how does xs in this function work?

Question

I am reading Hutton's book, Programming in Haskell. Here is a function:

pairs :: [a] -> [(a,a)]

pairs xs = zip xs (tail xs)

e.g.

>pairs [1,2,3,4]

>[(1,2),(2,3),(3,4)] --result

Problem is how to read this function ? from left to right?

I am confused how "tail" leave 1 element and then combine it with next element using "zip" Since "tail" suppose to get all remaining elements from the list right?

Answer 1

I haven't read the book you mentioned, but I'll try to explain what I know.

You're right about the tail function returning everything except the first element of the list. Let's see how zip works,

zip [1, 2, 3, 4] [5, 6, 7, 8]

gives,

[(1, 5), (2, 6), (3, 7), (4, 8)]

Now, consider the output we need from the input we have, observe the transformation required from input to output,

[1, 2, 3, 4] -> [(1,2),(2,3),(3,4)]

From the above application of zip, we can see the output we need can be obtained by calling zip with,

zip [1, 2, 3] [2, 3, 4]

Now, from the docs on zip function, we can see that if the two given lists are of unequal length, the extra items in the longer list are discarded. So, we'd get the same result with,

zip [1, 2, 3, 4] [2, 3, 4]

in which the last 4 in the first input list would be discarded and we get the result we want.

This can be written in a function as,

pairs xs = zip xs (tail xs)

If it is something else you are confused about, do let me know.