我在cuda中调用dgetrf遇到麻烦。从我发现的内容来看,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,返回的错误值为7,但我找不到该错误代码的对应枚举。下面是我的代码,任何帮助将不胜感激;
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");
}
cublas的错误代码7为CUBLAS_STATUS_INVALID_VALUE
。cublas中的矩阵求逆仅适用于平方矩阵,因此我认为m == n
在您的情况下。话虽这么说,函数cublas<t>getrfBatched
要求数据透视表数组的长度必须n
适合每个矩阵,所以您应该分配d_pivot_array
为:
error = cudaMalloc((void **)&d_pivot_array, n * sizeof(int));
为了更通用,将其分配为:
error = cudaMalloc((void **)&d_pivot_array, n * batchSize * sizeof(int));
这是我在测试CUBLAS函数时编写的方阵求逆代码。函数输入和输出是float
设备上分配的类型平方矩阵。
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