Accessing the module name that implements a protocol

[eksperimental]

I am working on a protocol, and I need to have access to the module
that implements the protocol. For example, `Enumerable.List`.

It is actually mentioned in the documentation, but I would like to know
whether this is an implementation detail or a non-documented standard
that I can rely on. In case it is an implementation detail, a function
like Protocol.impl(protocol_module, implementation_module) would be
desired.

iex> Protocol.impl(Enumerable, List)
Enumerable.List

Thank you.

[José Valim]

You should use `Protocol.impl_for(data)`, which is optimized to build the module without constructing the module name manually (which can be expensive so we want to avoid it).

If you only have the module name for some reason, then it is far to assume it will be Protocol + Impl name.

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[eksperimental]

> If you only have the module name for some reason, then it is far to
> assume it will be Protocol + Impl name.

Do you mean "it is fair" ?

I want to avoid the cost of protocol dispatching by calling the
functions directly,
Thank you

[José Valim]

The protocols are already optimized to dispatch as efficiently as possible. How can you be sure that, by doing your own protocol dispatch, you won't be doing that protocols do but slower?

In any case, impl_for returns the impl for a given data.

[eksperimental]

This protocol has some behaviour callbacks, one of them is `empty/0`. So
I cannot do MyProtocol.impl_for(term) because in that callback there is
no term (I am returning a term though).
https://github.com/eksperimental/buildable/blob/a2aa7658e58820472f56e7b1b7e3b3e2293c7950/lib/buildable/delegation.ex#L14

So what are my options for delegating to MyProtocol.MyImplementation ?

[José Valim]

So it seems you should not be using protocols in the first place and model it with behaviors. Or build the module name by hand as first mentioned. :)

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