przemyxe0p

przemyxe0p

Scope of variable passed to function

Hi,
I am wondering why my solution wil not work

1. source: Perfect Numbers in Elixir on Exercism

2. solution attempt:

  @spec classify(number :: integer) :: {:ok, atom} | {:error, String.t()}
  def classify(number) do
    case aliquot_sum(number) do
      number -> ok(:perfect)
      x when x > number -> ok(:abundant)
      x when x < number -> ok(:deficient)
    end
  end

  defp ok(item), do: {:ok, item}
  defp aliquot_sum(number), do: get_divisors(number) |> Enum.sum()

  defp get_divisors(number, divisor \\ 1, acc \\ [])
  defp get_divisors(number, divisor, acc) when divisor > div(number,2), do: acc
  defp get_divisors(number, divisor, acc) when rem(number, divisor) == 0,
    do: get_divisors(number, divisor + 1, [divisor | acc])
  defp get_divisors(number, divisor, acc), do: get_divisors(number, divisor + 1, acc)
end

3. question:

Complier is saying:

this clause cannot match because a previous clause at line 15
always matchesElixir
perfect_numbers.ex(15, 7): related

and

variable "number" is unused (there is a variable with the same name in the
context, use the pin operator (^) to match on it or prefix this variable 
with underscore if it is not meant to be used)Elixir

Why is that, why it cannot match with number passed to function?

Most Liked

al2o3cr

al2o3cr

This form rebinds the name number inside the case branch:

def some_fun(number) do
  case 2*number do
    number ->
      # number here is bound to a new value, shadowing the argument named number

This will match ANY integer passed to some_fun.

ON THE OTHER HAND

Using the pin operator says “match this existing value”:

def some_fun(number) do
  case 2*number do
    ^number ->
      # number here is still bound to the argument

That will only match if some_fun is passed 0, since 2*0 == 0

krasenyp

krasenyp

Oh, I get you now. You can use the pin operator ^number.

kokolegorille

kokolegorille

No, your logic is wrong and number might not be what You think…

As soon as there is a match, the rest of the case is discarded

the number in the case is not the number parameter, it is a new number

You probably should write this with cond

aliquot = aliquot_sum(number)
cond do
  aliquot > number -> ...
  aliquot < number -> ...
  aliquot == number -> ...
end

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