oidmap: make oidmap_free independent of struct layout

c8e424c (hashmap: introduce hashmap_free_entries, 2019-10-06)
introduced hashmap_free_entries(), which can free any struct
pointer, regardless of the hashmap_entry field offset.

oidmap does not make use of this flexibilty, hardcoding the offset to
zero instead. Let's fix this by passing struct type and member to
hashmap_free_entries().

Additionally, remove an erroneous semi-colon at the end of
hashmap_free_entries() macro.

Signed-off-by: Abhishek Kumar <[email protected]>
Signed-off-by: Junio C Hamano <[email protected]>

Author: abhishekkumar2718

hashmap.h

@@ -245,7 +245,7 @@ void hashmap_free_(struct hashmap *map, ssize_t offset);
  * where @member is the hashmap_entry struct used to associate with @map
  */
 #define hashmap_free_entries(map, type, member) \
-	hashmap_free_(map, offsetof(type, member));
+	hashmap_free_(map, offsetof(type, member))
 
 /* hashmap_entry functions */
 

oidmap.c

@@ -26,8 +26,10 @@ void oidmap_free(struct oidmap *map, int free_entries)
 	if (!map)
 		return;
 
-	/* TODO: make oidmap itself not depend on struct layouts */
-	hashmap_free_(&map->map, free_entries ? 0 : -1);
+	if (free_entries)
+		hashmap_free_entries(&map->map, struct oidmap_entry, internal_entry);
+	else
+		hashmap_free(&map->map);
 }
 
 void *oidmap_get(const struct oidmap *map, const struct object_id *key)