Cublas矩阵LU分解

Question

我给库达的dgetrf打电话有点麻烦。根据我所发现的，我只能称之为批处理版本(http://docs.nvidia.com/cuda/cublas/#cublas-lt-t-gt-getrfbatched)。当我调用它时，我得到了一个返回的错误值7，但没有找到对应的该错误代码的枚举。下面是我的代码，任何帮助都将不胜感激；

void cuda_matrix_inverse (int m, int n, double* a){

    cublasHandle_t handle;
    cublasStatus_t status;
    double **devPtrA = 0;
    double **devPtrA_dev = NULL;
    int *d_pivot_array;
    int *d_info_array;
    int rowsA = m;
    int colsA = n;
    int matrixSizeA;
    cudaError_t error;

    fprintf(stderr,"starting cuda inverse\n");

    error = cudaMalloc((void **)&d_pivot_array, sizeof(int));
    if (error != cudaSuccess) fprintf(stderr,"\nError: %s\n",cudaGetErrorString(error));
    error = cudaMalloc((void **)&d_info_array, sizeof(int));
    if (error != cudaSuccess) fprintf(stderr,"\nError: %s\n",cudaGetErrorString(error));

    fprintf(stderr,"malloced pivot and info\n");

    status = cublasCreate(&handle);
    if (status != CUBLAS_STATUS_SUCCESS) fprintf(stderr,"error %i\n",status);

    matrixSizeA = rowsA * colsA;

    devPtrA =(double **)malloc(1 * sizeof(*devPtrA));

    fprintf(stderr,"malloced devPtrA\n");

    error = cudaMalloc((void **)&devPtrA[0], matrixSizeA * sizeof(devPtrA[0][0]));
    if (error != cudaSuccess) fprintf(stderr,"\nError: %s\n",cudaGetErrorString(error));

    error = cudaMalloc((void **)&devPtrA_dev, 1 * sizeof(*devPtrA));
    if (error != cudaSuccess) fprintf(stderr,"\nError: %s\n",cudaGetErrorString(error));

    fprintf(stderr,"malloced device variables\n");

    error = cudaMemcpy(devPtrA_dev, devPtrA, 1 * sizeof(*devPtrA), cudaMemcpyHostToDevice);
    if (error != cudaSuccess) fprintf(stderr,"\nError: %s\n",cudaGetErrorString(error));

    fprintf(stderr,"copied from devPtrA to d_devPtrA\n");

    status = cublasSetMatrix(rowsA, colsA, sizeof(a[0]), a, rowsA, devPtrA[0], rowsA);
    if (status != CUBLAS_STATUS_SUCCESS) fprintf(stderr,"error %i\n",status);


    status = cublasDgetrfBatched(handle, m, devPtrA_dev,m,d_pivot_array,d_info_array,1); //cannot get this to work
    if (status != CUBLAS_STATUS_SUCCESS) fprintf(stderr,"error in dgetrf %i\n",status);


    fprintf(stderr,"done with cuda inverse\n");
}

Answer 1

cublas的错误代码7表示CUBLAS_STATUS_INVALID_VALUE。矩阵在cublas中的反演对于方阵是可能的，所以我假设在您的情况下是m == n。尽管如此，函数cublas<t>getrfBatched要求每个矩阵的枢轴数组的长度为n，因此您应该将d_pivot_array分配为：

error = cudaMalloc((void **)&d_pivot_array, n * sizeof(int));

为了更通用，它被分配为：

error = cudaMalloc((void **)&d_pivot_array, n * batchSize * sizeof(int));

Here是我在测试CUBLAS函数时编写的平方矩阵逆代码。功能输入和输出是分配在设备上的float型方阵。