Function calling another function

Hi @grangerelixx, can you make this more concrete? You aren’t actually posting valid elixir code, which makes this hard to understand.

So initially I had this function definition

function_1(arg1,arg2,arg3) do 
  does something
end

and I used this function in many parts of my code like this

val1 = 3
val2 = 6

result1 =
val1
|> function_1(4,5)

result2 =
val2
|>function_1(7,5)

result3 = 
val1
|> function_1(3,10)

result4 = 
val2
|> function_1(9,10)

Since the arg3 is repetitive in some cases, I would like to create two functions which call fucntion_1, say function _with_arg3_equals_5 and function _with_arg3_equals_10.

So now I can use these new functions

result1 =
val1
|> function _with_arg3_equals_5(4)

result2 =
val2
|>function _with_arg3_equals_5(7)

result3 = 
val1
|> function _with_arg3_equals_10(3)

result4 = 
val2
|> function _with_arg3_equals_10(9)

Of cause you can do that. However, I’d suggest you to finish some elixir tutorial on your own first, otherwise it is hard for other people to offer you help without knowing you have a baseline knowledge.

If you have prior experience with functional programming, the official Elixir tutorial should be enough. If you do not have any experience of functional programming, you can consider to purchase a book or some online tutorial that have more details than the official elixir tutorial.

So your question is?

It will help if you can post real code.
My advice is not to create extra functions, just be explicit when you call a function. It will make your code more readable, and lean.

Thank you all for the inputs. I was able to execute what I intended to.

def function_1(arg1,arg2,arg3) do 
  does something
end

def function_with_arg3_equals_5(arg1,arg2,arg3 \\ 5) do
  function_1(arg1,arg2,arg3)
end

def function_with_arg3_equals_10(arg1,arg2,arg3 \\ 10) do
  function_1(arg1,arg2,arg3)
end

IMHO, with no risk if being unfaithful with the functions names, you can skip completely the third optional argument:

I will replace “can” with “must”. Otherwise, if you call: function_with_arg3_equals_5(1, 2, 10000) it defeats the purpose of the funtion name, and you should be calling function_1/3