tpucci
Good morning,
First, I am a new learner and this is my first post ! I am really happy to join the community.
I am discovering Elixir syntax and I had a quick question:
Why can’t I use this syntax ?
Enum.all?([1, 30], is_number)
If I do this I get the following error
** (CompileError) iex:9: undefined function is_number/0
(stdlib) lists.erl:1354: :lists.mapfoldl/3
(stdlib) lists.erl:1355: :lists.mapfoldl/3
I expected Elixir to find is_number/1
Instead, this works:
Enum.all?([1, 30], &is_number(&1))
Enum.all?([1, "hi"], fn x -> is_number(x) end)
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kokolegorille
The question is more about why You need to use & than function arity.
Probably because
has the same result as
and You might wonder why wrapping this inside fn → end.
But as soon as there is more than one params… You see it starts to be clearer why the 2 forms are differents.
There is another reason, fn allows to capture the present value of variables in the scope (Closure).
This last example is like having
It’s really fun to create functions from other functions so easily
david_ex
Functions in Elixir have both a name AND an arity (number of arguments it takes). For Elixir to find the function, it needs both. Therefore, if you use the
&capture operator, you also need to specify the function’s arity:Enum.all?([1, 30], &is_number/1)With your previous syntax of
Enum.all?([1, 30], is_number), Elixir doesn’t know the function’s arity. In this case, there is only one version ofis_number, but other functions with the same name can have different arities, e.g.List.flatten/1andList.flatten/2.In Elixir, a function’s “identity” is composed of BOTH its name and arity. This is why you’ll see functions referred to as
is_number/1and not justis_number.