Bug 1760674. r=mak

Differential Revision: https://phabricator.services.mozilla.com/D141958

docshell/base/BaseHistory.cpp

@@ -108,10 +108,6 @@ void BaseHistory::RegisterVisitedCallback(nsIURI* aURI, Link* aLink) {
   // This will not catch a case where it is registered for two different URIs.
   MOZ_DIAGNOSTIC_ASSERT(!links->mLinks.Contains(aLink),
                         "Already tracking this Link object!");
-  // FIXME(emilio): We should consider changing this (see the entry.Remove()
-  // call in NotifyVisitedInThisProcess).
-  MOZ_DIAGNOSTIC_ASSERT(links->mStatus != VisitedStatus::Visited,
-                        "We don't keep tracking known-visited links");
 
   links->mLinks.AppendElement(aLink);
 
@@ -199,15 +195,6 @@ void BaseHistory::NotifyVisitedInThisProcess(nsIURI* aURI,
   for (Link* link : links.mLinks.BackwardRange()) {
     link->VisitedQueryFinished(visited);
   }
-
-  // We never go from visited -> unvisited.
-  //
-  // FIXME(emilio): It seems unfortunate to remove a link to a visited uri and
-  // then re-add it to the document to trigger a new visited query. It shouldn't
-  // if we keep track of mStatus.
-  if (visited) {
-    entry.Remove();
-  }
 }
 
 void BaseHistory::SendPendingVisitedResultsToChildProcesses() {

dom/base/Link.cpp

@@ -68,12 +68,6 @@ void Link::VisitedQueryFinished(bool aVisited) {
   // Set our current state as appropriate.
   mState = newState;
 
-  // We will be no longer registered if we're visited, as it'd be pointless, we
-  // never transition from visited -> unvisited.
-  if (aVisited) {
-    mRegistered = false;
-  }
-
   MOZ_ASSERT(LinkState() == NS_EVENT_STATE_VISITED ||
                  LinkState() == NS_EVENT_STATE_UNVISITED,
              "Unexpected state obtained from LinkState()!");