gearnode
Hello everyone !
How to reproduce
case Repo.one(Client) do
%Client{} = client -> client
nil -> {:error, "some error message"}
end
When Repo.one(Client) returns to me a %Client{...} struct it matches, no problem with that.
Now if I reverse this match operation.
case Repo.one(Client) do
client = %Client{} -> client
nil -> {:error, "some error message"}
end
When Repo.one(Client) returns to me a %Client{...} struct, I expect it matches and returns new empty %Client{} struct. But this returns same result as the first example.
Troubleshooting
I don’t understand how Elixir resolve this match. I have tried to reproduce this in a IEx outside a case context, without any success.
Anybody know if this behavior is valid ? And how Elixir resolve this context and condition ?
Thanks,
Bryan!
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benwilson512
You aren’t reversing the match operation, because matches do not have an order. WITHIN a match
foo = baris exactly the same asbar = foo. Specifically,is in no way whatsoever different than
This is true of every match. So for example:
The right hand most
=is necessary in order to establish that the left hand side is a match, but once you’re within a match order does not matter.Also Liked
Qqwy
The reason this works is because how
%Struct{}(and, more generally, any kind of map:%{}) matches.The idea is that, for structs and maps, you often are only interested in one or a few of the fields. Therefore, matching
%{foo: _}will match any map that has a key:foo. Likewise,%{}will match any map at all.When not matching but constructing a value, then, yes,
%{}means ‘empty map’ and%Client{}means ‘empty client (whose struct fields are set to their defaults)’.The order in which
=is used (i.e. which side contains the structure to match, and which side contains the name to bind the matched part to) inside a match does not matter. Some people prefer to do it in front, (client = %Client{}), some in the back (%Client{} = client). Both mean the same.To change your code so it returns a new client, write this: