How to exclude debug code

Question

Let\'s say I have simple logger:

void main() {
  var logger = new MyLogger();
  logger.log(\"hello Dart\");
}
         


        

Answer 1

I put @GünterZöchbauer "assert trick" inside the factory constructor:

class _ProductionPlug implements DebugClass{
  const _ProductionPlug();
  noSuchMethod(_) {} //do nothing
}

class DebugClass{
  static final DebugClass _plug = const _ProductionPlug();
  log(msg){print(msg);}
  DebugClass._(){}
  factory DebugClass(){
    DebugClass instance;
    assert((){
    instance = new DebugClass._();
    return true;
        });
    return instance != null ?  instance :  _plug;
  }
}
void main() {
  print("hello");
  new DebugClass()
    ..log("debugging");
}

This way nothing sticks out.