Binding operator - is it possible to add the ‘:=’ operator for binding assignments and keep the ‘=’ operator for pattern matching

Chat & Discussions Proposals: Ideas
1

Good morning to the Elixir team.

Please excuse me if I’m commenting in the wrong place; please indicate where I should discuss this.

I would like to know if it’s possible to add the ‘:=’ operator for binding assignments and keep the ‘=’ operator for pattern matching.

This would make teaching the features of Elixir/Erlang easier (as a language promoter).

Thank you very much for your attention.

2

Can you give an example of what you mean?

You can already do stuff like this:

data = {:ok, %{name: "Alice"}}

case data do
    {:ok = status, user} -> 
        // here we have both status and user variable available
end
3

But there is no such thing as “binding assignment operator” in Elixir. There is only pattern matching (with potential rebinding). So I do not really get how these operators would differ.

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4

Pattern-matching IS binding assignments, there’s no difference. An example:

{:ok, something} = MyFunc.call(params)

response = MyFunc.call(params) # Returns same, now `response` is bound as {:ok, something}

If you are a “language promoter” with the goal of teaching Elixir, you should probably spend some more time writing it, so you’re comfortable with how the language works.

3 Likes
5

I’m not following what’s gained by distinguishing these two uses of =. Can you provide some examples?

1 Like
6

Thank you for your feedback.

I asked this question because my team and I come from non-functional languages.

I am learning and really enjoying the practicality and ergonomics of Elixir and the capabilities of Erlang/OTP.

Without a doubt, I must and need to improve my understanding of the language further.

I thought that differentiating ‘Pattern matching’ from ‘binding assignments/rebinding’ would be simpler for understanding the Functional Paradigm, which is the heart of functional languages.

It was not my intention to modify the language into something meaningless.

I greatly appreciate the answers.

7

At first I also had a hard time accepting that there was no difference between pattern matching and assignment, especially when you’re told = is not “assign” but “match” and then get an example of foo = "bar" # "bar"

The guides have been improved a bit since I initially read them by immediately showing the “inverse” expression:

iex> x = 1
1
iex> 1 = x
1

1 = x is an brow-raiser for newbies but I think they must get some idea that it’s a special case to put a variable on the right hand side or something? I’m only guessing, but I find that showing them this last example really tends to drive it home that = really indeed not an assignment operator:

iex> 1 = 1
1

I always explain it as assignment being a side effect of matching. I know that’s sorta overloading a term, but it seems to help in the beginning, lol.

7 Likes
8

Think of it this way: ‘=’ is binding assignment within a pattern matching. There is no pattern matching without binding operator because it would be a useless thing to do, other than being an assertion. If you must, you can wrap a scope around it to avoid pollution:

iex(13)> x = 1
1
iex(14)> if true, do: x = 2
warning: variable "x" is unused (there is a variable with the same name in the context, use the pin operator (^) to match on it or prefix this variable with underscore if it is not meant to be used)
  iex:14

2
iex(15)> x
1
1 Like