C语言 多次执行后 CUDA 程序的结果不一致

Question

说明

我试图在 GPU 上使用 2D 网格和 2D 块执行矩阵求和，并在多次执行程序后得到不同的结果。对此行为的任何解释或修复都将有所帮助，谢谢。

细节

大多数情况下，CPU 上的结果与 GPU 上的结果一致。但有时(例如，在操作系统启动后(程序会告诉结果不一致。但是之后的所有执行都会产生一致的结果(并且运行速度似乎更快(。我还没有找到一种有保证的方式来重现这种行为。我尝试再次重新启动操作系统，但程序的第一次执行产生了一致的结果。

法典

main 函数对 CPU 和 GPU 上的两个 2^10 x 2^10 矩阵(具有 2^5 x 2^5 网格和 2^5

x 2^5 块(执行求和并比较结果。

#include "stdio.h"
#define FALSE 0
#define TRUE !FALSE
double *mallocMatrix(const int row, const int column)
{
    return (double*)malloc(row*column*sizeof(double));
}
void matrixInit(double *matrix, const int row, const int column)
{
    ;
}

int matEqual(double *mat1, double *mat2, const int row, const int column)
{
    for(int i=0;i<row;i++)
    {
        for(int j=0;j<column;j++)
        {
            int k=i*column+j;
            if(mat1[k]!=mat2[k])
            {
                printf("Entry %d doens't match.n",k);
                return FALSE;
            }
        }
    }
    return TRUE;
}
void matrixSumCpu(double *m1, double *m2, double *n, const int row, const int column)
{
    for(int i=0; i<row; i++)
    {
        for(int j=0; j<column; j++)
        {
            int k = i * column + j;
            n[k]=m1[k]+m2[k];
        }
    }
}
__global__ void _2dGrid2dBlockMatSum(double *m1, double *m2, double *n, const int row, const int column)
{
    int rowIndex=blockIdx.x*blockDim.x+threadIdx.x;
    int columnIndex=blockIdx.y*blockDim.y+threadIdx.y;
    if(rowIndex<row&&columnIndex<column)
    {
        int i=rowIndex*column+columnIndex;//flatten
        n[i]=m1[i]+m2[i];
    }
}

void checkGpuMalloc(cudaError_t code)
{
    if(code != cudaSuccess)
    {
        exit(-1);
        printf("CUDA ERROR occured. ");
    }
}
void printMatrix(double *mat, const int row, const int column)
{
    const int rowToPrint=3;
    const int columnToPrint=6;
    for(int i=0;i<rowToPrint;i++)
    {
        for(int j=0;j<columnToPrint;j++)
            printf("%lf", mat[i*column+j]);
        if(column>columnToPrint)
            printf("...");
        printf("n");
    }
    if(row>rowToPrint)
        printf("...n");
}
int main()
{
    int row=1<<10, column=1<<10;
    double *h_m1=NULL, *h_m2=NULL,*h_n1=NULL, *h_n2=NULL;//n=m1+m2
    h_m1=mallocMatrix(row, column);
    h_m2=mallocMatrix(row, column);
    h_n1=mallocMatrix(row, column);
    h_n2=mallocMatrix(row, column);
    if(h_m1==NULL||h_m2==NULL||h_n1==NULL||h_n2==NULL)
    {
        printf("Unable to allocate enough memory on CPUn");
        exit(-1);
    }
    matrixInit(h_m1,row,column);
    matrixInit(h_m2,row,column);
    printf("Summing matrices on CPU...n");
    matrixSumCpu(h_m1,h_m2,h_n1,row,column);
    double *d_m1=NULL, *d_m2=NULL, *d_n=NULL;
    checkGpuMalloc(cudaMalloc((void**)&d_m1, row*column*sizeof(double)));
    checkGpuMalloc(cudaMalloc((void**)&d_m2, row*column*sizeof(double)));
    checkGpuMalloc(cudaMalloc((void**)&d_n, row*column*sizeof(double)));
    cudaMemcpy(d_m1, h_m1, row*column*sizeof(double), cudaMemcpyHostToDevice);
    cudaMemcpy(d_m2, h_m2, row*column*sizeof(double), cudaMemcpyHostToDevice);
    printf("Summing matrices on GPU with 2D grid and 2D blocks.n");
    _2dGrid2dBlockMatSum<<<(1<<5,1<<5),(1<<5, 1<<5)>>>(d_m1, d_m2, d_n, row, column);
    cudaDeviceSynchronize();    
    cudaMemcpy(h_n2, d_n, row*column*sizeof(double), cudaMemcpyDeviceToHost);
    if(matEqual(h_n1, h_n2, row, column))
        printf("Matrices match.n");
    else
    {
        printf("Matrices don't match.nResult on CPU:n");
        printMatrix(h_n1, row, column);
        printf("Result on GPU:");
        printMatrix(h_n2, row, column);
    }
    free(h_m1);
    free(h_m2);
    free(h_n1);
    free(h_n2);
    cudaFree(d_m1);
    cudaFree(d_m2);
    cudaFree(d_n);
    cudaDeviceReset();
    return 0;
}

Answer 1

这不会做你认为它做的事情，当我编译你的代码时，编译器会在下面一行发出警告：

_2dGrid2dBlockMatSum<<<(1<<5,1<<5),(1<<5, 1<<5)>>>(d_m1, d_m2, d_n, row, column);

你应该做这样的事情：

_2dGrid2dBlockMatSum<<<dim3(1<<5,1<<5),dim3(1<<5, 1<<5)>>>(d_m1, d_m2, d_n, row, column);

这：

dim3(1<<5,1<<5)

与此不同：

(1<<5,1<<5)

C++编译器计算最后一个表达式，产生某种您意想不到的垃圾(标量数量 32，而不是 2D 数量 (32,32((。

为什么您的matrixInit函数为空？

如果要强制代码始终失败，请添加一些矩阵初始化：

void matrixInit(double *matrix, const int row, const int column)
{
    for (int i = 0; i < row; i++)
      for (int j = 0; j < column; j++)
        matrix[i*column+j] = 1;
}

并在内核调用之前添加此行：

cudaMemset(d_n, 0, row*column*sizeof(double));

然后编译并运行它，它将失败。

之后，按照我的建议进行dim3更改，它将修复它。

下面是固定示例：

#include "stdio.h"
#define FALSE 0
#define TRUE !FALSE
double *mallocMatrix(const int row, const int column)
{
    return (double*)malloc(row*column*sizeof(double));
}
void matrixInit(double *matrix, const int row, const int column)
{
    for (int i = 0; i < row; i++)
      for (int j = 0; j < column; j++)
        matrix[i*column+j] = 1;
}

int matEqual(double *mat1, double *mat2, const int row, const int column)
{
    for(int i=0;i<row;i++)
    {
        for(int j=0;j<column;j++)
        {
            int k=i*column+j;
            if(mat1[k]!=mat2[k])
            {
                printf("Entry %d doens't match.n",k);
                return FALSE;
            }
        }
    }
    return TRUE;
}
void matrixSumCpu(double *m1, double *m2, double *n, const int row, const int column)
{
    for(int i=0; i<row; i++)
    {
        for(int j=0; j<column; j++)
        {
            int k = i * column + j;
            n[k]=m1[k]+m2[k];
        }
    }
}
__global__ void _2dGrid2dBlockMatSum(double *m1, double *m2, double *n, const int row, const int column)
{
    int rowIndex=blockIdx.x*blockDim.x+threadIdx.x;
    int columnIndex=blockIdx.y*blockDim.y+threadIdx.y;
    if(rowIndex<row&&columnIndex<column)
    {
        int i=rowIndex*column+columnIndex;//flatten
        n[i]=m1[i]+m2[i];
    }
}

void checkGpuMalloc(cudaError_t code)
{
    if(code != cudaSuccess)
    {
        exit(-1);
        printf("CUDA ERROR occured. ");
    }
}
void printMatrix(double *mat, const int row, const int column)
{
    const int rowToPrint=3;
    const int columnToPrint=6;
    for(int i=0;i<rowToPrint;i++)
    {
        for(int j=0;j<columnToPrint;j++)
            printf("%lf", mat[i*column+j]);
        if(column>columnToPrint)
            printf("...");
        printf("n");
    }
    if(row>rowToPrint)
        printf("...n");
}
int main()
{
    int row=1<<10, column=1<<10;
    double *h_m1=NULL, *h_m2=NULL,*h_n1=NULL, *h_n2=NULL;//n=m1+m2
    h_m1=mallocMatrix(row, column);
    h_m2=mallocMatrix(row, column);
    h_n1=mallocMatrix(row, column);
    h_n2=mallocMatrix(row, column);
    if(h_m1==NULL||h_m2==NULL||h_n1==NULL||h_n2==NULL)
    {
        printf("Unable to allocate enough memory on CPUn");
        exit(-1);
    }
    matrixInit(h_m1,row,column);
    matrixInit(h_m2,row,column);
    printf("Summing matrices on CPU...n");
    matrixSumCpu(h_m1,h_m2,h_n1,row,column);
    double *d_m1=NULL, *d_m2=NULL, *d_n=NULL;
    checkGpuMalloc(cudaMalloc((void**)&d_m1, row*column*sizeof(double)));
    checkGpuMalloc(cudaMalloc((void**)&d_m2, row*column*sizeof(double)));
    checkGpuMalloc(cudaMalloc((void**)&d_n, row*column*sizeof(double)));
    cudaMemcpy(d_m1, h_m1, row*column*sizeof(double), cudaMemcpyHostToDevice);
    cudaMemcpy(d_m2, h_m2, row*column*sizeof(double), cudaMemcpyHostToDevice);
    cudaMemset(d_n, 0, row*column*sizeof(double));
    printf("Summing matrices on GPU with 2D grid and 2D blocks.n");
    printf("%dn", (1<<5,1<<5));
    _2dGrid2dBlockMatSum<<<(1<<5,1<<5),(1<<5, 1<<5)>>>(d_m1, d_m2, d_n, row, column);
    cudaDeviceSynchronize();
    cudaMemcpy(h_n2, d_n, row*column*sizeof(double), cudaMemcpyDeviceToHost);
    if(matEqual(h_n1, h_n2, row, column))
        printf("Matrices match.n");
    else
    {
        printf("Matrices don't match.nResult on CPU:n");
        printMatrix(h_n1, row, column);
        printf("Result on GPU:");
        printMatrix(h_n2, row, column);
    }
    free(h_m1);
    free(h_m2);
    free(h_n1);
    free(h_n2);
    cudaFree(d_m1);
    cudaFree(d_m2);
    cudaFree(d_n);
    cudaDeviceReset();
    return 0;
}