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git/merge-base.c
Junio C Hamano 9585e40622 Try to find the optimum merge base while resolving.
The merge-base command acquires a new option, '--all', that causes it
to output all the common ancestor candidates.  The "git resolve"
command then uses it to pick the optimum merge base by picking the one
that results in the smallest number of nontrivial merges.

Signed-off-by: Junio C Hamano <junkio@cox.net>
2005-08-23 21:08:59 -07:00

172 lines
4.0 KiB
C

#include <stdlib.h>
#include "cache.h"
#include "commit.h"
#define PARENT1 1
#define PARENT2 2
#define UNINTERESTING 4
static struct commit *interesting(struct commit_list *list)
{
while (list) {
struct commit *commit = list->item;
list = list->next;
if (commit->object.flags & UNINTERESTING)
continue;
return commit;
}
return NULL;
}
/*
* A pathological example of how this thing works.
*
* Suppose we had this commit graph, where chronologically
* the timestamp on the commit are A <= B <= C <= D <= E <= F
* and we are trying to figure out the merge base for E and F
* commits.
*
* F
* / \
* E A D
* \ / /
* B /
* \ /
* C
*
* First we push E and F to list to be processed. E gets bit 1
* and F gets bit 2. The list becomes:
*
* list=F(2) E(1), result=empty
*
* Then we pop F, the newest commit, from the list. Its flag is 2.
* We scan its parents, mark them reachable from the side that F is
* reachable from, and push them to the list:
*
* list=E(1) D(2) A(2), result=empty
*
* Next pop E and do the same.
*
* list=D(2) B(1) A(2), result=empty
*
* Next pop D and do the same.
*
* list=C(2) B(1) A(2), result=empty
*
* Next pop C and do the same.
*
* list=B(1) A(2), result=empty
*
* Now it is B's turn. We mark its parent, C, reachable from B's side,
* and push it to the list:
*
* list=C(3) A(2), result=empty
*
* Now pop C and notice it has flags==3. It is placed on the result list,
* and the list now contains:
*
* list=A(2), result=C(3)
*
* We pop A and do the same.
*
* list=B(3), result=C(3)
*
* Next, we pop B and something very interesting happens. It has flags==3
* so it is also placed on the result list, and its parents are marked
* uninteresting, retroactively, and placed back on the list:
*
* list=C(7), result=C(7) B(3)
*
* Now, list does not have any interesting commit. So we find the newest
* commit from the result list that is not marked uninteresting. Which is
* commit B.
*/
static int show_all = 0;
static int merge_base(struct commit *rev1, struct commit *rev2)
{
struct commit_list *list = NULL;
struct commit_list *result = NULL;
if (rev1 == rev2) {
printf("%s\n", sha1_to_hex(rev1->object.sha1));
return 0;
}
parse_commit(rev1);
parse_commit(rev2);
rev1->object.flags |= 1;
rev2->object.flags |= 2;
insert_by_date(rev1, &list);
insert_by_date(rev2, &list);
while (interesting(list)) {
struct commit *commit = list->item;
struct commit_list *tmp = list, *parents;
int flags = commit->object.flags & 7;
list = list->next;
free(tmp);
if (flags == 3) {
insert_by_date(commit, &result);
/* Mark parents of a found merge uninteresting */
flags |= UNINTERESTING;
}
parents = commit->parents;
while (parents) {
struct commit *p = parents->item;
parents = parents->next;
if ((p->object.flags & flags) == flags)
continue;
parse_commit(p);
p->object.flags |= flags;
insert_by_date(p, &list);
}
}
if (!result)
return 1;
while (result) {
struct commit *commit = result->item;
result = result->next;
if (commit->object.flags & UNINTERESTING)
continue;
printf("%s\n", sha1_to_hex(commit->object.sha1));
if (!show_all)
return 0;
commit->object.flags |= UNINTERESTING;
}
return 0;
}
static const char merge_base_usage[] =
"git-merge-base [--all] <commit-id> <commit-id>";
int main(int argc, char **argv)
{
struct commit *rev1, *rev2;
unsigned char rev1key[20], rev2key[20];
while (1 < argc && argv[1][0] == '-') {
char *arg = argv[1];
if (!strcmp(arg, "-a") || !strcmp(arg, "--all"))
show_all = 1;
else
usage(merge_base_usage);
argc--; argv++;
}
if (argc != 3 ||
get_sha1(argv[1], rev1key) ||
get_sha1(argv[2], rev2key))
usage(merge_base_usage);
rev1 = lookup_commit_reference(rev1key);
rev2 = lookup_commit_reference(rev2key);
if (!rev1 || !rev2)
return 1;
return merge_base(rev1, rev2);
}