Is this expected behavior or am I missing something conceptually? Erlang/OTP 24 [erts-12.1.2] [source] [64-bit] [smp:12:12] [ds:12:12:10] [async-threads:1] [jit] Interactive Elixir (1.12.2) - press Ctrl+C to exit (type h() ENTER for help) iex(1)> l = %{a: 1} %{a: 1} iex(2)> %{a: 1} = %{a: 1, b: 2}…

[image] olivierdevon: I am sorry to carry on the issue but I understand that the match is exact. My question is why is the semantics of ^l = %{a: 1, b: 2} different from %{a: 1} = %{a: 1, b: 2} even though ^l refers to the value of l which is %{a: 1}? The mismatch here is that the runtime…

[image] olivierdevon: I expected the last match to be successful. What misconception do I have? You are trying to pattern match within the pin operator. This is what is expected. iex> ^l = %{a: 1} %{a: 1}

I believe what it is trying to do is an exact match, so ^l = %{a: 1, b: 2} could be thought of like: %{a: 1} == %{a: 1, b: 2}

Pinning match is always exact equality (otherwise there would be no way to match %{} for example

I realized that it was performing an exact match. However, I cannot see why this is necessary. I thought that after the i = %{a: 1} match and binding that ^i = %{a: 1, b: 2} would involve a substitution of %{a: 1} for ^i resulting an the match %{a: 1} = %{a: 1, b: 2}, which should succeed. Thank yo…

Thank you for your reply. I still do not understand why an exact match is needed. Why would there be no way to match %{}?

In %{a: 1} = %{a: 1, b: 2} We are matching the values inside the map. a in this case, and since they are both 1, it matches. In ^l = %{a: 1, b: 2} We are matching the entire map. Not only the k/v inside

I am sorry to carry on the issue but I understand that the match is exact. My question is why is the semantics of ^l = %{a: 1, b: 2} different from %{a: 1} = %{a: 1, b: 2} even though ^l refers to the value of l which is %{a: 1}? If this is the semantics by design, then I think it should be mad…

Map matching is commonly used for destructuring a map, ie: %{key1: value} = %{key1: 1, key2: 2} will bind value to 1. So this has to be a partial match. With a pin operator, there is no new binding of variable so it does not make sense to do partial match anymore. Think this way, the partial match…

to be fair, this is SUPER confusing because maps are the only datastructure which afford you inexplicit partial matching in elixir. Occasionally this causes problems when refactoring because in tests i often assert with matching instead asserting with equality. in general, the map partial match %{…

Thanks @LostKobrakai that’s the best answer I’ve seen to this question so far. So the reason why this works: iex(2)> one = 1 1 iex(3)> %{a: ^one} = %{a: 1, b: 2} %{a: 1, b: 2} But this one doesn’t: iex(1)> map_with_one = %{a: 1} %{a: 1} iex(2)> ^map_with_one = %{a: 1, b: 2} ** (MatchError) no ma…

Elixir Programming Language Forum

Pattern matching with variables and the pin operator

Questions & Help Questions

trisolaran November 23, 2021, 8:49am 14

Thanks! Excellent explanation