我试图使用缓冲库来求解一些线性方程组,但却提出了一个异常,这是非常奇怪的。代码只使用库中的一个函数,其余的是内存分配和内存复制。功能是
cusolverSpScsrlsvcholHost(
cusolverSpHandle_t handle, int m, int nnz,
const cusparseMatDescr_t descrA, const float *csrVal,
const int *csrRowPtr, const int *csrColInd, const float *b,
float tol, int reorder, float *x, int *singularity); 我认为我的问题可能在于tol - reorder -奇异参数,其余的是矩阵参数,这是代码:
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <cuda.h>
#include <cusparse.h>
#include <cublas_v2.h>
#include <stdio.h>
#include <cusolverSp.h>
int main()
{
//initialize our test cases
const int size = 3;
int nnz = 6 ;
int sing = -1 ;
//float values[] = {0,0,0,0} ;
float values[] = {1,2,3,4,5,6} ;
int colIdx[] = {0,0,1,0,1,2};
int rowPtr[] = {0, 1,3,7};
float x[] = {4,-6,7};
float y[3]= {0,0,0} ;
float *dev_values = 0 ;
int *dev_rowPtr = 0 ;
int *dev_colIdx = 0 ;
float *dev_x = 0 ;
float *dev_y = 0 ;
cusolverSpHandle_t solver_handle ;
cusolverSpCreate(&solver_handle) ;
cusparseMatDescr_t descr = 0;
cusparseCreateMatDescr(&descr);
cusparseSetMatType(descr,CUSPARSE_MATRIX_TYPE_GENERAL);
cusparseSetMatIndexBase(descr,CUSPARSE_INDEX_BASE_ZERO);
// Choose which GPU to run on, change this on a multi-GPU system.
cudaSetDevice(0);
cudaEvent_t start, stop;
float time;
cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaEventRecord(start, 0);
// Allocate GPU buffers for three vectors (two input, one output) .
cudaMalloc((void**)&dev_x, size * sizeof(float));
cudaMalloc((void**)&dev_y, size * sizeof(float));
cudaMalloc((void**)&dev_values, nnz * sizeof(float));
cudaMalloc((void**)&dev_rowPtr, (size + 1) * sizeof(int));
cudaMalloc((void**)&dev_colIdx, nnz * sizeof(int));
//Memcpy
cudaMemcpyAsync(dev_x, x, size * sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpyAsync(dev_values, values, nnz * sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpyAsync(dev_rowPtr, rowPtr, (size + 1) * sizeof(int), cudaMemcpyHostToDevice);
cudaMemcpyAsync(dev_colIdx, colIdx, nnz * sizeof(int), cudaMemcpyHostToDevice);
cudaMemcpyAsync(dev_y, y, size * sizeof(float), cudaMemcpyHostToDevice);
cusolverSpScsrlsvluHost(solver_handle, size, nnz, descr, dev_values, dev_rowPtr, dev_colIdx, dev_y, 0,0, dev_x, &sing);
cudaMemcpyAsync(y, dev_y, size*sizeof(float), cudaMemcpyDeviceToHost );
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
cudaEventElapsedTime(&time, start, stop);
printf ("Time for the kernel: %f ms\n", time);
printf("%f\n",y[0]);
printf("%f\n",y[1]);
printf("%f\n",y[2]);
// cudaDeviceReset must be called before exiting in order for profiling and
// tracing tools such as Nsight and Visual Profiler to show complete traces.
cudaDeviceReset();
cudaFree(dev_x);
cudaFree(dev_y);
cudaFree(dev_values);
cudaFree(dev_rowPtr);
cudaFree(dev_colIdx);
return 1;
}发布于 2015-11-22 21:24:16
在您的代码中至少有三个问题:
cusolverSpScsrlsvluHost().主机变体:如果您查看文献资料 for cusolverSpScsrlsvluHost,您会发现对于主机MemSpace,该函数希望所有参数和指针参数都是基于主机的。但是,您正在传递指向该函数的设备指针。通过这样做,你将得到节段错误。对于所有像dev_values这样的参数,您需要用等效的主机数据指针(例如,values代替dev_values)替换那些参数。y和x错误地传递给cusolverSpScsrlsvluHost()。既然你已经在x中放置了非零值,那么大概你打算把它作为你的RHS向量。该向量在文档中使用b名称,并且它是要传递的第一个向量。那么,您的y向量大概就是解向量,它是按照参数的顺序传递的最后一个向量(它在文档中使用了名称x )。以下代码解决了上述问题,并产生了合理的结果:
$ cat t979.cu
#include <cusparse.h>
#include <stdio.h>
#include <cusolverSp.h>
#include <assert.h>
int main()
{
//initialize our test cases
const int size = 3;
const int nnz = 6 ;
int sing = 0;
//float values[] = {0,0,0,0} ;
float values[nnz] = {1,2,3,4,5,6} ;
int colIdx[nnz] = {0,0,1,0,1,2};
int rowPtr[size+1] = {0, 1,3,6};
float x[size] = {4,-6,7};
float y[size]= {0,0,0} ;
cusolverStatus_t cso;
cusolverSpHandle_t solver_handle ;
cso = cusolverSpCreate(&solver_handle) ;
assert(cso == CUSOLVER_STATUS_SUCCESS);
cusparseStatus_t csp;
cusparseMatDescr_t descr = 0;
csp = cusparseCreateMatDescr(&descr);
assert(csp == CUSPARSE_STATUS_SUCCESS);
csp = cusparseSetMatType(descr,CUSPARSE_MATRIX_TYPE_GENERAL);
assert(csp == CUSPARSE_STATUS_SUCCESS);
csp = cusparseSetMatIndexBase(descr,CUSPARSE_INDEX_BASE_ZERO);
assert(csp == CUSPARSE_STATUS_SUCCESS);
cso = cusolverSpScsrlsvluHost(solver_handle, size, nnz, descr, values, rowPtr, colIdx, x, 0.0,0, y, &sing);
assert(cso == CUSOLVER_STATUS_SUCCESS);
printf("%f\n",y[0]);
printf("%f\n",y[1]);
printf("%f\n",y[2]);
return 0;
}
$ nvcc -o t979 t979.cu -lcusolver -lcusparse
$ ./t979
4.000000
-4.666667
2.388889
$还请注意,有一个完全工作的CUDA 样本代码,它演示了正确使用此函数。
https://stackoverflow.com/questions/33859363
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