Yes, but from C side, a StablePtr is already a void*. (See HsFFI.h which typedefs HsStablePtr to void*)
So its sufficient for a use as an opaque pointer, no need to cast it.

So what is the use of casting it to a Ptr () if this doesn't allow to access the memory space addressed?


2012/2/12 Antoine Latter <aslatter@gmail.com>
On Sun, Feb 12, 2012 at 8:18 AM, Yves Parès <yves.pares@gmail.com> wrote:
> Hello,
>
> According to the documentation
> (http://hackage.haskell.org/packages/archive/base/4.5.0.0/doc/html/Foreign-StablePtr.html),
> StablePtrs aims at being opaque on C-side.
> But they provide functions to be casted to/from regular void*'s.
> Does that mean if for instance you have a StablePtr CInt you can cast it to
> Ptr () and alter it on C-side?
>
> void alter(void* data)
> {
>     int* x = (int*)data;
>     *x = 42;
> }
>
> --------------------------------------------------
>
> -- using 'unsafe' doesn't change anything.
> foreign import ccall safe "alter"
>     alter :: Ptr () -> IO ()
>
> main = do
>     sptr <- newStablePtr (0 :: CInt)
>     deRefStablePtr sptr >>= print
>     alter (castStablePtrToPtr sptr)  -- SEGFAULTS!
>     deRefStablePtr sptr >>= print
>     freeStablePtr sptr
>
>
> But I tried it, and it doesn't work: I got a segfault when 'alter' is
> called.
>

I think that 'castStablePtrToPtr' exists because many C APIs use
'void*' to mean 'opaque lump of data', and these exist to conform to
that sort of API.

Antoine