Question

我开始阅读＆＃34; Cuda By Example＆＃34;书和我使用＆＃34;共享内存＆＃34;的点示例存在问题。我从书中复制粘贴示例并设置：N = x * 1024; threadsPerBlock = 32; blocksPerGrid = 8.我在哪里测试＆＃34; x＆＃34;值为2,3,4,5。如果我设置x = 3，结果很糟糕，但是当我使用x = 2,4,5时一切正常。我不明白问题出在哪里。代码是：

#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <stdio.h>

#define imin(a, b) (a<b?a:b)
#define sum_squares(x) (x*(x+1)*(2*x+1)/6)

const int x = 3;
const int N = 3 * 1024;
const int threadsPerBlock = 32;
const int blocksPerGrid = 8;

__global__ void dot(float *a, float *b, float *c)
{
    __shared__ float cache[threadsPerBlock];
    int tid = threadIdx.x + blockIdx.x * blockDim.x;
    int cacheIndex = threadIdx.x;
    float temp = 0;

    while (tid < N)
    {
        temp += a[tid] * b[tid];
        tid += blockDim.x * gridDim.x;
    }

    cache[cacheIndex] = temp;

    __syncthreads();

    int i = blockDim.x / 2;
    while (i != 0)
    {
        if (cacheIndex < i)
            cache[cacheIndex] += cache[cacheIndex + i];
        __syncthreads();
        i /= 2;
    }

    if (cacheIndex == 0) 
        c[blockIdx.x] = cache[0];
}

int main()
{
    float *a, *b, *partial_c, result;
    float *d_a, *d_b, *d_partial_c;

    a = (float *)malloc(N * sizeof(float));
    b = (float *)malloc(N * sizeof(float));
    partial_c = (float *)malloc(blocksPerGrid * sizeof(float));

    cudaMalloc((void **)&d_a, N * sizeof(float));
    cudaMalloc((void **)&d_b, N * sizeof(float));
    cudaMalloc((void **)&d_partial_c, blocksPerGrid * sizeof(float));

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

    cudaMemcpy(d_a, a, N * sizeof(float), cudaMemcpyHostToDevice);
    cudaMemcpy(d_b, b, N * sizeof(float), cudaMemcpyHostToDevice);

    dot << <blocksPerGrid, threadsPerBlock >> >(d_a, d_b, d_partial_c);

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

    result = 0;
    for (int i = 0; i < blocksPerGrid; i++)
        result += partial_c[i];

    if (2 * sum_squares((float)(N - 1)) == result)
        printf(":)\n");
    else
        printf(":(\n");

    cudaFree(d_a);
    cudaFree(d_b);
    cudaFree(d_partial_c);

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

    getchar();
    return 0;
}

Answer 1

因为float没有足够的精度，只有~7 decimal digits。但对于x=3;，您的预期结果是

19317916672

包含11位数字。

对于x=4,5，我的计算机上的结果也很糟糕。

以Cuda为例的Dot产品对我不起作用

1 个答案: