从传入的FORTRAN数组生成CUSP coo_matrix
Generating CUSP coo_matrix from passed FORTRAN arrays
我正致力于将CUSP求解器集成到现有的FORTRAN代码中。作为第一步,我只是试图从FORTRAN传递一对整数数组和一个浮点数(FORTRAN中的real*4),这将用于构建然后打印COO格式CUSP矩阵。
到目前为止,我已经能够遵循这个线程,并得到编译和链接的一切:使用IFORT与nvcc和CUSP
未解决的引用不幸的是,程序显然是在向CUSP矩阵发送垃圾,并最终崩溃,出现以下错误:
$./fort_cusp_test
testing 1 2 3
sparse matrix <1339222572, 1339222572> with 1339222568 entries
libc++abi.dylib: terminating with uncaught exception of type thrust::system::system_error: invalid argument
Program received signal SIGABRT: Process abort signal.
Backtrace for this error:
#0 0x10ff86ff6
#1 0x10ff86593
#2 0x7fff8593ff19
Abort trap: 6
cuda和fortran源代码的代码如下:
cusp_runner.cu
#include <stdio.h>
#include <cusp/coo_matrix.h>
#include <iostream>
#include <cusp/krylov/cg.h>
#include <cusp/print.h>
#if defined(__cplusplus)
extern "C" {
#endif
void test_coo_mat_print_(int * row_i, int * col_j, float * val_v, int n, int nnz ) {
//wrap raw input pointers with thrust::device_ptr
thrust::device_ptr<int> wrapped_device_I(row_i);
thrust::device_ptr<int> wrapped_device_J(col_j);
thrust::device_ptr<float> wrapped_device_V(val_v);
//use array1d_view to wrap individual arrays
typedef typename cusp::array1d_view< thrust::device_ptr<int> > DeviceIndexArrayView;
typedef typename cusp::array1d_view< thrust::device_ptr<float> > DeviceValueArrayView;
DeviceIndexArrayView row_indices(wrapped_device_I, wrapped_device_I + n);
DeviceIndexArrayView column_indices(wrapped_device_J, wrapped_device_J + nnz);
DeviceValueArrayView values(wrapped_device_V, wrapped_device_V + nnz);
//combine array1d_views into coo_matrix_view
typedef cusp::coo_matrix_view<DeviceIndexArrayView,DeviceIndexArrayView,DeviceValueArrayView> DeviceView;
//construct coo_matrix_view from array1d_views
DeviceView A(n,n,nnz,row_indices,column_indices,values);
cusp::print(A);
}
#if defined(__cplusplus)
}
#endif
fort_cusp_test.f90
program fort_cuda_test
implicit none
interface
subroutine test_coo_mat_print_(row_i,col_j,val_v,n,nnz) bind(C)
use, intrinsic :: ISO_C_BINDING, ONLY: C_INT,C_FLOAT
implicit none
integer(C_INT) :: n, nnz, row_i(:), col_j(:)
real(C_FLOAT) :: val_v(:)
end subroutine test_coo_mat_print_
end interface
integer*4 n
integer*4 nnz
integer*4, target :: rowI(9),colJ(9)
real*4, target :: valV(9)
integer*4, pointer :: row_i(:)
integer*4, pointer :: col_j(:)
real*4, pointer :: val_v(:)
n = 3
nnz = 9
rowI = (/ 1, 1, 1, 2, 2, 2, 3, 3, 3/)
colJ = (/ 1, 2, 3, 1, 2, 3, 1, 2, 3/)
valV = (/ 1, 2, 3, 4, 5, 6, 7, 8, 9/)
row_i => rowI
col_j => colJ
val_v => valV
write(*,*) "testing 1 2 3"
call test_coo_mat_print_(row_i,col_j,val_v,n,nnz)
end program fort_cuda_test
如果你想自己试一试,这是我的makefile(相当不优雅):
Test:
nvcc -Xcompiler="-fPIC" -shared cusp_runner.cu -o cusp_runner.so -I/Developer/NVIDIA/CUDA-6.5/include/cusp
gfortran -c fort_cusp_test.f90
gfortran fort_cusp_test.o cusp_runner.so -L/Developer/NVIDIA/CUDA-6.5/lib -lcudart -o fort_cusp_test
clean:
rm *.o *.so
库路径当然需要适当地修改。
谁能指出我在正确的方向如何正确地传递所需的数组从fortran代码?
删除接口块并在C函数开始处添加print语句后,我可以看到数组被正确传递,但是n和nnz引起了问题。我得到以下输出:
$ ./fort_cusp_test
testing 1 2 3
n: 1509677596, nnz: 1509677592
i, row_i, col_j, val_v
0, 1, 1, 1.0000e+00
1, 1, 2, 2.0000e+00
2, 1, 3, 3.0000e+00
3, 2, 1, 4.0000e+00
4, 2, 2, 5.0000e+00
5, 2, 3, 6.0000e+00
6, 3, 1, 7.0000e+00
7, 3, 2, 8.0000e+00
8, 3, 3, 9.0000e+00
9, 0, 32727, 0.0000e+00
...
etc
...
Program received signal SIGSEGV: Segmentation fault - invalid memory reference.
Backtrace for this error:
#0 0x105ce7ff6
#1 0x105ce7593
#2 0x7fff8593ff19
#3 0x105c780a2
#4 0x105c42dbc
#5 0x105c42df4
Segmentation fault: 11
fort_cusp_test
interface
subroutine test_coo_mat_print_(row_i,col_j,val_v,n,nnz) bind(C)
use, intrinsic :: ISO_C_BINDING, ONLY: C_INT,C_FLOAT
implicit none
integer(C_INT),value :: n, nnz
integer(C_INT) :: row_i(:), col_j(:)
real(C_FLOAT) :: val_v(:)
end subroutine test_coo_mat_print_
end interface
integer*4 n
integer*4 nnz
integer*4, target :: rowI(9),colJ(9)
real*4, target :: valV(9)
integer*4, pointer :: row_i(:)
integer*4, pointer :: col_j(:)
real*4, pointer :: val_v(:)
n = 3
nnz = 9
rowI = (/ 1, 1, 1, 2, 2, 2, 3, 3, 3/)
colJ = (/ 1, 2, 3, 1, 2, 3, 1, 2, 3/)
valV = (/ 1, 2, 3, 4, 5, 6, 7, 8, 9/)
row_i => rowI
col_j => colJ
val_v => valV
write(*,*) "testing 1 2 3"
call test_coo_mat_print_(rowI,colJ,valV,n,nnz)
end program fort_cuda_test
cusp_runner.cu
#include <stdio.h>
#include <cusp/coo_matrix.h>
#include <iostream>
// #include <cusp/krylov/cg.h>
#include <cusp/print.h>
#if defined(__cplusplus)
extern "C" {
#endif
void test_coo_mat_print_(int * row_i, int * col_j, float * val_v, int n, int nnz ) {
printf("n: %d, nnz: %dn",n,nnz);
printf("%6s, %6s, %6s, %12s n","i","row_i","col_j","val_v");
for(int i=0;i<n;i++) {
printf("%6d, %6d, %6d, %12.4en",i,row_i[i],col_j[i],val_v[i]);
}
if ( false ) {
//wrap raw input pointers with thrust::device_ptr
thrust::device_ptr<int> wrapped_device_I(row_i);
thrust::device_ptr<int> wrapped_device_J(col_j);
thrust::device_ptr<float> wrapped_device_V(val_v);
//use array1d_view to wrap individual arrays
typedef typename cusp::array1d_view< thrust::device_ptr<int> > DeviceIndexArrayView;
typedef typename cusp::array1d_view< thrust::device_ptr<float> > DeviceValueArrayView;
DeviceIndexArrayView row_indices(wrapped_device_I, wrapped_device_I + n);
DeviceIndexArrayView column_indices(wrapped_device_J, wrapped_device_J + nnz);
DeviceValueArrayView values(wrapped_device_V, wrapped_device_V + nnz);
//combine array1d_views into coo_matrix_view
typedef cusp::coo_matrix_view<DeviceIndexArrayView,DeviceIndexArrayView,DeviceValueArrayView> DeviceView;
//construct coo_matrix_view from array1d_views
DeviceView A(n,n,nnz,row_indices,column_indices,values);
cusp::print(A); }
}
#if defined(__cplusplus)
}
#endif
从Fortran传递参数到C例程有两种方法:第一种是使用接口块(在现代Fortran中是一种新方法),第二种是不使用接口块(一种即使在Fortran77中也有效的旧方法)。
首先,下面是关于使用接口块的第一种方法。因为C例程期望接收C指针(row_i, col_j和val_v),我们需要从Fortran端传递这些变量的地址。为此,我们必须在接口块中使用星号(*)而不是冒号(:),如下所示。(如果我们使用冒号,那么这会告诉Fortran编译器发送Fortran指针对象的地址[1],这不是期望的行为。)此外,由于C例程中的n和nnz被声明为值(而不是指针),接口块需要为这些变量具有VALUE属性,以便Fortran编译器发送n和nnz的值而不是它们的地址。总而言之,在第一种方法中,C和Fortran例程是这样的:
Fortran routine:
...
interface
subroutine test_coo_mat_print_(row_i,col_j,val_v,n,nnz) bind(C)
use, intrinsic :: ISO_C_BINDING, ONLY: C_INT,C_FLOAT
implicit none
integer(C_INT) :: row_i(*), col_j(*)
real(C_FLOAT) :: val_v(*)
integer(C_INT), value :: n, nnz !! see note [2] below also
end subroutine test_coo_mat_print_
end interface
...
call test_coo_mat_print_( rowI, colJ, valV, n, nnz )
C routine:
void test_coo_mat_print_ (int * row_i, int * col_j, float * val_v, int n, int nnz )
下面是关于没有接口块的第二种方法。在这种方法中,首先完全删除接口块和数组指针,并将Fortran代码修改如下
Fortran routine:
integer rowI( 9 ), colJ( 9 ), n, nnz !! no TARGET attribute necessary
real valV( 9 )
! ...set rowI etc as above...
call test_coo_mat_print ( rowI, colJ, valV, n, nnz ) !! "_" is dropped
和C例程如下
void test_coo_mat_print_ ( int* row_i, int* col_j, float* val_v, int* n_, int* nnz_ )
{
int n = *n_, nnz = *nnz_;
printf( "%d %d n", n, nnz );
for( int k = 0; k < 9; k++ ) {
printf( "%d %d %10.6f n", row_i[ k ], col_j[ k ], val_v[ k ] );
}
// now go to thrust...
}
注意n_和nnz_在C例程中被声明为指针,因为没有接口块,Fortran编译器总是将实际参数的地址发送给C例程。还要注意,在上面的C例程中,将打印row_i等的内容,以确保正确传递参数。如果打印的值是正确的,那么我猜问题将更有可能在调用推力例程(包括如何传递大小信息,如n和nnz)。
[1]声明为"real, pointer:: a(:)"的Fortran指针实际上代表了类似于数组视图类(用c++术语来说)的东西,这与实际指向的数据不同。这里需要发送的是实际数据的地址,而不是这个数组视图对象的地址。此外,接口块中的星号(a(*))表示一个假定大小的数组,这是在Fortran中传递数组的旧方法。在本例中,如预期的那样,传递数组第一个元素的地址。
[2]如果n和nnz在C例程中被声明为指针(如第二种方法),则不应该附加这个VALUE属性,因为C例程需要实际参数的地址而不是它们的值。