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