diff A...B: give one possible diff when there are more than one merge-base

We instead showed a combined diff that explains one of the randomly
chosen merge-base as if it were the result of merging all the other
merge bases and two tips given, which made no sense at all.

An alternative is to simply fail such a request, telling the user that
there are criss-cross merges, but it wouldn't be so helpful.

Noticed by James Pickens.

Signed-off-by: Junio C Hamano <gitster@pobox.com>

diff --git a/builtin-diff.c b/builtin-diff.c
index 2e51f408f9f3399195604fd23d430c180a7f20a4..62b483bd5fa66e6828f18c9081157cfeb68a9a5f 100644 (file)
--- a/builtin-diff.c
+++ b/builtin-diff.c
@@ -405,17 +405,32 @@ int cmd_diff(int argc, const char **argv, const char *prefix)
                result = builtin_diff_index(&rev, argc, argv);
        else if (ents == 2)
                result = builtin_diff_tree(&rev, argc, argv, ent);
-       else if ((ents == 3) && (ent[0].item->flags & UNINTERESTING)) {
-               /* diff A...B where there is one sane merge base between
-                * A and B.  We have ent[0] == merge-base, ent[1] == A,
-                * and ent[2] == B.  Show diff between the base and B.
+       else if (ent[0].item->flags & UNINTERESTING) {
+               /*
+                * Perhaps the user gave us A...B, which expands
+                * to a list of negative merge bases followed by
+                * A (symmetric-left) and B?  Let's make sure...
                 */
-               ent[1] = ent[2];
+               for (i = 1; i < ents; i++)
+                       if (!(ent[i].item->flags & UNINTERESTING))
+                               break;
+               if (ents != i + 2 ||
+                   (ent[i+1].item->flags & UNINTERESTING) ||
+                   (!(ent[i].item->flags & SYMMETRIC_LEFT)) ||
+                   (ent[i+1].item->flags & SYMMETRIC_LEFT))
+                       die("what do you mean by that?");
+               /*
+                * 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.
+                */
+               ent[1] = ent[ents-1];
                result = builtin_diff_tree(&rev, argc, argv, ent);
-       }
-       else
+       } else
                result = builtin_diff_combined(&rev, argc, argv,
-                                            ent, ents);
+                                              ent, ents);
        result = diff_result_code(&rev.diffopt, result);
        if (1 < rev.diffopt.skip_stat_unmatch)
                refresh_index_quietly();