On 2018-06-13 10:19, Dima Stepanov wrote:
> The qemu_memfd_alloc_check() routine allocates the fd variable on stack.
> This variable is initialized inside the qemu_memfd_alloc() function.
> There are several cases when *fd will be left unintialized which can
> lead to the unexpected close() in the qemu_memfd_free() call.
>
> Set file descriptor to -1 before calling the qemu_memfd_alloc routine.
>
> Signed-off-by: Dima Stepanov <dimas...@yandex-team.ru>
> ---
> util/memfd.c | 1 +
> 1 file changed, 1 insertion(+)
>
> diff --git a/util/memfd.c b/util/memfd.c
> index d248a53..6287946 100644
> --- a/util/memfd.c
> +++ b/util/memfd.c
> @@ -187,6 +187,7 @@ bool qemu_memfd_alloc_check(void)
> int fd;
(in case you respin: You could also write "int fd = -1;" in above line)
> void *ptr;
>
> + fd = -1;
> ptr = qemu_memfd_alloc("test", 4096, 0, &fd, NULL);
> memfd_check = ptr ? MEMFD_OK : MEMFD_KO;
> qemu_memfd_free(ptr, 4096, fd);
>
Reviewed-by: Thomas Huth <th...@redhat.com>
(FWIW: GCC complains with -O3 about this uninitialized variable, too, so
the build breaks with -Werror here - just noticed this today)
