Binary search implemented in coffeescript.

binary_search.coffee

binary_search = (val, L) ->
  return false if L.length == 0
  mid = Math.floor(L.length / 2)
  if L[mid] == val
    return mid
  else if val > L[mid]
    binary_search(val, L[(mid + 1)..(L.length)])
  else
    binary_search(val, L[0..(mid - 1)])

list = [1,2,3,4,5,6,7,8]
alert binary_search(18, list) # false
alert binary_search(3, list)  # 2
alert binary_search(6, list)  # 5

Wouldn't line 5's return mid just constantly return 1?

Here's an example at each level:
Assume val = 2

1. L.length = 9
   mid = floor ( 9 / 2 ) = 4
2. L.length = 4
   mid = floor ( 4 / 2 ) = 2
3. L.length = 2
   mid = floor ( 2 / 2 ) = 1
   if L[mid] == val matches and we now return 1