c-cuBlas的不同结果

我已经实现了以下CUDA代码,但是我对此行为有些困惑.

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <cuda_runtime.h>
#include "cublas_v2.h"
#include <ctime>
#include <chrono>
#include <string>

#define IDX2F(i,j,ld) ((((j)-1)*(ld))+((i)-1)) 

void PrintMatrix(float* a, int n)
{
    int j, i;
    for (j = 1; j <= n; j++)
    {
        for (i = 1; i <= n; i++)
        {
            printf("%7.0f", a[IDX2F(i, j, n)]);
        }
        printf("\n");
    }
}

float* CreateMatrix(int n)
{
    float* matrix = static_cast<float *>(malloc(n * n * sizeof(float)));
    if (!matrix)
    {
        printf("host memory allocation failed");
        return nullptr;
    }

    for (int j = 1; j <= n; j++)
    {
        for (int i = 1; i <= n; i++)
        {
            matrix[IDX2F(i, j, n)] = 2;
        }
    }

    return matrix;
}

long CudaMatrixMultiply(float* matrix, int n)
{
    cudaError_t cudaStat;
    cublasStatus_t status;
    cublasHandle_t handle;
    float* deviceMatrix;

    cudaStat = cudaMalloc(reinterpret_cast<void**>(&deviceMatrix), n * n * sizeof(float));
    if (cudaStat != cudaSuccess)
    {
        printf("device memory allocation failed");
        return EXIT_FAILURE;
    }

    status = cublasCreate(&handle);
    if (status != CUBLAS_STATUS_SUCCESS)
    {
        printf("CUBLAS initialization failed\n");
        return EXIT_FAILURE;
    }

    status = cublasSetMatrix(n, n, sizeof(float), matrix, n, deviceMatrix, n);
    if (status != CUBLAS_STATUS_SUCCESS)
    {
        printf("data download failed");
        cudaFree(deviceMatrix);
        cublasDestroy(handle);
        return EXIT_FAILURE;
    }

    float alpha = 1;
    float beta = 0;
    cublasSgemm(handle, CUBLAS_OP_N, CUBLAS_OP_N, n, n, n, &alpha, deviceMatrix, n, deviceMatrix, n, &beta, deviceMatrix, n);

    status = cublasGetMatrix(n, n, sizeof(float), deviceMatrix, n, matrix, n);
    if (status != CUBLAS_STATUS_SUCCESS)
    {
        printf("data upload failed");
        cudaFree(deviceMatrix);
        cublasDestroy(handle);
        return EXIT_FAILURE;
    }

    cudaFree(deviceMatrix);
    cublasDestroy(handle);
    return EXIT_SUCCESS;
}

float* CpuMatrixMultiply(float* matrix, int size)
{
    float* result = new float[size * size]();

    // Copied from https://msdn.microsoft.com/en-us/library/hh873134.aspx
    for (int row = 1; row <= size; row++) 
    {
        for (int col = 1; col <= size; col++) 
        {
            // Multiply the row of A by the column of B to get the row, column of product.
            for (int inner = 1; inner <= size; inner++) 
            {
                // result[row][col] += matrix[row][inner] * matrix[inner][col];
                result[IDX2F(col, row, size)] += matrix[IDX2F(inner, row, size)] * matrix[IDX2F(col, inner, size)];
            }
        }
    }

    free(matrix);
    return result;
}

int main(void)
{
    // printf("Matrix * Matrix Test\n");
    int size = 1000;
    int runs = 10;

    for (int run = 0; run != runs; run++)
    {
        printf("=== Test %d (Matrix * Matrix, Size = %d) ===\n\n", run + 1, size);
        printf("RAM usage is: %f GB\n", size * size * sizeof(float) / 1000000000.0);

        float* cpuMatrix = CreateMatrix(size);
        cpuMatrix = CpuMatrixMultiply(cpuMatrix, size);

        PrintMatrix(cpuMatrix, 5);

        float* gpuMatrix = CreateMatrix(size);
        CudaMatrixMultiply(gpuMatrix, size);
        PrintMatrix(gpuMatrix, 5);

        free(cpuMatrix);
        free(gpuMatrix);
    }
    getchar();
    return EXIT_SUCCESS;
}

预期的MatrixMultiplication的CPU版本的输出如下：

4000 4000 4000 4000 4000
4000 4000 4000 4000 4000
4000 4000 4000 4000 4000
4000 4000 4000 4000 4000
4000 4000 4000 4000 4000

但是GPU的计算结果有时是正确的(见上文)或错误的random(？).第一次执行循环时,结果总是正确的.

我找不到我的代码中的错误,如果您能帮助我,那将是很棒的.

另外,如果我将尺寸(int主要方法)设置为例如16000然后我的驱动程序崩溃了,我收到一条错误消息.为此,我写了一份错误报告给NVidea,因为我的电脑崩溃了两次.但是也许这是我的编程错误？

司机：364.72(最新的)
SDK：CUDA工具包7.5
显卡：NVidia GeForce GTX 960(4GB)
Windows 10 64位

驱动错误

Display driver NVIDIA Windows kernel Mode Driver, Version 362.72 stopped responding and has successfully recovered.

编辑：在社区的帮助下,我发现这是看门狗计时器的问题.请参阅下面的答案.

解决方法:

关于问题的第二部分,按照njuffa的说明,您可以更改驱动程序行为的设置,以避免在增大大小时出错.打开NSIGHT监视器,然后在“选项”,“常规”,“ Microsoft显示驱动程序”中,将“启用WDDM TDR”字段更改为“否”.

从spec开始,单精度的32位FPU触发器应约为2.4 TFLOPS,因此,对于16000大小的矩阵,您的操作至少需要3.5秒.因此,驱动程序在2秒后恢复.

标签：cublas,c,cuda
来源： https://codeday.me/bug/20191012/1903517.html