Question

Below recursive method sums the integer values between a range

  def sumInts(a: Int, b: Int): Int = {
    if(a > b) 0
    else {
      println(a +"," + b)
      a + sumInts(a + 1 , b)
    }
   } 

So sumInts(2 , 5) returns 14

I’m confused about how the recursive call to sumInts sums the integer range. Can explain textually how this method works ?

How does sumInts return the incremented value ?? Perhaps I am missing something fundamental to recursion here

Answer 1

It calculates the sum of values in the range [a, b] by first calculating the sum of the range [a+1, b] (by recursively calling sumInts(a + 1 , b)) then adding a to it.

[Update] In Scala, the return statement is optional; functions return the value of the last expression evaluated. Thus the above function body is equivalent to

if(a > b) return 0
else {
  println(a +"," + b)
  return a + sumInts(a + 1 , b)
}

[/Update]

Which for the range [2, 5] it would do the following (I removed the println call for the sake of simplicity, and added brackets to mark recursive calls):

1. if(2 > 5) 0 else 2 + sumInts(2 + 1, 5) which, the condition being false, evaluates to
2. 2 + sumInts(3, 5)
3. 2 + (if(3 > 5) 0 else 3 + sumInts(3 + 1, 5)) which evaluates to
4. 2 + (3 + sumInts(4, 5))
5. 2 + (3 + (if(4 > 5) 0 else 4 + sumInts(4 + 1, 5))) which evaluates to
6. 2 + (3 + (4 + sumInts(5, 5)))
7. 2 + (3 + (4 + (if(5 > 5) 0 else 5 + sumInts(5 + 1, 5)))) which evaluates to
8. 2 + (3 + (4 + (5 + sumInts(6, 5))))
9. 2 + (3 + (4 + (5 + (if(6 > 5) 0 else 6 + sumInts(6 + 1, 5))))) which, the condition being true, evaluates to
10. 2 + (3 + (4 + (5 + (0))))