Defining my own concat

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I'm reading a book and it asked me to define my own concat function. I defined it correctly here:

concat :: [[a]] -> [a]
concat [] = []
-- concat [[]] = []
concat (xs:xss) = xs ++ (Main.concat xss)

I have two questions about this.

  1. Why don't I need that line I commented out?
  2. When I call this with Main.concat [[]], step by step, how is it evaluated? The way I think about it, it goes into the 2nd definition, but I can't make sense of that. If I'm right and it goes to the 2nd definition, what are the values of xs and xss?

If I'm right and it goes to the 2nd definition, what are the values of xs and xss?

Let's ask:

> f (xs:xss) = (xs, xss)
> f [[]]
([],[])

So now try substituting these values into your definition:

concat (xs:xss) = xs ++ (Main.concat xss)
                = [] ++ (Main.concat [] )
                = [] ++ []
                = []

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[[]] can also be written as [] : [] (using the (:) constructor) -- that is, the list whose head is the empty list and whose tail is the empty list. So this can successfully pattern match against xs:xss.

So after matching xs with [] and xss with [], we effectively get the following behavior:

let xs = [] in
  let xss = [] in
    xs ++ (Main.concat xss)

And with substitution, we get

[] ++ (Main.concat [])

That recursive call will hit the base case and thus return [], giving [] ++ [] and eventually [].

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TL/DR: [[]] = [] : [] just as [1] = 1 : [], so it matches the x:xs pattern.

In Haskell a list is defined as so:

data [a]    -- a list of `a`s is...
  = []      -- an empty list...
  | a : [a] -- or an `a`, then a list of `a`s.

When we write [1,2,3], this is just syntactic sugar for 1 : (2 : (3 : [])), and since : is what we call right-associative, we can also just write 1:2:3:[].

For this reason there are only two cases that are normally required when pattern matching on lists: [] and (x:xs). [] is obviously the empty list, and (x:xs) is a non-empty list, where x is the first element and xs is the rest of the list.

Let's examine that definition, ignoring the commented line. I've removed some unnecessary brackets, by the way:

concat :: [[a]] -> [a]
concat []     = []                  -- The empty list case
concat (x:xs) = x ++ Main.concat xs -- The nonempty list case

We see that you've actually covered all the cases: this is now defined for all lists, so at least in theory the case concat [[]] = [] is redundant.

In practice this is also true, because concat [[]] = [] ++ concat [] = [] anyway, which falls into the second line's case, since [[]] = [] : [] which matches the (x:xs) pattern.

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Comments
  • hint: [] ++ [] == []
  • Hint 2: [x1,x2,x3,x4,...,xn] = x1 : x2 : x3 : x4 : ... : xn : []