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cmd_diff(): use an object_array for holding trees

Change cmd_diff() to use a (struct object_array) for holding the trees
that it accumulates, rather than rolling its own equivalent.

Incidentally, this change removes a hard-coded limit of 100 trees in
combined diff, not that it matters in practice.

Signed-off-by: Michael Haggerty <mhagger@alum.mit.edu>
Signed-off-by: Junio C Hamano <gitster@pobox.com>
This commit is contained in:
Michael Haggerty 2013-05-25 11:08:04 +02:00 committed by Junio C Hamano
parent 91de344d76
commit 33055fa823

View File

@ -253,8 +253,8 @@ int cmd_diff(int argc, const char **argv, const char *prefix)
{
int i;
struct rev_info rev;
struct object_array_entry ent[100];
int ents = 0, blobs = 0, paths = 0;
struct object_array ent = OBJECT_ARRAY_INIT;
int blobs = 0, paths = 0;
const char *path = NULL;
struct blobinfo blob[2];
int nongit;
@ -351,13 +351,8 @@ int cmd_diff(int argc, const char **argv, const char *prefix)
if (obj->type == OBJ_COMMIT)
obj = &((struct commit *)obj)->tree->object;
if (obj->type == OBJ_TREE) {
if (ARRAY_SIZE(ent) <= ents)
die(_("more than %d trees given: '%s'"),
(int) ARRAY_SIZE(ent), name);
obj->flags |= flags;
ent[ents].item = obj;
ent[ents].name = name;
ents++;
add_object_array(obj, name, &ent);
continue;
}
if (obj->type == OBJ_BLOB) {
@ -381,7 +376,7 @@ int cmd_diff(int argc, const char **argv, const char *prefix)
/*
* Now, do the arguments look reasonable?
*/
if (!ents) {
if (!ent.nr) {
switch (blobs) {
case 0:
result = builtin_diff_files(&rev, argc, argv);
@ -402,22 +397,26 @@ int cmd_diff(int argc, const char **argv, const char *prefix)
}
else if (blobs)
usage(builtin_diff_usage);
else if (ents == 1)
else if (ent.nr == 1)
result = builtin_diff_index(&rev, argc, argv);
else if (ents == 2)
result = builtin_diff_tree(&rev, argc, argv, &ent[0], &ent[1]);
else if (ent[0].item->flags & UNINTERESTING) {
else if (ent.nr == 2)
result = builtin_diff_tree(&rev, argc, argv,
&ent.objects[0], &ent.objects[1]);
else if (ent.objects[0].item->flags & UNINTERESTING) {
/*
* diff A...B where there is at least one merge base
* between A and B. We have ent[0] == merge-base,
* ent[ents-2] == A, and ent[ents-1] == B. Show diff
* between the base and B. Note that we pick one
* merge base at random if there are more than one.
* between A and B. We have ent.objects[0] ==
* merge-base, ent.objects[ents-2] == A, and
* ent.objects[ents-1] == B. Show diff between the
* base and B. Note that we pick one merge base at
* random if there are more than one.
*/
result = builtin_diff_tree(&rev, argc, argv, &ent[0], &ent[ents-1]);
result = builtin_diff_tree(&rev, argc, argv,
&ent.objects[0],
&ent.objects[ent.nr-1]);
} else
result = builtin_diff_combined(&rev, argc, argv,
ent, ents);
ent.objects, ent.nr);
result = diff_result_code(&rev.diffopt, result);
if (1 < rev.diffopt.skip_stat_unmatch)
refresh_index_quietly();