Re: Function returning uint16

[Richard Henderson <rth at redhat dot com>]

On 06/16/2016 08:03 AM, StÃphane Glondu wrote:
> Hello,
> 
> I am trying to debug
> 
>   https://github.com/ocamllabs/ocaml-ctypes/issues/404
> 
> and I realized that the following code (on amd64):
> 
> #include <stdio.h>
> #include <stdint.h>
> #include <ffi.h>
> 
> uint16_t retrieve() {
>   return 0x4242;
> }
> 
> int main() {
>   uint16_t r[2] = { 0xdead, 0xbeef };
>   ffi_cif cif;
>   printf("r = {%x, %x}\n", r[0], r[1]);
>   ffi_prep_cif(&cif, FFI_DEFAULT_ABI, 0, &ffi_type_uint16, NULL);
>   ffi_call(&cif, FFI_FN(retrieve), &r[0], NULL);
>   printf("r = {%x, %x}\n", r[0], r[1]);
>   return 0;
> }
> 
> returns:
> 
> r = {dead, beef}
> r = {4242, 0}
> 
> Is that expected? I don't expect r[1] to be overwritten...

It is expected.  In the documentation,

In most situations, @samp{libffi} will handle promotion according to
the ABI.  However, for historical reasons, there is a special case
with return values that must be handled by your code.  In particular,
for integral (not @code{struct}) types that are narrower than the
system register size, the return value will be widened by
@samp{libffi}.  @samp{libffi} provides a type, @code{ffi_arg}, that
can be used as the return type.  For example, if the CIF was defined
with a return type of @code{char}, @samp{libffi} will try to store a
full @code{ffi_arg} into the return value.


r~