FFI: how to pass pointers in C to C functions?

Question

I am struggling with libffi (Function Foreign Interface library), probably, due to the lack of C experience. I have the following program that calls a function my_func dynamically using libffi:

#include <stdio.h>
#include <stdlib.h>
#include <ffi.h>

unsigned my_func(double a, double b, double *res) {
    res[0] = a + b;
    res[1] = a - b;

    return 0;
}


int main(int argc, char *argv[])
{
    ffi_cif cif;
    ffi_type *arg_types[3] = {
        &ffi_type_double,
        &ffi_type_double,
        &ffi_type_pointer
    };
    ffi_type *rettype = &ffi_type_uint;

    if (ffi_prep_cif(&cif, FFI_DEFAULT_ABI, 2, rettype, arg_types) != FFI_OK) {
        fprintf(stderr, "ffi_prep_cif is not successful\n");
        exit(EXIT_FAILURE);
    }

    double a = 3.0;
    double b = 2.0;
    double res[2] = {99.0, 15.0};
    void *arg_values[3] = {
        &a,
        &b,
        res
    };

    unsigned status;
    ffi_call(&cif, FFI_FN(my_func), &status, arg_values);

    printf("Function return status code %u\n", status);
    printf("Values in res array: \n");
    printf("[0] = %f\n", res[0]);
    printf("[1] = %f\n", res[1]);
    
    return 0;
}

When I debug the program, it seems that all the arg_values are set correctly:

(gdb) p ((double *) arg_values[2])[0]
$1 = 99
(gdb) p ((double *) arg_values[2])[1]
$2 = 15

You can see above that the arg_values third values is correctly set to the res array.

However, when I am inside the function my_func, the third argument (the res array) becomes NULL:

Breakpoint 2, my_func (a=3, b=2, res=0x0) at test_ffi.c:6
6           res[0] = a + b;

I do not really understand what is wrong here. Could somebody explain to me, why this happens and how to fix it? Thank you!

Update

Thanks to the comments by @selbie, the following modified program works (note that we obtain a second pointer to the array res and pass its address to FFI):

#include <stdio.h>
#include <stdlib.h>
#include <ffi.h>

unsigned my_func(double a, double b, double *res) {
    res[0] = a + b;
    res[1] = a - b;

    return 0;
}

int main(int argc, char *argv[]) {
    ffi_cif cif;
    ffi_type *arg_types[3] = {
        &ffi_type_double,
        &ffi_type_double,
        &ffi_type_pointer
    };
    ffi_type *rettype = &ffi_type_uint;

    if (ffi_prep_cif(&cif, FFI_DEFAULT_ABI, 3, rettype, arg_types) != FFI_OK) {
        fprintf(stderr, "ffi_prep_cif is not successful\n");
        exit(EXIT_FAILURE);
    }

    double a = 3.0;
    double b = 2.0;
    double res[2] = {99.0, 15.0};
    double *p_res = res;
    void *arg_values[3] = {
        &a,
        &b,
        // res
        &p_res
    };

    unsigned status;
    ffi_call(&cif, FFI_FN(my_func), &status, arg_values);

    printf("Function return status code %u\n", status);
    printf("Values in res array: \n");
    printf("[0] = %f\n", res[0]);
    printf("[1] = %f\n", res[1]);
    
    return 0;
}

Answer 1

Shouldn't this:

if (ffi_prep_cif(&cif, FFI_DEFAULT_ABI, 2, rettype, arg_types) != FFI_OK)

Be this:

if (ffi_prep_cif(&cif, FFI_DEFAULT_ABI, 3, rettype, arg_types) != FFI_OK)

Given that you are invoking a function that expects three parameters, but only initialized the ffi stuff to expect 2 parameters, it shouldn't be a surprised that one parameter goes missing.