Question

我是CUDA C中的新手......我想在1个块中对数组元素（使用reduce）求和，267个线程使用共享内存。我通过示例读了一本书＆＃34; CUDA，介绍了GPU编程的通用＆＃34;。根据她的一些推荐，我写了我的程序版本：

__global__ void
conva(int* a, int* out) 
{
    __shared__ int cache[534];
    int cacheIndex = threadIdx.x;
    for(int n=0; n<2;++n) {
        cache[cacheIndex+n] = a[cacheIndex+n];
        int i = blockDim.x/2;
        while (i != 0) {
            if (cacheIndex < i)
                cache[cacheIndex + n] += cache[cacheIndex + n + i];
            __syncthreads();
            i /= 2;
        } 
    }
    //need or not this __syncthreads(), I don't know
    __syncthreads();
     if (cacheIndex == 0)
         out = &cache[0];
}

int main(int argc, char** argv)
{
    //enter array for sum
    int convolution[534];
    for(int i=0; i<534; ++i) 
        convolution[i] = 1;
    //variable in which we take a sum from device
    int summa = 0;
    //it we copy on device from host
    int* tash;

    int* convolution_gpu;
    cudaMalloc((void**)(&convolution_gpu), 534*sizeof(int));
    cudaMalloc((void**)(&tash), sizeof(int));

    cudaMemcpy(convolution_gpu, convolution, 534*sizeof(int),  cudaMemcpyHostToDevice );
    //call core with 1 block and 267 threads
    conva<<<1, 267>>>(convolution_gpu, tash);

    cudaMemcpy(&summa, tash, sizeof(int), cudaMemcpyDeviceToHost);
    //and here I want 534 but I have garbage(may be)
    std::cout<<summa<<std::endl;

    cudaFree(convolution_gpu);
    cudaFree(tash);
    getchar();
}

请告诉我，这里有错误并帮我解决她... （对不起我的英文）

Answer 1

在你的内核中，这个：

System.out.print("Name: ");

几乎肯定是错的。当然你想要这样的东西：

  try (Scanner sc = new Scanner(System.in)) {
        System.out.print("Name: ");
        while (sc.hasNextLine()) {
            String name = sc.nextLine();
            System.out.println("Name is \"" + name + "\"");
        }
    }

简单减少CUDA时遇到错误

1 个答案: