undef/git-commit-vandalism
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Do a very simple "merge-base" that finds the most recent common

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parent of two commits.

The question of "best" commit can probably be tweaked almost arbitrarily.
In particular, trying to take things like how big the tree differences
are into account migt be a good idea. This one is just very simple.
This commit is contained in:
Linus Torvalds 2005-04-17 12:18:17 -07:00
parent 15000d7899
commit 6683463ed6
2 changed files with 61 additions and 1 deletions
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5
Makefile
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@ -14,7 +14,7 @@ CC=gcc
PROG= update-cache show-diff init-db write-tree read-tree commit-tree \
cat-file fsck-cache checkout-cache diff-tree rev-tree show-files \
check-files ls-tree
check-files ls-tree merge-base
all: $(PROG)
@ -64,6 +64,9 @@ check-files: check-files.o read-cache.o
ls-tree: ls-tree.o read-cache.o
$(CC) $(CFLAGS) -o ls-tree ls-tree.o read-cache.o $(LIBS)
merge-base: merge-base.o read-cache.o
$(CC) $(CFLAGS) -o merge-base merge-base.o read-cache.o $(LIBS)
read-cache.o: cache.h
show-diff.o: cache.h

57
merge-base.c Normal file
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@ -0,0 +1,57 @@
#include "cache.h"
#include "revision.h"
/*
* This is stupid. We could have much better heurstics, I bet.
*/
static int better(struct revision *new, struct revision *old)
{
return new->date > old->date;
}
static struct revision *common_parent(struct revision *rev1, struct revision *rev2)
{
int i;
struct revision *best = NULL;
mark_reachable(rev1, 1);
mark_reachable(rev2, 2);
for (i = 0; i < nr_revs ;i++) {
struct revision *rev = revs[i];
if ((rev->flags & 3) != 3)
continue;
if (!best) {
best = rev;
continue;
}
if (better(rev, best))
best = rev;
}
return best;
}
int main(int argc, char **argv)
{
unsigned char rev1[20], rev2[20];
struct revision *common;
if (argc != 3 || get_sha1_hex(argv[1], rev1) || get_sha1_hex(argv[2], rev2))
usage("merge-base <commit1> <commit2>");
/*
* We will eventually want to include a revision cache file
* that "rev-tree.c" has generated, since this is going to
* otherwise be quite expensive for big trees..
*
* That's some time off, though, and in the meantime we know
* that we have a solution to the eventual expense.
*/
parse_commit(rev1);
parse_commit(rev2);
common = common_parent(lookup_rev(rev1), lookup_rev(rev2));
if (!common)
die("no common parent found");
printf("%s\n", sha1_to_hex(common->sha1));
return 0;
}