PE>#include 

PE>struct mystruct {
PE>  int x;
PE>};

PE>struct mystruct __far16 *func(void);

a 32 bit function that returns a 16 bit pointer?  How does it find a 16
bit pointer?  It should be
struct mystruct __far16 * __far16 __cdecl func(void);
to be a 16 bit function returning a 16 bit pointer.  However, if it
really is a 32 bit function, leave it as it is.

PE>int main(void)
PE>{
PE>    struct mystruct __far16 *a;

This is the problem. By the time "a" gets the pointer it should have
been thunked back to 32 bits.  It should be
struct mystruct *a;
The important thing is not that "a" is a 16 bit pointer.  You don't want
a 16 bit pointer.  You want a 32 bit pointer to a 16 bit data segment.
The important thing is that since the compiler knows that the function
returns a 16 bit pointer, it automatically converts it to the
appropriate 32 bit pointer.

PE>    a = func();
PE>    printf("x is %d\n", a->x);
PE>    return (0);
PE>}

I'm not sure exactly what you are doing here.  I have a solution that
works if my assumptions about your intentions are correct.  I have an
assembly routine that must be assembled into a 16 bit .OBJ file and
linked to the 32 bit .OBJ file that BCC creates from my .C code.

#include 

struct mystruct {
  int x;
};

struct mystruct  __far16 * __far16 __cdecl func(void);

int main(void)
{
    struct mystruct *a;

    a = func();
    printf("x is %d\n", a->x);
    return (0);
}
/*
        .386
_mine   segment word    use16   public  'DATA'
_var    dd      5               ;dword data in a 16 bit data segment
_mine   ends

_minetoo        segment word    use16   public  'CODE'
        assume  cs:_minetoo
        public  _func
_func   proc    far             ;returns a 16 bit pointer to data in 16
        mov     dx,seg _var     ;bit data segment
        mov     ax,offset _var
        ret
_func   endp
_minetoo        ends
        end
*/


.
--- QScan v1.14b / 01-0169