Question:

I'm wondering how would the following function look without any syntactic sugar:

``````tails1 :: [a] -> [[a]]
tails1 [] = [[]]
tails1 xs@(x:xs') = xs:tails1 xs'
``````

I'm mostly concerned with the usage of the @ operator, I've tried what follows, but that's obviously not the correct way

``````tails1 ((:) x ((:) xs' [])) = xs:tails1 xs'
``````

First you need to understand the list data type. Here are the list data constructors:

``````[] :: [a]
(:) :: a -> [a] -> [a]
``````

where:

1. [] create an empty list
2. (:) takes an `a` , a list of `a` and returns a list of `a` with the new element appedned

Let say you have a list `[1,2,3,4]`. This can be written as `(:) 1 ((:) 2 ((:) 3 ((:) 4 [])))` In expression `x:xs'`, `x` hold `1` and `xs'` will hold `(:) 2 ((:) 3 ((:) 4 []))`. In other words `:` takes an element and a list, appending this element to the list.

The equivalent expression for your example is:

``````tails1 ((:) x xs') = ((:)x xs'):tails1 xs'
``````

Where `x` is holding the first element of the list and `xs'` the rest of the list. `xs'` cannot hold multiple elements. In your example `tails1 ((:) x ((:) xs' [])) = xs:tails1 xs'`, `xs'` should hold everything except first element and `[]`. (in my examle it should be 2:3:4 which is not a valid list because is not ended by []).

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