How does &(&1 + &2) work?

I was trying to write this Enum.reduce([1,2,3], &Kernel.+(&1, &2)) more succinctly and I discovered I could do Enum.reduce([1,2,3]), &(&1 + &2)).

What is this doing in the background? What kind of expressions can you put in &()? What is the term for this so I can google more or find the docs?

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I have heard this mentioned as shorthand notation for anonymous function. They are mentioned in the elixir getting started guide under Function capturing:

The & is a capture operator (

In your example &(&1 + &2) translates to fun (x, y) -> x + y end The &() “creates” the function and the &1 and &2 means the position of the argument to the function.

If you google elixir shorthand anonymous function you’ll get better explanations and examples :smiley:


Ah, Doh. I was awake enough to know + maps to Kernel.+ but didn’t even consider that Kernel.& would obviously map to &…

This is great, I can trim down all those one liner fn (val) -> val == search_target end functions! That always seemed awkward that Elixir didn’t have a nice clean short hand!


The expression could have been even shorter

Enum.reduce([1, 2, 3], &+/2)

For this specific case, the more succint option is Enum.sum/1.

The capture is very powerful, you can make things like &"Hello #{&1}", also with charlists, tuples, maps, structs… :slightly_smiling_face: