CUDA共享內存的使用示例

CUDA共享內存使用示例如下：參考教材《GPU高性能編程CUDA實戰》。P54-P65

教材下載地址：http://download.csdn.net/download/yizhaoyanbo/10150300。如果沒有下載分可以評論區留下郵箱，我發你。

#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 = 33 * 1024;
const int threadsPerBlock = 256;
const int blocksPerGrid = imin(32, (N + threadsPerBlock - 1) / 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 = 0;
    //每個線程負責計算的點乘，再加和
    while (tid<N)
    {
        temp += a[tid] * b[tid];
        tid += blockDim.x*gridDim.x;
    }
    
    //每個線程塊中線程計算的加和保存到緩沖區cache，一共有blocksPerGrid個緩沖區副本
    cache[cacheIndex] = temp;
    //對線程塊中的線程進行同步
    __syncthreads();

    //歸約運算，將每個緩沖區中的值加和，存放到緩沖區第一個元素位置
    int i = blockDim.x / 2;
    while (i != 0)
    {
        if (cacheIndex < i)
        {
            cache[cacheIndex] += cache[cacheIndex + i];
        }
        __syncthreads();
        i /= 2;
    }
    //使用第一個線程取出每個緩沖區第一個元素賦值到C數組
    if (cacheIndex == 0)
    {
        c[blockIdx.x] = cache[0];
    }
}

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

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

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

    float sum = 0;
    for (int i = 0; i < N; i++)
    {
        a[i] = i;
        b[i] = i * 2;
    }

    //將數組上傳到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 完成最終求和
    c = 0;
    for (int i = 0; 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, 2 * sum_squares((float)(N - 1)));

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

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

 

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