudaMemcpy(dev_a, a, N * sizeof(float), cudaMemcpyHostToDevice);
76     cudaMemcpy(dev_b, b, N * sizeof(float), cudaMemcpyHostToDevice);
77 
78     dot << <blocksPerGrid, threadsPerBlock >> > (dev_a, dev_b, dev_partial_c);
79 
80     cudaMemcpy(partial_c, dev_partial_c, blocksPerGrid * sizeof(float), cudaMemcpyDeviceToHost);
81     
82     //CPU 完成最终求和
83     c = 0;
84     for (int i = 0; i < blocksPerGrid; i++)
85     {
86         c += partial_c[i];
87     }
88 
89 #define sum_squares(x) (x*(x+1)*(2*x+1)/6)
90     printf("does GPU value %.6g = %.6g?\n", c, 2 * sum_squares((float)(N - 1)));
91 
92     cudaFree(dev_a);
93     cudaFree(dev_b);
94     cudaFree(dev_partial_c);
95 
96     free(a);
97     free(b);
98     free(partial_c);
99 }

 

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posted @ 2017-12-07 22:23  一度逍遥  阅读(4616)  评论(1)  编辑  收藏  举报