astery

astery

Here is the list and it is at the bottom of the table.

I would expect that <- would have a right associativity like =. Because in both operators when we match statements:

{:ok, _} = Some.function()

and

with {:ok, _} <- Some.function() do
  # ...
end

Pattern is on the left side and one evaluated is on the right. So I’ve expected Right associativity.

What’s wrong with my reasoning?

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IvanIvanoff

IvanIvanoff

Associativity plays a role when multiple operators are chained. It controls the order of grouping operands to operators with the same precedence.

= being with Right associativity means that a = b = 10 will be evaluated as (a = (b = 10)) and not as ((a = b) = 10).

- is also left associative, meaning that 5 - 4 - 3 is evaluated as ((5 - 4) - 3) == -2. If it was right associative it would have been evaluated as (5 - (4 - 3)) == 4.

When the operator is used a single time, the associativity does not play a role – 5 - 3 == 2 no matter if - is left or right associative

<- is left associative not only in the docs, but also in the parser. So there is no error in the documentation.

I was not able to actually come up with an example where multiple <- are chained, something like this does not work:

with a <- b <- 10 do
  {a, b}
end

If I had to guess before looking it up, I would have guessed that <- is non-associative. Example for non associative operator is the unary not. A single not true is valid, but chaining multiple nots is not: not true not false is a syntax error.

cevado

cevado

i belive that <- is left associative due to how it works when you use several of them in sequence, like

for {x, a} <- something, {b, c} <- a do
...
end
#or 
when {:ok, something} <- Some.function(),
     {:ok, result <- Other.function(something) do
...
end

since I can use the result of the first <- in the next <- statement. At least is how I understand it.

IvanIvanoff

IvanIvanoff

In these examples of with and for the comma , separates the expressions. The comma itself is left associative, which is what controls the order of evaluation.

You can check that by replacing the left associative <- with the right associative = and still being able to use the result of the previous results:

with a = 5,
     b = a + 10,
     c = b + 20 do
  {a, b, c}
end
# => {5, 15, 35}
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