+/* If a referenced allocation is explicitly freed the new owner
+ should be the same as if the same allocation is implicitly freed
+ (because it's owner was freed).
+ Traditionally in talloc an explicit free will free the top non-child reference
+ but an implicit free will move the top (any) reference to be the new owner */
+static bool test_implicit_explicit_free(void)
+{
+ void *root, *p1, *p2, *p3, *ref, *r1;
+ int e, i;
+
+ printf("test: test_implicit_explicit_free\n# SINGLE REFERENCE IMPLICIT FREE\n");
+
+ root = talloc_named_const(NULL, 0, "root");
+ p1 = talloc_named_const(root, 1, "p1");
+ p2 = talloc_named_const(p1, 1, "p2");
+ /* Now root owns p1, and p1 owns p2 */
+
+ r1 = talloc_named_const(root, 1, "r1");
+ ref = talloc_reference(r1, p2);
+ /* now r1 has ref reference to p2 */
+ talloc_report_full(root, stderr);
+
+ CHECK_BLOCKS(__FUNCTION__, p1, 2);
+ CHECK_BLOCKS(__FUNCTION__, p2, 1);
+ CHECK_BLOCKS(__FUNCTION__, r1, 2);
+
+ fprintf(stderr, "Freeing p2\n");
+ talloc_free(p2);
+ /* how many blocks is r1 taking against p2 ? */
+ e=talloc_total_blocks(r1);
+
+ talloc_report_full(root, stderr);
+ talloc_free(root);
+
+ /* now repeat, but this time free p1 */
+ printf("test: test_implicit_explicit_free\n# SINGLE REFERENCE EXPLICIT FREE\n");
+
+ root = talloc_named_const(NULL, 0, "root");
+ p1 = talloc_named_const(root, 1, "p1");
+ p2 = talloc_named_const(p1, 1, "p2");
+ /* Now root owns p1, and p1 owns p2 */
+
+ r1 = talloc_named_const(root, 1, "r1");
+ ref = talloc_reference(r1, p2);
+ /* now r1 has ref reference to p2 */
+ talloc_report_full(NULL, stderr);
+
+ CHECK_BLOCKS(__FUNCTION__, p1, 2);
+ CHECK_BLOCKS(__FUNCTION__, p2, 1);
+ CHECK_BLOCKS(__FUNCTION__, r1, 2);
+
+ fprintf(stderr, "Freeing p1\n");
+ talloc_free(p1);
+ /* how many blocks is r1 taking against p2 ? */
+ i=talloc_total_blocks(r1);
+ talloc_report_full(NULL, stderr);
+
+ CHECK_BLOCKS(__FUNCTION__,r1, e);
+
+ talloc_free(root);
+
+ printf("success: ref1\n");
+ return true;
+}
+