Binding vs Assignment - ruby

Question

I've read a number of articles on the difference between assignment and binding, but it hasn't clicked yet (specifically in the context of an imperative language vs one without mutation).
I asked in IRC, and someone mentioned these 2 examples illustrate the difference, but then I had to go and I didn't see the full explanation.
Can someone please explain how/why this works in a detailed way, to help illustrate the difference?
Ruby
x = 1; f = lambda { x }; x = 2; f.call
#=> 2
Elixir
x = 1; f = fn -> x end; x = 2; f.()
#=> 1

Answer 1

I've heard this explanation before and it seems pretty good:
You can think of binding as a label on a suitcase, and assignment as a
suitcase.
In other languages, where you have assignment, it is more like putting a value in a suitcase. You actually change value that is in the suitcase and put in a different value.
If you have a suitcase with a value in it, in Elixir, you put a label on it. You can change the label, but the value in the suitcase is still the same.
So, for example with:
iex(1)> x = 1
iex(2)> f = fn -> x end
iex(3)> x = 2
iex(4)> f.()
1
You have a suitcase with 1 in it and you label it x.
Then you say, "Here, Mr. Function, I want you to tell me what is in this suitcase when I call you."
Then, you take the label off of the suitcase with 1 in it and put it on another suitcase with 2 in it.
Then you say "Hey, Mr. Function, what is in that suitcase?"
He will say "1", because the suitcase hasn't changed. Although, you have taken your label off of it and put it on a different suitcase.

Answer 2

After a while, I came up with the answer that is probably the best explanation of the difference between “binding” and “assignment”; it has nothing in common with what I have written in another answer, hence it’s posted as a separate answer.
In any functional language, where everything is immutable, there is no meaningful difference between terms “binding” and “assignment.” One might call it either way; the common pattern is to use the word “binding,“ explicitly denoting that it’s a value bound to a variable. In Erlang, for instance, one can not rebound a variable. In Elixir this is possible (why, for God’s sake, José, what for?)
Consider the following example in Elixir:
iex> x = 1
iex> 1 = x
The above is perfectly valid Elixir code. It is evident, one cannot assign anything to one. It is neither assignment nor binding. It is matching. That is how = is treated in Elixir (and in Erlang): a = b fails if both are bound to different values; it returns RHO if they match; it binds LHO to RHO if LHO is not bound yet.
In Ruby it differs. There is a significant difference between assignment (copying the content,) and binding (making a reference.)

Answer 3

Elixir vs Ruby might not be the best contrast for this. In Elixir, we can readily "re-assign" the value of a previously assigned named variable. The two anonymous-function examples you provided demonstrate the difference in how the two languages assign local variables in them. In Ruby, the variable, meaning the memory reference, is assigned, which is why when we change it, the anonymous function returns the current value stored in that memory-reference. While in Elixir, the value of the variable at the time the anonymous function is defined (rather than the memory reference) is copied and stored as the local variable.
In Erlang, Elixir's "parent" language, however, variables as a rule are "bound." Once you've declared the value for the variable named X, you are not allowed to alter it for the remainder of the program and any needed alterations would need to be stored in new named variables. (There is a way to reassign a named variable in Erlang but it is not the custom.)

Answer 4

Binding refers to particular concept used in expression-based languages that may seem foreign if you're used to statement-based languages. I'll use an ML-style example to demonstrate:
let x = 3 in
let y = 5 in
x + y
val it : int = 8
The let... in syntax used here demonstrates that the binding let x = 3 is scoped only to the expression following the in. Likewise, the binding let y = 5 is only scoped to the expression x + y, such that, if we consider another example:
let x = 3 in
let f () =
x + 5
let x = 4 in
f()
val it : int = 8
The result is still 8, even though we have the binding let x = 4 above the call to f(). This is because f itself was bound in the scope of the binding let x = 3.
Assignment in statement-based languages is different, because the variables being assigned are not scoped to a particular expression, they are effectively 'global' for whatever block of code they're in, so reassigning the value of a variable changes the result of an evaluation that uses the same variable.

Answer 5

The easiest way to understand the difference, would be to compare the AST that is used by the language interpreter/compiler to produce machine-/byte-code.
Let’s start with ruby. Ruby does not provide the AST viewer out of the box, so I will use RubyParser gem for that:
> require 'ruby_parser'
> RubyParser.new.parse("x = 1; f = -> {x}; x = 2; f.()").inspect
#=> "s(:block, s(:lasgn, :x, s(:lit, 1)),
# s(:lasgn, :f, s(:iter, s(:call, nil, :lambda), 0, s(:lvar, :x))),
# s(:lasgn, :x, s(:lit, 2)), s(:call, s(:lvar, :f), :call))"
The thing we are looking for is the latest node in the second line: there is x variable inside the proc. In other words, ruby expects the bound variable there, named x. At the time the proc is evaluated, x has a value of 2. Hence the the proc returns 2.
Let’s now check Elixir.
iex|1 ▶ quote do
...|1 ▶ x = 1
...|1 ▶ f = fn -> x end
...|1 ▶ x = 2
...|1 ▶ f.()
...|1 ▶ end
#⇒ {:__block__, [],
# [
# {:=, [], [{:x, [], Elixir}, 1]},
# {:=, [], [{:f, [], Elixir}, {:fn, [], [{:->, [], [[], {:x, [], Elixir}]}]}]},
# {:=, [], [{:x, [], Elixir}, 2]},
# {{:., [], [{:f, [], Elixir}]}, [], []}
# ]}
Last node in the second line is ours. It still contains x, but during a compilation stage this x will be evaluated to it’s currently assigned value. That said, fn -> not_x end will result in compilation error, while in ruby there could be literally anything inside a proc body, since it’ll be evaluated when called.
In other words, Ruby uses a current caller’s context to evaluate proc, while Elixir uses a closure. It grabs the context it encountered the function definition and uses it to resolve all the local variables.