unpack-trees: simplify 'all other failures' case

In the 'if (current)' block of twoway_merge, we handle the boring
errors by checking if the entry from the old tree, current index, and
new tree are present, to get a pathname for the error message from one
of them:

	if (oldtree)
		return o->gently ? -1 : reject_merge(oldtree, o);
	if (current)
		return o->gently ? -1 : reject_merge(current, o);
	if (newtree)
		return o->gently ? -1 : reject_merge(newtree, o);
	return -1;

Since this is guarded by 'if (current)', the second test is guaranteed
to succeed.  Moreover, any of the three entries, if present, would
have the same path because there is no rename detection in this code
path.   Even if some day in the future the entries' paths differ, the
'current' path used in the index and worktree would presumably be the
most recognizable for the end user.

Simplify by just using 'current'.

Noticed by coverity, Id:290002

Signed-off-by: Stefan Beller <[email protected]>
Improved-by: Junio C Hamano <[email protected]>
Signed-off-by: Jonathan Nieder <[email protected]>
Signed-off-by: Junio C Hamano <[email protected]>

Author: stefanbeller

unpack-trees.c

@@ -1791,16 +1791,8 @@ int twoway_merge(const struct cache_entry * const *src,
 			/* 20 or 21 */
 			return merged_entry(newtree, current, o);
 		}
-		else {
-			/* all other failures */
-			if (oldtree)
-				return o->gently ? -1 : reject_merge(oldtree, o);
-			if (current)
-				return o->gently ? -1 : reject_merge(current, o);
-			if (newtree)
-				return o->gently ? -1 : reject_merge(newtree, o);
-			return -1;
-		}
+		else
+			return o->gently ? -1 : reject_merge(current, o);
 	}
 	else if (newtree) {
 		if (oldtree && !o->initial_checkout) {