[email protected] writes:
> + /*
> + * Fetch the tree from the ODB for each peer directory in the
> + * n commits.
> + *
> + * For 2- and 3-way traversals, we try to avoid hitting the
> + * ODB twice for the same OID. This should yield a nice speed
> + * up in checkouts and merges when the commits are similar.
> + *
> + * We don't bother doing the full O(n^2) search for larger n,
> + * because wider traversals don't happen that often and we
> + * avoid the search setup.
> + *
> + * When 2 peer OIDs are the same, we just copy the tree
> + * descriptor data. This implicitly borrows the buffer
> + * data from the earlier cell.
This "borrowing" made me worried, but it turns out that this is
perfectly OK.
fill_tree_descriptor() uses read_sha1_file() to give a tree_desc its
own copy of the tree object data, so the code that calls into the
tree traversal API is responsible for freeing the buffer returned
from fill_tree_descriptor(). The updated code avoids double freeing
by setting buf[i] to NULL for borrowing [i].
> + */
> for (i = 0; i < n; i++, dirmask >>= 1) {
> - const unsigned char *sha1 = NULL;
> - if (dirmask & 1)
> - sha1 = names[i].oid->hash;
> - buf[i] = fill_tree_descriptor(t+i, sha1);
> + if (i > 0 && are_same_oid(&names[i], &names[i - 1])) {
> + t[i] = t[i - 1];
> + buf[i] = NULL;
> + } else if (i > 1 && are_same_oid(&names[i], &names[i - 2])) {
> + t[i] = t[i - 2];
> + buf[i] = NULL;
> + } else {
> + const unsigned char *sha1 = NULL;
> + if (dirmask & 1)
> + sha1 = names[i].oid->hash;
> + buf[i] = fill_tree_descriptor(t+i, sha1);
> + }
> }
>
> bottom = switch_cache_bottom(&newinfo);
