In the code below I get the expected result:
(the point of the learning exercise I'm working on is to write code to modify the original array rather than returning a new array)
def filter_out!(array, &prc)
array.uniq.each { |el| array.delete(el) if prc.call(el) } #my block of code
end
arr_2 = [11, 17, 13, 15 ]
filter_out!(arr_2) { |x| x.odd? }
p arr_2 # []
However, if I remove the .uniq and only utilize array.each the output changes to [17, 15].
I believe the difference is that when only using array.each the index is being cycled through and when deleting 11 (because its odd) at the zero index, it looks at the next index, 1, but 17 is no longer at that index (13 is now) so the element is skipped for testing against the block. Same for 15 which is why it and 17 remain.
Is my assumption correct? If so, how does the underlying functionality of .uniq bypass this? I would assume that chaining .uniq in before .each would simply return the same 'incorrect answer' of [17, 15] since all values are already unique and .each would once again be performed on [11, 17, 13, 15 ] .