spammy

spammy

Say we have a function:

def func(a, b \\ 0) do

I believe this generates func/1 and func/2. Is it possible to make func/2 private without spinning out a helper/implementation/do_func/2 equivalent?

First 7 of 7 Posts Switch mode

dimitarvp

dimitarvp

Does not seem to be possible, no.

al2o3cr

al2o3cr

You can mix def and defp with different arities:

defmodule Foo do
  def bar(arg), do: bar(arg, 42)
  defp bar(arg, arg2), do: {arg, arg2}
end

iex(6)> Foo.bar(1)
{1, 42}

iex(7)> Foo.bar(1, 2)
** (UndefinedFunctionError) function Foo.bar/2 is undefined or private. Did you mean:

      * bar/1

    Foo.bar(1, 2)
    iex:7: (file)
spammy

spammy OP

That is where I landed, that the default parameter has to be expressed differently.

Would it be more idiomatic to change the name of the private function?

LostKobrakai

LostKobrakai

I’d say yes. It’s just accidental that the private one happens to have a different arity, which allows for it being named the same.

rvirding

rvirding

Creator of Erlang

This is just because what defines a function is its module, name and arity. So foo/1 and foo/2 are 2 different functions that just happen to have the same name.

spammy

spammy OP

I guess my more general question was “is it possible to affect implied functions created by default parameters”. An example that doesn’t work:

defmodule Foo do
  def bar(arg)
  defp bar(arg, arg2 \\ 0), do: {arg, arg2}
end

error: defp bar/1 already defined as def in iex:5

sodapopcan

sodapopcan

I’ve always found it helpful to think of \\ as a simple convenience rather than “default arguments.” All it’s doing is turning this:

def bar(a, b \\ 0) do
  a + b
end

into this:

def bar(a) do
  bar(a, 0)
end

def bar(a, b) do
  a + b
end

The easiest and clearest way to get what you want it to just write it out manually as suggested.

So the short answer is no, but technically yes if you consider redefining def and changing how \\ works :wink:

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