Issue 6916 - Avoid Set object for filters with only one subscription

lib/filterClasses.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 23 matching lines @@
 /**
  * Abstract base class for filters
  *
  * @param {string} text   string representation of the filter
  * @constructor
  */
 function Filter(text)
 {
   this.text = text;
 
+  /**
+   * Subscriptions to which this filter belongs.
+   * @type {(Subscription|Set.<Subscription>)?}
+   * @private
+   */
   this._subscriptions = null;
 }
 exports.Filter = Filter;
 
 Filter.prototype =
 {
   /**
    * String representation of the filter
    * @type {string}
    */
   text: null,
 
   /**
    * Filter type as a string, e.g. "blocking".
    * @type {string}
    */
   get type()
   {
     throw new Error("Please define filter type in the subclass");
   },
 
   /**
-   * Yields filter subscriptions the filter belongs to
+   * Yields subscriptions to which the filter belongs.
    * @yields {Subscription}

[Jon Sonesen, 2018/09/02 17:22:50]

Should this be {(Subscription|Set.<Subscription>)?}

[Manish Jethani, 2018/09/03 19:03:42]

No, the function only ever yields a single Subscription object. Note that we're
using `yield*` if it's a set, so each entry in the set is yielded individually.
This way the caller doesn't need to know about the internal optimization and
always gets a consistent value.

    */
   *subscriptions()
   {
     if (this._subscriptions)
     {
       if (this._subscriptions instanceof Set)
         yield* this._subscriptions;
       else
         yield this._subscriptions;
     }
   },
 
+  /**
+   * The number of subscriptions to which the filter belongs.
+   * @type {number}
+   */
+  get subscriptionCount()
+  {
+    if (this._subscriptions instanceof Set)
+      return this._subscriptions.size;
+
+    return this._subscriptions ? 1 : 0;
+  },
+
+  /**
+   * Adds a subscription to the set of subscriptions to which the filter
+   * belongs.
+   * @param {Subscription} subscription
+   */
   addSubscription(subscription)
   {
+    // Since we use truthy checks in our logic, we must avoid adding a
+    // subscription that isn't a non-null object.

[Jon Sonesen, 2018/09/02 17:22:50]

the double negative in the comment is sort of confusing, perhaps 

"Since we use truthy checks in our logic, we must avoid adding a subscription
that is null or not an object"

is a bit more clear?

[Manish Jethani, 2018/09/03 19:03:42]

The non-null part is a bit of a distraction: the problem is that `typeof null`
is also "object" in JavaScript, otherwise we could have just had the typeof
check. I can remove the "non-null" in that sentence if it makes it more
readable, it's obvious from the line that follows that we're checking for
`null`. What do you think?

+    if (subscription === null || typeof subscription != "object")
+      return;
+
     if (this._subscriptions)
     {
       if (this._subscriptions instanceof Set)
         this._subscriptions.add(subscription);
       else if (subscription != this._subscriptions)
         this._subscriptions = new Set([this._subscriptions, subscription]);
     }
     else
     {
       this._subscriptions = subscription;
     }
   },
 
+  /**
+   * Removes a subscription from the set of subscriptions to which the filter
+   * belongs.
+   * @param {Subscription} subscription
+   */
   removeSubscription(subscription)
   {
     if (this._subscriptions)
     {
       if (this._subscriptions instanceof Set)
       {
         this._subscriptions.delete(subscription);
 
         if (this._subscriptions.size == 1)
           this._subscriptions = [...this._subscriptions][0];

[Jon Sonesen, 2018/09/02 17:22:50]

Just asking, not so much  a comment but this line changes _subscriptions to
whichever the remaining member of the set is?

[Manish Jethani, 2018/09/03 19:03:42]

Yes, that's exactly what it does. We can't do `this._subscriptions[0]` because
it's not an array, so basically we convert it to an array first using the spread
operator.

       }
       else if (subscription == this._subscriptions)
       {
         this._subscriptions = null;
       }
     }
   },
 
   /**
    * Serializes the filter to an array of strings for writing out on the disk.
@@ skipping 1105 matching lines @@
 
   /**
    * Script that should be executed
    * @type {string}
    */
   get script()
   {
     return this.body;
   }
 });