Issue 4127 - [emscripten] Convert subscription classes to C++ - Part 1

compiled/subscription/Subscription.cpp

 /*
  * This file is part of Adblock Plus <https://adblockplus.org/>,
  * Copyright (C) 2006-2017 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 49 matching lines @@
 {
   // TODO
   return OwnedString();
 }
 
 Subscription* Subscription::FromID(const String& id)
 {
   if (id.empty())
   {
     // Generate a new random ID
-    unsigned seed = knownSubscriptions.size();

[sergei, 2017/04/12 12:04:33]

it would be better to put "unsigned int" (int is added).

[Wladimir Palant, 2017/04/13 13:04:41]

Done.

+    unsigned int seed = knownSubscriptions.size();
     OwnedString randomID(u"~user~000000"_str);
     do
     {
       int number = rand_r(&seed);
       for (int i = randomID.length() - 6; i < randomID.length(); i++)
       {
         randomID[i] = '0' + (number % 10);
         number /= 10;
       }
     } while (knownSubscriptions.find(randomID));
     return FromID(randomID);
   }
 
   auto knownSubscription = knownSubscriptions.find(id);
   if (knownSubscription)
   {
     knownSubscription->second->AddRef();
     return knownSubscription->second;
   }
 
   SubscriptionPtr subscription;
   if (!id.empty() && id[0] == '~')

[sergei, 2017/04/12 12:04:33]

id cannot be empty here, though maybe it's still good to check it.

[Wladimir Palant, 2017/04/13 13:04:41]

No, it cannot - with the current logic. I just verified, the compiler knows that
as well and removes this check. It's still a good idea to have it here, this way
the code won't break should the logic change later in such a way that empty IDs
become possible.

     subscription = new UserDefinedSubscription(id);
   else
     subscription = new DownloadableSubscription(id);
 
   // This is a hack: we looked up the entry using id but create it using
   // subscription->mID. This works because both are equal at this point.
   // However, id refers to a temporary buffer which will go away.
   enter_context("Adding to known subscriptions");
   knownSubscription.assign(subscription->mID, subscription.get());
   exit_context();
 
   return subscription.release();
 }