[pypy-dev] Call rpython from outside

[Maciej Fijalkowski]

Hi Armin

We ended up (Aleksandr is here at pycon russia) using rffi_platform to
get the exact shape of the structure from the header file. There were
a few bugs how exactly this got mapped, so it ended up being a good
way to do it.

On Sat, Jul 15, 2017 at 8:02 AM, Armin Rigo <armin.rigo at gmail.com> wrote:
> Hi Aleksandr,
>
> On 11 July 2017 at 18:33, Aleksandr Koshkin <tinysnippets at gmail.com> wrote:
>> So ok, I have to specify headers containing my structs and somehow push it
>> to rpython toolchain, if I got you correctly.
>> 0. Why? This structures are already described in the vm file as a bunch of
>> ffi.CStruct objects.
>
> rffi.CStruct() is used to declare the RPython interface for structs
> that are originally defined in C.
>
> You can use lltype.Struct(), but it's not recommended in your case
> because lltype.Struct() is meant to define structs in RPython where
> you *don't* need a precise C-level struct; for example,
> lltype.Struct() could rename and reorder the fields in C if it is more
> efficient.
>
> We don't have a direct way to declare the struct in RPython but also
> force it to generate exactly the C struct declaration you want,
> because we never needed it.  You need to use rffi.CStruct() and write
> the struct in the .h file manually too.
>
>> 1. If I have to, how would I do that, is there any example of embedding
>> rpython into something?
>
> Not really.  Look maybe at tests using rpython.rlib.entrypoint, like
> rpython/translator/c/test/test_genc:test_entrypoints.
>
>
> A bientôt,
>
> Armin.