mirror of
https://github.com/git/git.git
synced 2024-12-12 03:14:11 +08:00
e7da938570
Use of the `the_repository` variable is deprecated nowadays, and we slowly but steadily convert the codebase to not use it anymore. Instead, callers should be passing down the repository to work on via parameters. It is hard though to prove that a given code unit does not use this variable anymore. The most trivial case, merely demonstrating that there is no direct use of `the_repository`, is already a bit of a pain during code reviews as the reviewer needs to manually verify claims made by the patch author. The bigger problem though is that we have many interfaces that implicitly rely on `the_repository`. Introduce a new `USE_THE_REPOSITORY_VARIABLE` macro that allows code units to opt into usage of `the_repository`. The intent of this macro is to demonstrate that a certain code unit does not use this variable anymore, and to keep it from new dependencies on it in future changes, be it explicit or implicit For now, the macro only guards `the_repository` itself as well as `the_hash_algo`. There are many more known interfaces where we have an implicit dependency on `the_repository`, but those are not guarded at the current point in time. Over time though, we should start to add guards as required (or even better, just remove them). Define the macro as required in our code units. As expected, most of our code still relies on the global variable. Nearly all of our builtins rely on the variable as there is no way yet to pass `the_repository` to their entry point. For now, declare the macro in "biultin.h" to keep the required changes at least a little bit more contained. Signed-off-by: Patrick Steinhardt <ps@pks.im> Signed-off-by: Junio C Hamano <gitster@pobox.com>
371 lines
8.7 KiB
C
371 lines
8.7 KiB
C
#define USE_THE_REPOSITORY_VARIABLE
|
|
|
|
#include "git-compat-util.h"
|
|
#include "hex.h"
|
|
#include "match-trees.h"
|
|
#include "strbuf.h"
|
|
#include "tree.h"
|
|
#include "tree-walk.h"
|
|
#include "object-store-ll.h"
|
|
|
|
static int score_missing(unsigned mode)
|
|
{
|
|
int score;
|
|
|
|
if (S_ISDIR(mode))
|
|
score = -1000;
|
|
else if (S_ISLNK(mode))
|
|
score = -500;
|
|
else
|
|
score = -50;
|
|
return score;
|
|
}
|
|
|
|
static int score_differs(unsigned mode1, unsigned mode2)
|
|
{
|
|
int score;
|
|
|
|
if (S_ISDIR(mode1) != S_ISDIR(mode2))
|
|
score = -100;
|
|
else if (S_ISLNK(mode1) != S_ISLNK(mode2))
|
|
score = -50;
|
|
else
|
|
score = -5;
|
|
return score;
|
|
}
|
|
|
|
static int score_matches(unsigned mode1, unsigned mode2)
|
|
{
|
|
int score;
|
|
|
|
/* Heh, we found SHA-1 collisions between different kind of objects */
|
|
if (S_ISDIR(mode1) != S_ISDIR(mode2))
|
|
score = -100;
|
|
else if (S_ISLNK(mode1) != S_ISLNK(mode2))
|
|
score = -50;
|
|
|
|
else if (S_ISDIR(mode1))
|
|
score = 1000;
|
|
else if (S_ISLNK(mode1))
|
|
score = 500;
|
|
else
|
|
score = 250;
|
|
return score;
|
|
}
|
|
|
|
static void *fill_tree_desc_strict(struct tree_desc *desc,
|
|
const struct object_id *hash)
|
|
{
|
|
void *buffer;
|
|
enum object_type type;
|
|
unsigned long size;
|
|
|
|
buffer = repo_read_object_file(the_repository, hash, &type, &size);
|
|
if (!buffer)
|
|
die("unable to read tree (%s)", oid_to_hex(hash));
|
|
if (type != OBJ_TREE)
|
|
die("%s is not a tree", oid_to_hex(hash));
|
|
init_tree_desc(desc, hash, buffer, size);
|
|
return buffer;
|
|
}
|
|
|
|
static int base_name_entries_compare(const struct name_entry *a,
|
|
const struct name_entry *b)
|
|
{
|
|
return base_name_compare(a->path, tree_entry_len(a), a->mode,
|
|
b->path, tree_entry_len(b), b->mode);
|
|
}
|
|
|
|
/*
|
|
* Inspect two trees, and give a score that tells how similar they are.
|
|
*/
|
|
static int score_trees(const struct object_id *hash1, const struct object_id *hash2)
|
|
{
|
|
struct tree_desc one;
|
|
struct tree_desc two;
|
|
void *one_buf = fill_tree_desc_strict(&one, hash1);
|
|
void *two_buf = fill_tree_desc_strict(&two, hash2);
|
|
int score = 0;
|
|
|
|
for (;;) {
|
|
int cmp;
|
|
|
|
if (one.size && two.size)
|
|
cmp = base_name_entries_compare(&one.entry, &two.entry);
|
|
else if (one.size)
|
|
/* two lacks this entry */
|
|
cmp = -1;
|
|
else if (two.size)
|
|
/* two has more entries */
|
|
cmp = 1;
|
|
else
|
|
break;
|
|
|
|
if (cmp < 0) {
|
|
/* path1 does not appear in two */
|
|
score += score_missing(one.entry.mode);
|
|
update_tree_entry(&one);
|
|
} else if (cmp > 0) {
|
|
/* path2 does not appear in one */
|
|
score += score_missing(two.entry.mode);
|
|
update_tree_entry(&two);
|
|
} else {
|
|
/* path appears in both */
|
|
if (!oideq(&one.entry.oid, &two.entry.oid)) {
|
|
/* they are different */
|
|
score += score_differs(one.entry.mode,
|
|
two.entry.mode);
|
|
} else {
|
|
/* same subtree or blob */
|
|
score += score_matches(one.entry.mode,
|
|
two.entry.mode);
|
|
}
|
|
update_tree_entry(&one);
|
|
update_tree_entry(&two);
|
|
}
|
|
}
|
|
free(one_buf);
|
|
free(two_buf);
|
|
return score;
|
|
}
|
|
|
|
/*
|
|
* Match one itself and its subtrees with two and pick the best match.
|
|
*/
|
|
static void match_trees(const struct object_id *hash1,
|
|
const struct object_id *hash2,
|
|
int *best_score,
|
|
char **best_match,
|
|
const char *base,
|
|
int recurse_limit)
|
|
{
|
|
struct tree_desc one;
|
|
void *one_buf = fill_tree_desc_strict(&one, hash1);
|
|
|
|
while (one.size) {
|
|
const char *path;
|
|
const struct object_id *elem;
|
|
unsigned short mode;
|
|
int score;
|
|
|
|
elem = tree_entry_extract(&one, &path, &mode);
|
|
if (!S_ISDIR(mode))
|
|
goto next;
|
|
score = score_trees(elem, hash2);
|
|
if (*best_score < score) {
|
|
free(*best_match);
|
|
*best_match = xstrfmt("%s%s", base, path);
|
|
*best_score = score;
|
|
}
|
|
if (recurse_limit) {
|
|
char *newbase = xstrfmt("%s%s/", base, path);
|
|
match_trees(elem, hash2, best_score, best_match,
|
|
newbase, recurse_limit - 1);
|
|
free(newbase);
|
|
}
|
|
|
|
next:
|
|
update_tree_entry(&one);
|
|
}
|
|
free(one_buf);
|
|
}
|
|
|
|
/*
|
|
* A tree "oid1" has a subdirectory at "prefix". Come up with a tree object by
|
|
* replacing it with another tree "oid2".
|
|
*/
|
|
static int splice_tree(const struct object_id *oid1, const char *prefix,
|
|
const struct object_id *oid2, struct object_id *result)
|
|
{
|
|
char *subpath;
|
|
int toplen;
|
|
char *buf;
|
|
unsigned long sz;
|
|
struct tree_desc desc;
|
|
unsigned char *rewrite_here;
|
|
const struct object_id *rewrite_with;
|
|
struct object_id subtree;
|
|
enum object_type type;
|
|
int status;
|
|
|
|
subpath = strchrnul(prefix, '/');
|
|
toplen = subpath - prefix;
|
|
if (*subpath)
|
|
subpath++;
|
|
|
|
buf = repo_read_object_file(the_repository, oid1, &type, &sz);
|
|
if (!buf)
|
|
die("cannot read tree %s", oid_to_hex(oid1));
|
|
init_tree_desc(&desc, oid1, buf, sz);
|
|
|
|
rewrite_here = NULL;
|
|
while (desc.size) {
|
|
const char *name;
|
|
unsigned short mode;
|
|
|
|
tree_entry_extract(&desc, &name, &mode);
|
|
if (strlen(name) == toplen &&
|
|
!memcmp(name, prefix, toplen)) {
|
|
if (!S_ISDIR(mode))
|
|
die("entry %s in tree %s is not a tree", name,
|
|
oid_to_hex(oid1));
|
|
|
|
/*
|
|
* We cast here for two reasons:
|
|
*
|
|
* - to flip the "char *" (for the path) to "unsigned
|
|
* char *" (for the hash stored after it)
|
|
*
|
|
* - to discard the "const"; this is OK because we
|
|
* know it points into our non-const "buf"
|
|
*/
|
|
rewrite_here = (unsigned char *)(desc.entry.path +
|
|
strlen(desc.entry.path) +
|
|
1);
|
|
break;
|
|
}
|
|
update_tree_entry(&desc);
|
|
}
|
|
if (!rewrite_here)
|
|
die("entry %.*s not found in tree %s", toplen, prefix,
|
|
oid_to_hex(oid1));
|
|
if (*subpath) {
|
|
struct object_id tree_oid;
|
|
oidread(&tree_oid, rewrite_here, the_repository->hash_algo);
|
|
status = splice_tree(&tree_oid, subpath, oid2, &subtree);
|
|
if (status)
|
|
return status;
|
|
rewrite_with = &subtree;
|
|
} else {
|
|
rewrite_with = oid2;
|
|
}
|
|
hashcpy(rewrite_here, rewrite_with->hash, the_repository->hash_algo);
|
|
status = write_object_file(buf, sz, OBJ_TREE, result);
|
|
free(buf);
|
|
return status;
|
|
}
|
|
|
|
/*
|
|
* We are trying to come up with a merge between one and two that
|
|
* results in a tree shape similar to one. The tree two might
|
|
* correspond to a subtree of one, in which case it needs to be
|
|
* shifted down by prefixing otherwise empty directories. On the
|
|
* other hand, it could cover tree one and we might need to pick a
|
|
* subtree of it.
|
|
*/
|
|
void shift_tree(struct repository *r,
|
|
const struct object_id *hash1,
|
|
const struct object_id *hash2,
|
|
struct object_id *shifted,
|
|
int depth_limit)
|
|
{
|
|
char *add_prefix;
|
|
char *del_prefix;
|
|
int add_score, del_score;
|
|
|
|
/*
|
|
* NEEDSWORK: this limits the recursion depth to hardcoded
|
|
* value '2' to avoid excessive overhead.
|
|
*/
|
|
if (!depth_limit)
|
|
depth_limit = 2;
|
|
|
|
add_score = del_score = score_trees(hash1, hash2);
|
|
add_prefix = xcalloc(1, 1);
|
|
del_prefix = xcalloc(1, 1);
|
|
|
|
/*
|
|
* See if one's subtree resembles two; if so we need to prefix
|
|
* two with a few fake trees to match the prefix.
|
|
*/
|
|
match_trees(hash1, hash2, &add_score, &add_prefix, "", depth_limit);
|
|
|
|
/*
|
|
* See if two's subtree resembles one; if so we need to
|
|
* pick only subtree of two.
|
|
*/
|
|
match_trees(hash2, hash1, &del_score, &del_prefix, "", depth_limit);
|
|
|
|
/* Assume we do not have to do any shifting */
|
|
oidcpy(shifted, hash2);
|
|
|
|
if (add_score < del_score) {
|
|
/* We need to pick a subtree of two */
|
|
unsigned short mode;
|
|
|
|
if (!*del_prefix)
|
|
return;
|
|
|
|
if (get_tree_entry(r, hash2, del_prefix, shifted, &mode))
|
|
die("cannot find path %s in tree %s",
|
|
del_prefix, oid_to_hex(hash2));
|
|
return;
|
|
}
|
|
|
|
if (!*add_prefix)
|
|
return;
|
|
|
|
splice_tree(hash1, add_prefix, hash2, shifted);
|
|
}
|
|
|
|
/*
|
|
* The user says the trees will be shifted by this much.
|
|
* Unfortunately we cannot fundamentally tell which one to
|
|
* be prefixed, as recursive merge can work in either direction.
|
|
*/
|
|
void shift_tree_by(struct repository *r,
|
|
const struct object_id *hash1,
|
|
const struct object_id *hash2,
|
|
struct object_id *shifted,
|
|
const char *shift_prefix)
|
|
{
|
|
struct object_id sub1, sub2;
|
|
unsigned short mode1, mode2;
|
|
unsigned candidate = 0;
|
|
|
|
/* Can hash2 be a tree at shift_prefix in tree hash1? */
|
|
if (!get_tree_entry(r, hash1, shift_prefix, &sub1, &mode1) &&
|
|
S_ISDIR(mode1))
|
|
candidate |= 1;
|
|
|
|
/* Can hash1 be a tree at shift_prefix in tree hash2? */
|
|
if (!get_tree_entry(r, hash2, shift_prefix, &sub2, &mode2) &&
|
|
S_ISDIR(mode2))
|
|
candidate |= 2;
|
|
|
|
if (candidate == 3) {
|
|
/* Both are plausible -- we need to evaluate the score */
|
|
int best_score = score_trees(hash1, hash2);
|
|
int score;
|
|
|
|
candidate = 0;
|
|
score = score_trees(&sub1, hash2);
|
|
if (score > best_score) {
|
|
candidate = 1;
|
|
best_score = score;
|
|
}
|
|
score = score_trees(&sub2, hash1);
|
|
if (score > best_score)
|
|
candidate = 2;
|
|
}
|
|
|
|
if (!candidate) {
|
|
/* Neither is plausible -- do not shift */
|
|
oidcpy(shifted, hash2);
|
|
return;
|
|
}
|
|
|
|
if (candidate == 1)
|
|
/*
|
|
* shift tree2 down by adding shift_prefix above it
|
|
* to match tree1.
|
|
*/
|
|
splice_tree(hash1, shift_prefix, hash2, shifted);
|
|
else
|
|
/*
|
|
* shift tree2 up by removing shift_prefix from it
|
|
* to match tree1.
|
|
*/
|
|
oidcpy(shifted, &sub2);
|
|
}
|