Is it possible to call a Fortran interface from C++

Question

I have the following code that does not compile. Is it possible to call the Fortran interface as overloaded functions in C++, as I try below?

This is the Fortran code:

module functions
    use, intrinsic :: iso_c_binding, only : c_double, c_int
    implicit none

    interface increment bind(c, name="increment")
        module procedure increment_int, increment_double
    end interface

contains
    subroutine increment_int(a, b)
        integer(c_int), intent(inout) :: a
        integer(c_int), value :: b

        a = a + b
    end subroutine

    subroutine increment_double(a, b)
        real(c_double), intent(inout) :: a
        real(c_double), value :: b

        a = a + b
    end subroutine
end module functions

And this is the C++ code:

#include <iostream>

namespace
{
    extern "C" void increment(int&, int);
    extern "C" void increment(double&, double);
}

int main()
{
    int a = 6;
    const int b = 2;

    double c = 6.;
    const int d = 2.;

    increment(a, b);
    increment(c, d);

    std::cout << "a = " << a << std::endl;
    std::cout << "c = " << c << std::endl;

    return 0;
}

Answer 1

No, it cannot be avoided. But you can instantiate a template to call the two. Or just make a C++ generic wrapper to those two extern C functions. You definitely cannot make Fortran to export the interface. Interface is just a description how to call the two subroutines under some name internally in Fortran.

This question importing interface module procedure from fortran into C is very similar. I even originally closed yours as a duplicate, but then I changed my mind because the attempted solution is slightly different.

But the principle is the same. Fortran generics and C++ generics are not compatible. And you have C between them which has no generics of similar type.

Note: As @RichardCritten suggest you also should not pass by reference in an extern C function. The compiler probably compiles it and implements as passing a pointer by value, but it is not guaranteed. Just pass a pointer. See C++ by-reference argument and C linkage for more.