memory: expose alignment used for allocating RAM as MemoryRegion API

introduce memory_region_get_alignment() that returns
underlying memory block alignment or 0 if it's not
relevant/implemented for backend.

Signed-off-by: Igor Mammedov <imammedo@redhat.com>
Reviewed-by: Michael S. Tsirkin <mst@redhat.com>
Signed-off-by: Michael S. Tsirkin <mst@redhat.com>

1 files changed, 2 insertions, 1 deletions

diff --git a/include/qemu/osdep.h b/include/qemu/osdep.h
index c032434..b3300cc 100644
--- a/include/qemu/osdep.h
+++ b/include/qemu/osdep.h
@@ -5,6 +5,7 @@
 #include <stdarg.h>
 #include <stddef.h>
 #include <stdbool.h>
+#include <stdint.h>
 #include <sys/types.h>
 #ifdef __OpenBSD__
 #include <sys/signal.h>
@@ -103,7 +104,7 @@ typedef signed int              int_fast16_t;
 int qemu_daemon(int nochdir, int noclose);
 void *qemu_try_memalign(size_t alignment, size_t size);
 void *qemu_memalign(size_t alignment, size_t size);
-void *qemu_anon_ram_alloc(size_t size);
+void *qemu_anon_ram_alloc(size_t size, uint64_t *align);
 void qemu_vfree(void *ptr);
 void qemu_anon_ram_free(void *ptr, size_t size);