return -1;
}
+/*
+ * This is a little subtle. We *only* have ->max_extent_size set if we actually
+ * searched through the bitmap and figured out the largest ->max_extent_size,
+ * otherwise it's 0. In the case that it's 0 we don't want to tell the
+ * allocator the wrong thing, we want to use the actual real max_extent_size
+ * we've found already if it's larger, or we want to use ->bytes.
+ *
+ * This matters because find_free_space() will skip entries who's ->bytes is
+ * less than the required bytes. So if we didn't search down this bitmap, we
+ * may pick some previous entry that has a smaller ->max_extent_size than we
+ * have. For example, assume we have two entries, one that has
+ * ->max_extent_size set to 4k and ->bytes set to 1M. A second entry hasn't set
+ * ->max_extent_size yet, has ->bytes set to 8k and it's contiguous. We will
+ * call into find_free_space(), and return with max_extent_size == 4k, because
+ * that first bitmap entry had ->max_extent_size set, but the second one did
+ * not. If instead we returned 8k we'd come in searching for 8k, and find the
+ * 8k contiguous range.
+ *
+ * Consider the other case, we have 2 8k chunks in that second entry and still
+ * don't have ->max_extent_size set. We'll return 16k, and the next time the
+ * allocator comes in it'll fully search our second bitmap, and this time it'll
+ * get an uptodate value of 8k as the maximum chunk size. Then we'll get the
+ * right allocation the next loop through.
+ */
static inline u64 get_max_extent_size(struct btrfs_free_space *entry)
{
- if (entry->bitmap)
+ if (entry->bitmap && entry->max_extent_size)
return entry->max_extent_size;
return entry->bytes;
}