问题
My main question is why arrays do such weird things and whether there is any way at all to do the following in a "clean" way.
I currently have a C program foo.c interfacing a Fortran program bar.f90 via dlopen/dlsym, roughly like in the code below:
foo.c:
#include <dlfcn.h>
#include <stdio.h>
int main()
{
int i, k = 4;
double arr[k];
char * e;
void * bar = dlopen("Code/Test/bar.so", RTLD_NOW | RTLD_LOCAL);
void (*allocArray)(int*);
*(void **)(&allocArray) = dlsym(bar, "__bar_MOD_allocarray");
void (*fillArray)(double*);
*(void **)(&fillArray) = dlsym(bar, "__bar_MOD_fillarray");
void (*printArray)(void);
*(void **)(&printArray) = dlsym(bar, "__bar_MOD_printarray");
double *a = (double*)dlsym(bar, "__bar_MOD_a");
for(i = 0; i < k; i++)
arr[i] = i * 3.14;
(*allocArray)(&k);
(*fillArray)(arr);
(*printArray)();
for(i = 0; i < 4; i++)
printf("%f ", a[i]);
printf("\n");
return 0;
}
bar.f90:
module bar
integer, parameter :: pa = selected_real_kind(15, 307)
real(pa), dimension(:), allocatable :: a
integer :: as
contains
subroutine allocArray(asize)
integer, intent(in) :: asize
as = asize
allocate(a(asize))
return
end subroutine
subroutine fillArray(values)
real(pa), dimension(as), intent(in) :: values
a = values
return
end subroutine
subroutine printArray()
write(*,*) a
return
end subroutine
end module
Running main yields
0.0000000000000000 3.1400000000000001 6.2800000000000002 9.4199999999999999
0.000000 -nan 0.000000 0.000000
which shows that Fortran allocates the array correctly and even correctly stores the given values but they are not accessible via dlsym anymore (working on that data results in segfaults). I also tried this for fixed-size arrays - the results stay the same.
Does anyone know the reason for this behaviour? I, personally, would have expected things to either work out bidirectional or alternatively not at all - this "Fortran accepting C arrays but not vice versa" makes me wonder whether there's some basic mistake I made on accessing the array from C in this fashion.
The other (and even more important) question is, how to do array accesses like these "the right way". Currently I'm not even sure if sticking to the "Fortran as .so" interface is a good way at all - I think it would also be possible to attempt mixed programming in this case. Nonetheless, the arrays issue remains - I read that this could be solved somehow using the ISO C Binding, but I couldn't figure out how, yet (I haven't worked a lot with Fortran, yet, especially not with said Binding), so help on this issue would be greatly appreciated.
Edit:
Okay, so I read into the ISO C Binding a little more and found a quite useful approach here. Using C_LOC I can get C pointers to my Fortran structures. Unfortunately, pointers to arrays seem to be pointers to pointers and need to be dereferenced in the C code before they can be treates as C arrays - or something like that.
Edit:
Got my program to work now using C binding the way Vladimir F pointed out, at least for the most part. The C file and Fortran files are now linked together, so I can avoid the libdl interface, at least for the Fortran part - I still need to load a dynamic C library, get a function pointer to one of the symbols in there and pass that as a function pointer to Fortran, which later calls that function as part of its calculation. As said function expects double*s [arrays], I couldn't manage to pass my Fortran arrays using C_LOC, strangely - neither C_LOC(array) nor C_LOC(array(1)) passed the correct pointers back to the C function. array(1) did the trick though. Sadly, this isn't the "cleanest" way to do this. If anyone got a hint for me how to do this using the C_LOC function, that would be great. Nonetheless I accept Vladimir F's answer, as I deem it to be the safer solution.
回答1:
In my opinion it is not good practice to try to access global data in Fortran library. It can be done using COMMON blocks, but they are evil and require statically sized arrays. Generally storage association is a bad thing.
Never access the module symbols as "__bar_MOD_a" they are compiler specific and not meant to be used directly. Pass poiters using functions and subroutines.
Pass the array as a subroutine argument. You can also allocate the array in C and pass it to Fortran. What can be also done is getting a pointer to the first element of the array. It will serve es the C pointer to the array.
My solution, for simplicity without the .so, it is trivial to add it:
bar.f90
module bar
use iso_C_binding
implicit none
integer, parameter :: pa = selected_real_kind(15, 307)
real(pa), dimension(:), allocatable,target :: a
integer :: as
contains
subroutine allocArray(asize,ptr) bind(C,name="allocArray")
integer, intent(in) :: asize
type(c_ptr),intent(out) :: ptr
as = asize
allocate(a(asize))
ptr = c_loc(a(1))
end subroutine
subroutine fillArray(values) bind(C,name="fillArray")
real(pa), dimension(as), intent(in) :: values
a = values
end subroutine
subroutine printArray() bind(C,name="printArray")
write(*,*) a
end subroutine
end module
main.c
#include <dlfcn.h>
#include <stdio.h>
int main()
{
int i, k = 4;
double arr[k];
char * e;
double *a;
void allocArray(int*,double**);
void fillArray(double*);
void allocArray();
for(i = 0; i < k; i++)
arr[i] = i * 3.14;
allocArray(&k,&a);
fillArray(arr);
printArray();
for(i = 0; i < 4; i++)
printf("%f ", a[i]);
printf("\n");
return 0;
}
compile and run:
gcc -c -g main.c
gfortran -c -g -fcheck=all bar.f90
gfortran main.o bar.o
./a.out
0.0000000000000000 3.1400000000000001 6.2800000000000002 9.4199999999999999
0.000000 3.140000 6.280000 9.420000
Note: There is no reason for the returns in your Fortran subroutines, they only obscure the code.
回答2:
Many Fortran compilers use internally something called array descriptors - structures that hold the shape of the array (that is the size and range of each dimension as well as pointer to the real data). It allows for the implementation of things like assumed-shape array arguments, array pointers and allocatable arrays to work. What you are accessing through the __bar_MOD_a symbol is the descriptor of the allocatable array and not its data.
Array descriptors are compiler-specific and code that relies on specific descriptor format is not portable. Example descriptors:
- GNU Fortran
- Intel Fortran
Note that even those are specific to some versions of those compilers. Intel, for example, states that their current descriptor format is not compatible with the format used in Intel Fortran 7.0.
If you look at both descriptors you would see that they are largerly similar and that the first element is a pointer to the array data. So you would be able to easily read the data using double ** instead of double *:
double **a_descr = (double**)dlsym(bar, "__bar_MOD_a");
...
for(i = 0; i < 4; i++)
printf("%f ", (*a_descr)[i]);
Once again, this is not portable as the format of those descriptors might change in the future (although I doubt that the data pointer would be moved somewhere else than at the beginning of the descriptor). There is a draft specification that tries to unify all descriptor formats but it is not clear how and when it will be adopted by the different compiler vendors.
Edit: Here is how to use an accessor function that uses C_LOC() from the ISO_C_BINDING module to portably obtain a pointer to the allocatable array:
Fortran code:
module bar
use iso_c_binding
...
! Note that the array should be a pointer target
real(pa), dimension(:), allocatable, target :: a
...
contains
...
function getArrayPtr() result(cptr)
type(c_ptr) :: cptr
cptr = c_loc(a)
end function
end module
C code:
...
void * (*getArrayPtr)(void);
*(void **)(&getArrayPtr) = dlsym(bar, "__bar_MOD_getarrayptr");
...
double *a = (*getArrayPtr)();
for(i = 0; i < 4; i++)
printf("%f ", a[i]);
...
Result:
$ ./prog.x
0.0000000000000000 3.1400000000000001 6.2800000000000002
9.4199999999999999
0.000000 3.140000 6.280000 9.420000
来源：https://stackoverflow.com/questions/11934822/how-to-access-dynamically-allocated-fortran-arrays-in-c