Function refs in guards

jgonet

Popcorn Core Team

I was creating simple RPN calculator function, but I stumbled upon a problem in matching functions in guards.

Code looks like so:

def compute_RPN_expr(value, stack) when is_number(value), do: [value | stack]

def compute_RPN_expr(operator, stack) when operator in [&+/2, &-/2, &*/2, &//2] do
  [a, b | stack] = stack
  stack = [operator.(b, a) | stack]
end

After trying to define it inside IEX I’m getting invalid expression in guard, & is not allowed in guards.

Why? I can do:

a = &//2
a === &//2

Without much hassle, but I can’t use the same equality test via in.
Is there any Elixir/Erlang limitation or this was simply overlooked?

9 comments

#guard-clauses #troubleshooting

1 1858 9

2018-08-26 19:07:16 UTC

Marked As Solved

Show Best Posts
Show All (oldest first)
Show All (newest first)

Qqwy

TypeCheck Core Team

One important reason is hot code reloading and the distributed kind of Erlang/Elixir programs: If you are running code on multiple nodes, then they both have separate versions of the functions that might be executed (and if you try to upgrade both nodes to a new version, they are not guaranteed to upgrade at exactly the same time). If you pass on a function reference from one node to the next, it has to be transformed to an intermediate representation on one side, and back on the other side.

In short: The reference on one node will not point to the same location as the one on the other node. This means that we cannot check for equality (which is stronger than, but would infer, equivalence, which is what your code actually requires) by comparing pointers.

In Erlang/Elixir, all data structures can be checked for structural equivalence by matching on them. However, functions are, for the reason outlined above, opaque, so we cannot check their internals.

And therefore, an extra symbolic intermediate step like what @NobbZ proposes is required.

In a more general sense, you can never prove the equivalence of two functions, because this would be the same as solving the Halting Problem. The only thing you could do is to prove that two functions are equivalent because they are equal (exactly the same instance), but this is something that is not possible in Elixir because of above-outlined restrictions imposed by the distributed nature of Elixir’s runtime system.

Post #4

Also Liked

benwilson512

Author of Craft GraphQL APIs in Elixir with Absinthe

The BEAM Book: Understanding the Erlang Runtime System is a good resource for looking at the internals of how the BEAM itself works. Decisions and capabilities like the one you’re describing are gonna either be present or absent on the basis of the BEAM itself, Elixir just gets compiled to BEAM byte code.

Post #7

Qqwy

TypeCheck Core Team

@jgonet There are not many cases in which you actually touch the Erlang/BEAM low-level stuff directly or have to keep them in mind (most of the time, Elixir’s abstractions are very leak-free). However, I would recommend watching Solid Ground because it will give you a good high-level foundation of understanding what trade-offs have been made in Erlang/BEAM .

Post #6

NobbZ

There are no function pointers in elixir.

&+/2 is different in every module you write this. Also, even though both of the following functions are obviously producing the same result for their respective domains, you can’t computationally compare them for equality without doing manual prooves:

def multiply_a(a, b), do: a * b

def multiply_b(0, _), do: 0
def multiply_b(_, 0), do: 0
def multiply_b(1, b), do: b
def multiply_b(a, b) when a > 0, do: b + multiply(a - 1, b)
def multiply_b(a, b) when a < 0, do: multiply(-a, -b)

Both of these functions are “equal” to */2, but for the latter it will be very hard to proof in a language like elixir.

Just rewrite your dispatcher to use atoms (or accept all functions with arity 2).

def f(op, [a, b | s]) when op in [:+, :-, :*, :/], do: [apply(:erlang, op, [a, b]) | s]
def f(v, s) when is_number(v), do: [v | s]

def g(op, [a, b | s]) when is_function(op, 2), do: [op.(a, b) | s]
def g(v, s), do: [v | s]

Post #3