From: Dima Stepanov <dimas...@yandex-team.ru>
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>
Reviewed-by: Marc-André Lureau <marcandre.lur...@redhat.com>
Reviewed-by: Thomas Huth <th...@redhat.com>
Signed-off-by: Thomas Huth <th...@redhat.com>
---
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;
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);
--
1.8.3.1
