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

You could embed the code in an assert. Assertions are ignored in production code and I'm sure not built to JS when pub build is run in release mode.

class X {
  X() {
    print('x created');
  }

  void log(String m) {
    print(m);
  }
}

bool log(String m) {
  new X()..log(m);
      return true;
}

void main() {
  assert(() {
    new X()..log('in Assert');
    return true;
  });

  assert(() => log('in Assert')); // use a wrapper function
}

When you create a wrapper method that returns true than you don't have to do it explicit each time.

You can also take a look at this question How to achieve precompiler directive like functionality

Answer 2

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.