Re: [Qemu-devel] [PATCH] linux-user: fix wait* syscall status returns

[Peter Maydell <peter.maydell@linaro.org>]

On 23 November 2011 22:03, Alexander Graf <agraf@suse.de> wrote:
> On 23.11.2011, at 22:55, Peter Maydell <peter.maydell@linaro.org> wrote:
>> If the problem is that waitpid() can return success without writing to
>> status, then this code is still not right because we will get the
>> initial target waitstatus into status, and then pass it through
>> host_to_target_waitstatus(), possibly modifying it, before writing
>> it back to guest memory.
>
> Yes. Maybe we should add a check if input_state != output_state and
> only then do the conversion?

I'm not sure this works (unless you go to the effort of implementing
target_to_host_waitstatus() which seems overkill to me) but I'm not
entirely sure what you're proposing to compare to what.

>> I think waitpid() will always and only write to status if the return
>> value is > 0 (ie it's a PID, not 0 or -1). So I think the right fix
>> for this problem is to have the if() protecting the put_user_s32()
>> read "if (ret && !is_error(ret) && ...".
>>
>> (ret == 0 is of course the WNOHANG-and-no-child-yet case you are hitting.)
>
> The man page wasn't really clear here. It sounded as if you can also
> get 0 as return value and still have status change. That's why I
> jumped through this hoop in the first place :)

I think POSIX is clear enough. Either
 * a child status is available, return pid and write to status
 * WNOHANG & not ready, return 0 (don't write to status)
 * -1 and set errno (don't write to status)

(strictly, whether status is written for EINTR is unpredictable, but I
think we can assume nobody relies on this and I bet the kernel folks
wouldn't worry too much about changing that between versions...)

-- PMM