Ok, I get it, but the question was, why it was not working with case this, matching why do we have to use this ^ pin operator?
przemyxe0p
sodapopcan
Oh sorry wasn’t responding specifically to your question.
But to answer it: Elixir allows variable re-binding, ie, you can say: number = number+ 1 and not get an error that number is already defined. Like =, -> is also a binding operator, so when you say number -> ok(:perfect), number is rebound. The match on the value of number, you need to use the pin operator.
1
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
2