builtin_diff_tree(): make it obvious that function wants two entries
Instead of accepting an array and using exactly two elements from the array, take two single (struct object_array_entry *) arguments. Signed-off-by: Michael Haggerty <mhagger@alum.mit.edu> Signed-off-by: Junio C Hamano <gitster@pobox.com>
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@ -153,7 +153,8 @@ static int builtin_diff_index(struct rev_info *revs,
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static int builtin_diff_tree(struct rev_info *revs,
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static int builtin_diff_tree(struct rev_info *revs,
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int argc, const char **argv,
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int argc, const char **argv,
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struct object_array_entry *ent)
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struct object_array_entry *ent0,
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struct object_array_entry *ent1)
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{
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{
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const unsigned char *(sha1[2]);
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const unsigned char *(sha1[2]);
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int swap = 0;
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int swap = 0;
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@ -161,13 +162,14 @@ static int builtin_diff_tree(struct rev_info *revs,
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if (argc > 1)
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if (argc > 1)
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usage(builtin_diff_usage);
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usage(builtin_diff_usage);
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/* We saw two trees, ent[0] and ent[1].
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/*
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* if ent[1] is uninteresting, they are swapped
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* We saw two trees, ent0 and ent1. If ent1 is uninteresting,
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* swap them.
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*/
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*/
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if (ent[1].item->flags & UNINTERESTING)
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if (ent1->item->flags & UNINTERESTING)
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swap = 1;
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swap = 1;
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sha1[swap] = ent[0].item->sha1;
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sha1[swap] = ent0->item->sha1;
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sha1[1-swap] = ent[1].item->sha1;
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sha1[1-swap] = ent1->item->sha1;
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diff_tree_sha1(sha1[0], sha1[1], "", &revs->diffopt);
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diff_tree_sha1(sha1[0], sha1[1], "", &revs->diffopt);
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log_tree_diff_flush(revs);
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log_tree_diff_flush(revs);
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return 0;
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return 0;
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@ -403,7 +405,7 @@ int cmd_diff(int argc, const char **argv, const char *prefix)
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else if (ents == 1)
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else if (ents == 1)
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result = builtin_diff_index(&rev, argc, argv);
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result = builtin_diff_index(&rev, argc, argv);
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else if (ents == 2)
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else if (ents == 2)
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result = builtin_diff_tree(&rev, argc, argv, ent);
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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[0].item->flags & UNINTERESTING) {
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/*
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/*
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* diff A...B where there is at least one merge base
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* diff A...B where there is at least one merge base
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@ -412,8 +414,7 @@ int cmd_diff(int argc, const char **argv, const char *prefix)
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* between the base and B. Note that we pick one
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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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* merge base at random if there are more than one.
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*/
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*/
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ent[1] = ent[ents-1];
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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, ent);
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} else
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} else
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result = builtin_diff_combined(&rev, argc, argv,
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result = builtin_diff_combined(&rev, argc, argv,
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ent, ents);
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ent, ents);
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