notmuch: keep track of all siblings in thread

Consider a thread with a root message and three child messages, each of
which has a single child message, like so:

    root
        child 1
            grandchild 1a
        child 2
            grandchild 2a
        child 3
            grandchild 3a

With the previous implementation, if child 2 (or child 3) is not shown
by the current query, then grandchild 2a (or grandchild 3a) will not be
shown either. Ensure that this bug does not occur; that is, ensure that
children of non-queried non-first-child messages are shown in threads.

A more complex implementation is possible that maintains only a single
loop over the messages, but thinking about it made my head hurt. Use a
second (at times redundant) loop instead for simplicity's sake.

Changelog-fixed: Don't lose child messages of non-queried parents in
 notmuch threads
Signed-off-by: Jason Cox <me@jasoncarloscox.com>
Acked-by: Robin Jarry <robin@jarry.cc>
Tested-by: Inwit <inwit@sindominio.net>

1 files changed, 3 insertions, 5 deletions

diff --git a/worker/notmuch/lib/database.go b/worker/notmuch/lib/database.go
index b2a22227..f6e49ae1 100644
--- a/worker/notmuch/lib/database.go
+++ b/worker/notmuch/lib/database.go
@@ -311,7 +311,6 @@ func (db *DB) KeyFromUid(uid uint32) (string, bool) {
 }
 
 func (db *DB) makeThread(parent *types.Thread, msgs *notmuch.Messages, threadContext bool) []*types.Thread {
-	var lastSibling *types.Thread
 	var siblings []*types.Thread
 	for msgs.Next() {
 		msg := msgs.Message()
@@ -345,12 +344,11 @@ func (db *DB) makeThread(parent *types.Thread, msgs *notmuch.Messages, threadCon
 		if parent != nil && parent.FirstChild == nil {
 			parent.FirstChild = node
 		}
-		if lastSibling != nil {
-			lastSibling.NextSibling = node
-		}
-		lastSibling = node
 		siblings = append(siblings, node)
 		db.makeThread(node, &replies, threadContext)
 	}
+	for i := 1; i < len(siblings); i++ {
+		siblings[i-1].NextSibling = siblings[i]
+	}
 	return siblings
 }