Given a method that accepts a call-by-name parameter, it means the method decides when and how many times the parameter evaluates.

Boolean.&& is one of that kind of methods. Technically, there is no way to pass a monadic expression to methods like that. You could not pass an Option[Boolean] to a => Boolean typed method because type mismatch.

Instead, the only thing monadic macro can do is evaluating the monadic expression before passing it to a method. The following code is generated by monadic for you example code.

{
  val monad$macro$1: scalaz.Traverse[Option] with scalaz.MonadPlus[Option] with scalaz.Each[Option] with scalaz.Index[Option] with scalaz.Length[Option] with scalaz.Cozip[Option] with scalaz.Zip[Option] with scalaz.Unzip[Option] with scalaz.Align[Option] with scalaz.IsEmpty[Option] with scalaz.Cobind[Option] with scalaz.Optional[Option]{def point[A](a: => A): Some[A]; def cojoin[A](a: Option[A]): Option[Some[A]]} = scalaz.std.option.optionInstance;
  monad$macro$1.bind(com.thoughtworks.each.Monadic.EachOps[Some, Int](scala.Some.apply[Int](1)).underlying)(((element$macro$2: Int) => {
    val i: Boolean = element$macro$2.<(0);
    monad$macro$1.map({
      scala.this.Predef.println("foo");
      monad$macro$1.map(com.thoughtworks.each.Monadic.EachOps[Some, Int](scala.Some.apply[Int](2)).underlying)(((element$macro$3: Int) => element$macro$3.>(0)))
    })(((element$macro$4: Boolean) => i.&&(element$macro$4)))
  }))
}

In brief, there is no way to pass a monadic expression to a method that accept call-by-name parameters and still keep the same behavior as the behavior when passing a direct expression. No way, even if you manually create the monadic expression without using Each library.

Strange behavior of && operator (maybe also exists with other logic operation) #37

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Strange behavior of `&&` operator (maybe also exists with other logic operation) #37