midx: do not require packs to be sorted in lexicographic order

The MIDX file format currently requires that pack files be identified by
the lexicographic ordering of their names (that is, a pack having a
checksum beginning with "abc" would have a numeric pack_int_id which is
smaller than the same value for a pack beginning with "bcd").

As a result, it is impossible to combine adjacent MIDX layers together
without permuting bits from bitmaps that are in more recent layer(s).

To see why, consider the following example:

          | packs       | preferred pack
  --------+-------------+---------------
  MIDX #0 | { X, Y, Z } | Y
  MIDX #1 | { A, B, C } | B
  MIDX #2 | { D, E, F } | D

, where MIDX #2's base MIDX is MIDX #1, and so on. Suppose that we want
to combine MIDX layers #0 and #1, to create a new layer #0' containing
the packs from both layers. With the original three MIDX layers, objects
are laid out in the bitmap in the order they appear in their source
pack, and the packs themselves are arranged according to the pseudo-pack
order. In this case, that ordering is Y, X, Z, B, A, C.

But recall that the pseudo-pack ordering is defined by the order that
packs appear in the MIDX, with the exception of the preferred pack,
which sorts ahead of all other packs regardless of its position within
the MIDX. In the above example, that means that pack 'Y' could be placed
anywhere (so long as it is designated as preferred), however, all other
packs must be placed in the location listed above.

Because that ordering isn't sorted lexicographically, it is impossible
to compact MIDX layers in the above configuration without permuting the
object-to-bit-position mapping. Changing this mapping would affect all
bitmaps belonging to newer layers, rendering the bitmaps associated with
MIDX #2 unreadable.

One of the goals of MIDX compaction is that we are able to shrink the
length of the MIDX chain *without* invalidating bitmaps that belong to
newer layers, and the lexicographic ordering constraint is at odds with
this goal.

However, packs do not *need* to be lexicographically ordered within the
MIDX. As far as I can gather, the only reason they are sorted lexically
is to make it possible to perform a binary search over the pack names in
a MIDX, necessary to make `midx_contains_pack()`'s performance
logarithmic in the number of packs rather than linear.

Relax this constraint by allowing MIDX writes to proceed with packs that
are not arranged in lexicographic order. `midx_contains_pack()` will
lazily instantiate a `pack_names_sorted` array on the MIDX, which will
be used to implement the binary search over pack names.

This change produces MIDXs which may not be correctly read with external
tools or older versions of Git. Though older versions of Git know how to
gracefully degrade and ignore any MIDX(s) they consider corrupt,
external tools may not be as robust. To avoid unintentionally breaking
any such tools, guard this change behind a version bump in the MIDX's
on-disk format.

Signed-off-by: Taylor Blau <me@ttaylorr.com>
Signed-off-by: Junio C Hamano <gitster@pobox.com>

1 files changed, 22 insertions, 4 deletions

diff --git a/midx-write.c b/midx-write.c
index 8a54644e42..5c8700065a 100644
--- a/midx-write.c
+++ b/midx-write.c
@@ -36,10 +36,13 @@ extern int cmp_idx_or_pack_name(const char *idx_or_pack_name,
 
 static size_t write_midx_header(const struct git_hash_algo *hash_algo,
 				struct hashfile *f, unsigned char num_chunks,
-				uint32_t num_packs)
+				uint32_t num_packs, int version)
 {
+	if (version != MIDX_VERSION_V1 && version != MIDX_VERSION_V2)
+		BUG("unexpected MIDX version: %d", version);
+
 	hashwrite_be32(f, MIDX_SIGNATURE);
-	hashwrite_u8(f, MIDX_VERSION);
+	hashwrite_u8(f, version);
 	hashwrite_u8(f, oid_version(hash_algo));
 	hashwrite_u8(f, num_chunks);
 	hashwrite_u8(f, 0); /* unused */
@@ -105,6 +108,8 @@ struct write_midx_context {
 
 	uint32_t preferred_pack_idx;
 
+	int version; /* must be MIDX_VERSION_V1 or _V2 */
+
 	int incremental;
 	uint32_t num_multi_pack_indexes_before;
 
@@ -410,7 +415,9 @@ static int write_midx_pack_names(struct hashfile *f, void *data)
 		if (ctx->info[i].expired)
 			continue;
 
-		if (i && strcmp(ctx->info[i].pack_name, ctx->info[i - 1].pack_name) <= 0)
+		if (ctx->version == MIDX_VERSION_V1 &&
+		    i && strcmp(ctx->info[i].pack_name,
+				ctx->info[i - 1].pack_name) <= 0)
 			BUG("incorrect pack-file order: %s before %s",
 			    ctx->info[i - 1].pack_name,
 			    ctx->info[i].pack_name);
@@ -1026,6 +1033,12 @@ static bool midx_needs_update(struct multi_pack_index *midx, struct write_midx_c
 		goto out;
 
 	/*
+	 * If the version differs, we need to update.
+	 */
+	if (midx->version != ctx->version)
+		goto out;
+
+	/*
 	 * Ignore incremental updates for now. The assumption is that any
 	 * incremental update would be either empty (in which case we will bail
 	 * out later) or it would actually cover at least one new pack.
@@ -1100,6 +1113,7 @@ static int write_midx_internal(struct write_midx_opts *opts)
 	struct tempfile *incr;
 	struct write_midx_context ctx = {
 		.preferred_pack_idx = NO_PREFERRED_PACK,
+		.version = MIDX_VERSION_V2,
 	 };
 	struct multi_pack_index *midx_to_free = NULL;
 	int bitmapped_packs_concat_len = 0;
@@ -1114,6 +1128,10 @@ static int write_midx_internal(struct write_midx_opts *opts)
 	ctx.repo = r;
 	ctx.source = opts->source;
 
+	repo_config_get_int(ctx.repo, "midx.version", &ctx.version);
+	if (ctx.version != MIDX_VERSION_V1 && ctx.version != MIDX_VERSION_V2)
+		die(_("unknown MIDX version: %d"), ctx.version);
+
 	ctx.incremental = !!(opts->flags & MIDX_WRITE_INCREMENTAL);
 
 	if (ctx.incremental)
@@ -1445,7 +1463,7 @@ static int write_midx_internal(struct write_midx_opts *opts)
 	}
 
 	write_midx_header(r->hash_algo, f, get_num_chunks(cf),
-			  ctx.nr - dropped_packs);
+			  ctx.nr - dropped_packs, ctx.version);
 	write_chunkfile(cf, &ctx);
 
 	finalize_hashfile(f, midx_hash, FSYNC_COMPONENT_PACK_METADATA,