Function returning uint16
Richard Henderson
rth@redhat.com
Thu Jun 16 19:15:00 GMT 2016
On 06/16/2016 08:03 AM, Stéphane 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~
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