From: Uno on
This is a bit of a repost. I wanted to clean up the context, refocus,
and thank parties who got me over the hump with my latest foray into
interop. Elliot helped me tremendously by giving me the proper query
syntax for google groups on c.l.f. I'll just show that again for
interested parties:

interop zax keyword site:groups.google.com/group/comp.lang.fortran

Erik managed to tell me -c without applying any of the flame opportunity
that existed. thx.

I've really been enjoying a chapter of MR&C that I hadn't read closely.
Now that I've gotten one good result with interop, I want more.

I believe that the declaration of pass on p. 259 is to enable the
passing of a 2-d array from C to fortran. Although a lot of this stuff
works in both directions, I am now focusing on C to fortran.

Critical to this treatment is this struct (Richard is of course correct
that it is no subroutine. It's funny how a person gets crossed up
between syntaxes.):

struct pass
{
int lenc, lenf;
float *c, f;

};

Since I got Michael Metcalf to straighten me out once on this, I'm
hoping to get another good break. I really don't know much about how
fortran arrays look like underneath the hood; I've never had the need to
look.

Anyways, this is what I've got now:

$ gcc -c -lm c_mm3.c -o caller.o
$ gfortran -c -Wall -Wextra f_mm2.f03 -o pass.o
$ gcc pass.o caller.o -lgfortran -o out
$ ./out
vector[0][0] is 0.785398
vector[0][1] is 2.107149
vector[0][2] is 3.249046
vector[0][3] is 4.325818
vector[0][4] is 5.373401
vector[0][5] is 6.405648
vector[0][6] is 7.428899
vector[1][0] is 2.107149
vector[1][1] is 3.249046
vector[1][2] is 4.325818
vector[1][3] is 5.373401
vector[1][4] is 6.405648
vector[1][5] is 7.428899
vector[1][6] is 8.446442
vector[2][0] is 3.249046
vector[2][1] is 4.325818
vector[2][2] is 5.373401
vector[2][3] is 6.405648
vector[2][4] is 7.428899
vector[2][5] is 8.446442
vector[2][6] is 9.460139
vector[3][0] is 4.325818
vector[3][1] is 5.373401
vector[3][2] is 6.405648
vector[3][3] is 7.428899
vector[3][4] is 8.446442
vector[3][5] is 9.460139
vector[3][6] is 10.471128
vector[4][0] is 5.373401
vector[4][1] is 6.405648
vector[4][2] is 7.428899
vector[4][3] is 8.446442
vector[4][4] is 9.460139
vector[4][5] is 10.471128
vector[4][6] is 11.480137
0.78539819 2.1071486 3.2490458 4.3258176
5.3734007
$

So, my C output doesn't look like my fortran output. In, particular,
the fortran output shows no dimensionality greater than one.

Here are the source listings:

$ cat c_mm3.c
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#define size1 5
#define size2 7
struct pass
{
int lenc, lenf;
float *c, f;

};
void simulation (struct pass *arrays);
int
main ()
{

float vector[size1][size2];
int i, j;
struct pass arrays;

for (i = 0; i < size1; ++i)
for (j = 0; j < size2; ++j)
{
{
vector[i][j] = atan (1 + i + j) + i + j;
printf ("vector[%d][%d] is %f\n", i, j, vector[i][j]);
}
}

arrays.lenc = size1;
arrays.lenf = size2;
arrays.c = &vector[0][0];
// arrays.f = NULL;

simulation (&arrays);
return 0;
}

// gcc -c -lm c_mm3.c -o caller.o
// gcc pass.o caller.o -lgfortran -o out
$ cat f_mm2.f03

subroutine simulation(arrays) bind(c)
use iso_c_binding
type, bind(c) :: pass
integer (c_int) :: lenc, lenf
type (c_ptr) :: c, f
end type pass
type (pass), intent(in) :: arrays
real (c_float), pointer :: c_array(:)

! associate c_array with an array allocated in C
call c_f_pointer( arrays%c, c_array, (/arrays%lenc/) )
print *, c_array
end subroutine simulation
! gfortran -c -Wall -Wextra f_mm2.f03 -o pass.o
$

Simple question: how do I get the fortran output to show the values
that the C output does *and* show up in the types of rows and columns
that a person wants who got an A in fortran 20 years ago and doesn't
feel like just taking a slopshot.

Thanks for your comments, and cheers,
--
Uno