println in scala for-comprehension

Question

In a for-comprehension, I can't just put a print statement:

def prod (m: Int) = {
  for (a <- 2 to m/(2*3);
    print (a + "  ");
    b <- (a+1) to m/a;
    c = (a*b) 
    if (c < m)) yield c
}

but I can circumvent it easily with a dummy assignment:

def prod (m: Int) = {
  for (a <- 2 to m/(2*3);
    dummy = print (a + "  ");
    b <- (a+1) to m/a;
    c = (a*b) 
    if (c < m)) yield c
}

Being a side effect, and only used (so far) in code under development, is there a better ad hoc solution?

Is there a serious problem why I shouldn't use it, beside being a side effect?

update showing the real code, where adapting one solution is harder than expected:

From the discussion with Rex Kerr, the necessity has risen to show the original code, which is a bit more complicated, but did not seem to be relevant for the question (2x .filter, calling a method in the end), but when I tried to apply Rex' pattern to it I failed, so I post it here:

  def prod (p: Array[Boolean], max: Int) = {
    for (a <- (2 to max/(2*3)).
        filter (p);
      dummy = print (a + "  ");
      b <- (((a+1) to max/a).
         filter (p));
      if (a*b <= max)) 
        yield (em (a, b, max)) }

Here is my attempt -- (b * a).filter is wrong, because the result is an int, not a filterable collection of ints:

  // wrong: 
  def prod (p: Array[Boolean], max: Int) = {
    (2 to max/(2*3)).filter (p).flatMap { a =>
      print (a + " ")
      ((a+1) to max/a).filter (p). map { b => 
        (b * a).filter (_ <= max).map (em (a, b, max))
      }
    }
  }

Part II belongs to the comments, but can't be read, if written there - maybe I delete it in the end. Please excuse.

Ok - here is Rex last answer in code layout:

  def prod (p: Array[Boolean], max: Int) = {
    (2 to max/(2*3)).filter (p).flatMap { a =>
      print (a + " ")
      ((a+1) to max/a).filter (b => p (b) 
        && b * a < max).map { b => (m (a, b, max))
      }
    }
  }