Implementing a counter, aka: how to keep state in a closure

[Travis]

Completely new to Elixir, but have an issue I am trying to figure out:

When interviewing candidates for Javascript roles I usually have them do a simple technical question:
counter1 = getCounter(1);
counter2 = getCounter(17);
counter1(); //gives 1
counter1(); //gives 2
counter2(); //gives17
counter1(); //gives 3

And the challenge is “Implement the function getCounter”. For candidates who understand functions and closures this takes 4-6 lines and less than a minute, and is generally impossible for everyone else, so is a very efficient sort. For new languages, I try building a version of this to get a sense of how anonymous functions and closures work (or don’t) in the language. I have not been able to solve this for Elixr. It is functional, it has closures, you CAN return functions, so it seems this should not only work, but be trivial…

Javascript:
getCounter = function(initVal) {
i = initVal - 1 //Fix off by 1
return function() {
i = i +1;
return i;
}
}

Go:
func getCounter(initVal int) func() int {
i := initVal - 1
return func() int {
i += 1
return i
}
}

Note that you do NOT know beforehand how many times your anonymous counter might be called (or when), nor what start value you might get. I have seen recursion mentioned quite a bit, but that would require that you know how many times you want to call your counter when you start. What is the Elixir way to do this properly?

[benwilson512]

The problem definition seems to assume mutable state so there’s a bit of a mismatch here. It’d be a bit like saying “In every language I want to learn I look at how to set register X0 to 1 and then increment it, how do I do that with JS?”. Javascript doesn’t manipulate registers, and Elixir doesn’t mutate values in closures.

Nonetheless, You can simulate mutable state with a process, and the Agent module provides a handy API in this case.

iex(6)> {:ok, pid} = Agent.start_link(fn -> 0  end)                     
{:ok, #PID<0.113.0>}
iex(7)> fun = fn -> Agent.get_and_update(pid, fn i -> {i, i + 1} end) end
#Function<20.127694169/0 in :erl_eval.expr/5>
iex(8)> fun.()                                                           
0
iex(9)> fun.()
1
iex(10)> fun.()
2
iex(11)> fun.()
3
iex(12)> fun.()
4

I’ll leave the code to make two counters as an exercise for the reader

[amarraja]

The attached article is a great read about state in Elixir. Interestingly, the examples used are pretty much the same as yours.

Storing state in Elixir with processes

[NobbZ]

Due to the way elixir works, you have to return an updated closure as well on each call (untested):

def get_counter(i \\1) do
  fn -> {i, get_counter(i + 1)} end
end

Usage is roughly like this:

c = get_counter(17)
{v, c} = c.()
#=> {17, ...}
{v, c} = c.()
#=> {18, ...}

[Nicd]

To expand a bit on the reason why you need to do it in a way that may not be intuitive for a first time functional programmer: Data in Elixir is immutable. Immutabililty means that once you have created a piece of data, it can’t be changed.

Trying to simply create an anonymous function that adds to a counter like this won’t work:

c = 0
f = fn -> c = c + 1; c end
IO.inspect(f.()) # 1
IO.inspect(f.()) # Still 1
IO.inspect(f.()) # Even still 1

That’s because the data in variable c is immutable. Inside the function the code can rebind the variable c to point to another piece of data, but the original c outside the function does not change. Even if you set c to point to a new value outside the function, it still would not affect any earlier usage of the variable. So the c that is bound to the context of the closure f is always 1 and will be.

So that is why you mainly have two avenues to solve this problem. Either you can create a new process (like the Agent suggested by benwilson512), or you can create a new function with the updated value on each call and use the updated function on the next call, like NobbZ suggested.

You might wonder how a process can hold state and update it if data is immutable. That’s because a process basically runs an infinitely recursing function that it gives the updated state to for the next iteration. So (very much simplified) a process looks like this:

def run_process(state) do
  receive do
    msg ->
      new_state = update_state(msg) # Do something based on the state and return the new state
      run_process(new_state) # Recurse with the new state, effectively updating the state of the process
  end
end

You can see there that no immutability guarantees are broken and still the state is updated.

Hope this helps and doesn’t confuse further.