Immutability explanation

The value of 1 hasn’t changed, though it’s not accessible under the name a anymore, as there is a new thing named a with the value 2 “shadowing” the old a.

You can try this code:

a = 1
f = fn -> a end
a = 2
IO.puts("a: #{a}")
IO.puts("f(): #{f.()}")

a: 2
f(): 1

But as often, it has nothing to do with immutability, it is just how scoping rules work in elixir.

So the first 1 is still in memory?

if you run this code only:

a = 1
a = 2

The 1 would be on the stack memory segment, so it would not exist anymore. But I think the compiler will just skip that value completely.

How can the first assignment a = 1 be immutable after the a = 2 rebind?

First you need to know that the = operator is called the match operator, it is not called the assignment operator.

So, Elixir can make the match a = 1 true by binding the a variable to the specified value.

Hence, what happens next in Elixir is that the variable is rebound to the new value when Elixir is making the match a = 2 true.

Take into account that Elixir will only bind variables on the left hand side of the match operator.

Yes, from the beams point of view, the first a is actually named Va_1, the second one is Va_2, though from elixir you can only ever “see” the last binding from the current scope.

Any way to access previous binding?

As I said, no.

As I always say, variables are mutable in Elixir, no matter what fancy explanations are given to me

In my point of view, whats matter, its what I can see from outside the box, not what’s going inside the box.

Outside the box, the developer will not have the guarantee that a variable keeps the value it was first assigned, thus for me they are mutable.

Inside the box the Beam just points the variable to other location in memory where the new value its stored, or something similar, and normally is referred as rebind, thus its why everyone says that they are immutable.

In the end of the day what matters for me as a developer is the public interface the language presents to me, and I don’t care how its done under the hood(inside the box), I just care how it looks without the need to look inside the box.

NOTE: And yes I already have read Comparing Elixir and Erlang variables « Plataformatec Blog

You do, though, absolutely have a guarantee that nothing inside the box can change a variable that you have assigned.