CUDA结果使用非常大的数组返回垃圾,但报告没有错误

Question

我正在创建一个测试程序,它将创建一个大小为n的设备和主机数组,然后启动一个内核,创建n个线程,为设备阵列中的每个位置分配常量值0​​.95f.完成后,将设备阵列复制到主机阵列,并汇总所有条目并显示最终总计.

下面的程序似乎可以很好地处理大约6000万个浮点数的数组并且很快返回正确的结果,但是当达到7000万时,程序似乎挂起了一段时间并最终返回总结果的NAN结果.在6000万次运行后检查主机阵列显示它已正确填充0.95f,但在7000万次运行后检查它显示它填充了NAN.据我所知,没有一个CUDA调用返回错误.

我使用的是2GB GT640m(Compute 3.0),最大块大小为1024,最大网格尺寸为2147483647.

我相信有更好的方法来实现类似的东西,我想听听建议.但我也想了解这里出了什么问题,以便我可以从中吸取教训.

#include "cuda_runtime.h"
#include "device_launch_parameters.h"

#include <stdio.h>
#include <fstream>

void cudaErrorHandler(cudaError_t status)
{
    // Cuda call returned an error, just print error for now
    if(status != cudaSuccess)
    {
        printf("Error");
    }
}

__global__ void addKernel(float* _Results, int _TotalCombinations)
{
    // Get thread Id
    unsigned int Id = (blockDim.x * blockDim.y * blockIdx.x) + (blockDim.x * threadIdx.y) + threadIdx.x;

    //If the Id is within simulation range, log it
    if(Id < _TotalCombinations)
    {
        _Results[Id] = 0.95f;
    }
}

#define BLOCK_DIM_X 32
#define BLOCK_DIM_Y 32
#define BLOCK_SIZE BLOCK_DIM_X * BLOCK_DIM_Y // Statc block size of 32*32 (1024)
#define CUDA_CALL(x) cudaErrorHandler(x)

int main()
{
    // The number of simulations to run
    unsigned int totalCombinations = 45000000;

    int gridsize = 1;

    // Work out how many blocks of size 1024 are required to perform all of totalCombinations
    for(unsigned int totalsize = gridsize * BLOCK_SIZE; totalsize < totalCombinations; 
        gridsize++, totalsize = gridsize * BLOCK_SIZE)
        ;

    // Allocate host memory
    float* host_results = new float[totalCombinations];
    memset(host_results, 0, sizeof(float) * totalCombinations);
    float *dev_results = 0;

    cudaSetDevice(0);

    // Allocate device memory
    CUDA_CALL(cudaMalloc((void**)&dev_results, totalCombinations * sizeof(float)));

    dim3 grid, block;

    block = dim3(BLOCK_DIM_X, BLOCK_DIM_Y);

    grid = dim3(gridsize);

    // Launch kernel
    addKernel<<<gridsize, block>>>(dev_results, totalCombinations);

    // Wait for synchronize
    CUDA_CALL(cudaDeviceSynchronize());

    // Copy device data back to host
    CUDA_CALL(cudaMemcpy(host_results, dev_results, totalCombinations * sizeof(float), cudaMemcpyDeviceToHost));

    double total = 0.0;

    // Total the results in the host array
    for(unsigned int i = 0; i < totalCombinations; i++)
        total+=host_results[i];

    // Print results to screen
    printf("Total %f\n", total);

    delete[] host_results;

    return 0;
}

Answer 1

正如您所发现的,您的错误处理方法无效.下面我已经使用我经常使用的错误检查方法粘贴了一段代码.事情在你的失败点不起作用的原因是你的网格化(你正在启动一维网格)超过了X维度中的最大网格大小(默认为65535,即计算能力高达2.x).如果要利用更大的网格大小(2 ^ 31 -1是计算能力3.0的限制),则需要使用-arch=sm_30交换机进行编译.

这里仅供参考,是一个代码版本,它显示了我经常使用的错误检查方法.

#include <stdio.h>
#include <fstream>


#define cudaCheckErrors(msg) \
    do { \
        cudaError_t __err = cudaGetLastError(); \
        if (__err != cudaSuccess) { \
            fprintf(stderr, "Fatal error: %s (%s at %s:%d)\n", \
                msg, cudaGetErrorString(__err), \
                __FILE__, __LINE__); \
            fprintf(stderr, "*** FAILED - ABORTING\n"); \
            exit(1); \
        } \
    } while (0)

__global__ void addKernel(float* _Results, int _TotalCombinations)
{
    // Get thread Id
    unsigned int Id = (blockDim.x * blockDim.y * blockIdx.x) + (blockDim.x * threadIdx.y) + threadIdx.x;

    //If the Id is within simulation range, log it
    if(Id < _TotalCombinations)
    {
        _Results[Id] = 0.95f;
    }
}

#define BLOCK_DIM_X 32
#define BLOCK_DIM_Y 32
#define BLOCK_SIZE BLOCK_DIM_X * BLOCK_DIM_Y // Statc block size of 32*32 (1024)

int main()
{
    // The number of simulations to run
    unsigned int totalCombinations = 65000000;

    int gridsize = 1;

    // Work out how many blocks of size 1024 are required to perform all of totalCombinations
    for(unsigned int totalsize = gridsize * BLOCK_SIZE; totalsize < totalCombinations;
        gridsize++, totalsize = gridsize * BLOCK_SIZE)
        ;
    printf("gridsize = %d, blocksize = %d\n", gridsize, BLOCK_SIZE);
    // Allocate host memory
    float* host_results = new float[totalCombinations];
    memset(host_results, 0, sizeof(float) * totalCombinations);
    float *dev_results = 0;

    cudaSetDevice(0);

    // Allocate device memory
    cudaMalloc((void**)&dev_results, totalCombinations * sizeof(float));
    cudaCheckErrors("cudaMalloc fail");

    dim3 grid, block;

    block = dim3(BLOCK_DIM_X, BLOCK_DIM_Y);

    grid = dim3(gridsize);

    // Launch kernel
    addKernel<<<gridsize, block>>>(dev_results, totalCombinations);
    cudaCheckErrors("kernel fail");
    // Wait for synchronize
    cudaDeviceSynchronize();
    cudaCheckErrors("sync fail");

    // Copy device data back to host
    cudaMemcpy(host_results, dev_results, totalCombinations * sizeof(float), cudaMemcpyDeviceToHost);
    cudaCheckErrors("cudaMemcpy 2 fail");

    double total = 0.0;

    // Total the results in the host array
    for(unsigned int i = 0; i < totalCombinations; i++)
        total+=host_results[i];

    // Print results to screen
    printf("Total %f\n", total);

    delete[] host_results;

    return 0;
}