CODEWITHSUNDEEP
erlangelixir
reknirt
reknirt

Reputation: 2254

Why do some variables automatically bind to other values in Elixir? Ex. case

I'm in the process of learning Elixir and I was curious why the following occurs:

iex(4)> case {:one, :two} do
...(4)>   {:four, :five} ->
...(4)>     "This won't match"
...(4)>   {:one, x} ->
...(4)>     "This will match and bind `x` to `:two`"
...(4)>   _ ->
...(4)>     "This will match any value"
...(4)> end
"This will match and bind `x` to `:two`"

So if in this 'Pattern Matching' example, why does the empty variable x automatically bind to the atom :two and provide a positive match? x doesn't equal :two when this case is first run.

I'm just not grasping what exactly is going on.

Thanks.

Upvotes: 1

Views: 166

Answers (2)

Pascal
Pascal

Reputation: 14042

Pattern matching in a clause (case or function) performs the same operation as {:one, x} = {:one, :two} (which is also pattern matching). In this second case it is obvious that you want to test if the 2 expressions match, and you want to bind the variable x if it was unbound previously. The only difference is that if the match fails in a clause (for example {:four, :five} = {:one, :two}) , it will try the next clause if any previous clause is throwing an exception.

It is a very powerful feature because it does a lot of operations with very few lines and keeps the code easy to read.

Upvotes: 6

mipadi
mipadi

Reputation: 410672

x doesn't equal :two when this case is first run.

Exactly. x is an unbound variable, so it can match against anything (and will then be bound to what it matched). In this case, {:one, x} successfully matches {:one, :two} (since x can match anything), and since x was unbound, it can now be bound to :two.

Upvotes: 3

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