使用 2D 倾斜阵列减少 CUB 总和

Question

我正在尝试使用 float/double 类型的 CUB 和二维数组执行总和约简。虽然它适用于行+列的某些组合，但对于相对较大的数组，我在上次传输期间遇到非法内存访问错误。一个最小的例子如下：

#include <stdio.h>
#include <stdlib.h>

#include <cub/device/device_reduce.cuh>
#include "cuda_runtime.h"

#ifdef DP
#define real double
#else
#define real float
#endif

void generatedata(const int num, real* vec, real start, real finish) {
    real rrange = finish - start;
    for (auto i = 0; i < num; ++i)
        vec[i] = rand() / float(RAND_MAX) * rrange + start;
}

real reduce_to_sum(const int num, const real* vec) {
    real total = real(0.0);
    for (auto i = 0; i < num; ++i)
        total += vec[i];

    return total;
}

int main() {
    int rows = 2001;
    int cols = 3145;
    size_t msize = rows * cols;

    real* data = (real*)malloc(msize * sizeof(real));
    if (!data)
        return -999;

    generatedata(msize, data, 0., 50.);
    real ref_sum = reduce_to_sum(msize, data);

    real* d_data_in = nullptr;
    real* d_data_out = nullptr;
    size_t pitch_in, pitch_out;
    cudaError_t err = cudaMallocPitch(&d_data_in, &pitch_in, cols * sizeof(real), rows);
    if (err != cudaSuccess) {
        printf("data_in :: %s \n", cudaGetErrorString(err));
        return -999;
    }
    err = cudaMallocPitch(&d_data_out, &pitch_out, cols * sizeof(real), rows);
    if (err != cudaSuccess) {
        printf("data_out :: %s \n", cudaGetErrorString(err));
        return -999;
    }

    err = cudaMemset(d_data_in, 0, rows * pitch_in);
    if (err != cudaSuccess) {
        printf("set data_in :: %s \n", cudaGetErrorString(err));
        return -999;
    }
    err = cudaMemcpy2D(d_data_in, pitch_in, data, cols * sizeof(real), cols * sizeof(real), rows, cudaMemcpyHostToDevice);
    if (err != cudaSuccess) {
        printf("copy data :: %s \n", cudaGetErrorString(err));
        return -999;
    }

    void* d_temp = nullptr;
    size_t   temp_bytes = 0;
    cub::DeviceReduce::Sum(d_temp, temp_bytes, d_data_in, d_data_out, rows * pitch_out);
    err = cudaMalloc(&d_temp, temp_bytes);
    if (err != cudaSuccess) {
        printf("temp :: %s \n", cudaGetErrorString(err));
        return -999;
    }

    err = cudaMemset(d_data_out, 0, rows * pitch_out);
    if (err != cudaSuccess) {
        printf("set temp :: %s \n", cudaGetErrorString(err));
        return -999;
    }
    // Run sum-reduction
    cub::DeviceReduce::Sum(d_temp, temp_bytes, d_data_in, d_data_out, rows * pitch_out);
    err = cudaGetLastError();
    if (err != cudaSuccess) {
        printf("reduction :: %s \n", cudaGetErrorString(err));
        return -999;
    }

    real gpu_sum = real(0.0);
    err = cudaMemcpy(&gpu_sum, d_data_out, sizeof(real), cudaMemcpyDeviceToHost);
    if (err != cudaSuccess) {
        printf("copy final :: %s \n", cudaGetErrorString(err));
        return -999;
    }

    printf("Difference in sum (h)%f - (d)%f = %f \n", ref_sum, gpu_sum, ref_sum - gpu_sum);

    if (data) free(data);
    if (d_data_in) cudaFree(d_data_in);
    if (d_data_out) cudaFree(d_data_out);
    if (d_temp) cudaFree(d_temp);
    cudaDeviceReset();
    return 0;
}

在“copy final ::”处抛出错误。我对为什么某些行 x 列有效而其他行无效感到困惑。我确实注意到它是导致它的较大值，但无法理解。任何建议将不胜感激。

Answer 1

cub::DeviceReduce::Sum的第5个参数应该是输入元素的个数。但是，rows * pitch_out 是以字节为单位的输出缓冲区的大小。

假设 pitch_in % sizeof(real) == 0，以下调用可能有效。

cub::DeviceReduce::Sum(d_temp, temp_bytes, d_data_in, d_data_out, rows * (pitch_in / sizeof(real)));

另请注意，cub::DeviceReduce::Sum 可能会 return 完成还原。在这种情况下，如果在执行过程中发生任何错误，cudaMemcpy 将报告此错误。

使用 2D 倾斜阵列减少 CUB 总和

CUB sum reduction with 2D pitched arrays

cuda

cub