Issue 7096 - Make it possible to lazy initialize a filter's subscriptions

lib/filterStorage.js

 /*
  * This file is part of Adblock Plus <https://adblockplus.org/>,
  * Copyright (C) 2006-present eyeo GmbH
  *
  * Adblock Plus is free software: you can redistribute it and/or modify
  * it under the terms of the GNU General Public License version 3 as
  * published by the Free Software Foundation.
  *
  * Adblock Plus is distributed in the hope that it will be useful,
  * but WITHOUT ANY WARRANTY; without even the implied warranty of
@@ skipping 110 matching lines @@
   /**
    * The number of subscriptions in the storage.
    * @type {number}
    */
   get subscriptionCount()
   {
     return this.knownSubscriptions.size;
   }
 
   /**
+   * Adds any subscriptions to which a filter belongs to the filter.
+   * @param {Filter} filter
+   */
+  addSubscriptionsToFilter(filter)
+  {
+    for (let subscription of this.subscriptions())
+    {
+      if (subscription.filters.indexOf(filter) != -1)

[Manish Jethani, 2018/11/03 21:31:25]

This is super inefficient, but currently it is not needed in practice, and in
the future (a couple of patches later) it's going to be a set lookup.

See the big idea here: https://codereview.adblockplus.org/29934588/

+        filter.addSubscription(subscription);
+    }
+
+    filter.subscriptionsAssigned = true;
+  }
+
+  /**
    * Finds the filter group that a filter should be added to by default. Will
    * return <code>null</code> if this group doesn't exist yet.
    * @param {Filter} filter
    * @returns {?SpecialSubscription}
    */
   getGroupForFilter(filter)
   {
     let generalSubscription = null;
     for (let subscription of this.knownSubscriptions.values())
     {
@@ skipping 522 matching lines @@
  *   disconnected from its filters.
  */
 function disconnectSubscriptionFilters(subscription)
 {
   if (!filterStorage.knownSubscriptions.has(subscription.url))
     return;
 
   for (let filter of subscription.filters)
     filter.removeSubscription(subscription);
 }