Nested-if in Haskell

Question

I follow the real world haskell, and there is an exercise in chapter 2.

My solution is

lastButOne xs = if null xs || null (tail xs)
                then []
                else if null (tail (tail xs))
                     then head xs
                     else lastButOne (tail xs)

but it doesn't work other than [], and produces such an error.

*Main> lastButOne []
[]
*Main> lastButOne [1, 2]

<interactive>:5:13:
    No instance for (Num [a0]) arising from the literal `1'
    Possible fix: add an instance declaration for (Num [a0])
    In the expression: 1
    In the first argument of `lastButOne', namely `[1, 2]'
    In the expression: lastButOne [1, 2]

I am fairly a newbie and don't understand the cryptic error messages. Any ideas?

You'll find this much easier to do if you pattern match on the list instead of trying to deconstruct it using null, head and tail. — Lee
– Lee, Commented Jun 6, 2013 at 14:29
Lee//Thank you for the suggestion, but I didn't see any mention about pattern matching in chapter 1 and chapter 2. So I don't know about it. — kReoSo
– kReoSo, Commented Jun 6, 2013 at 14:39

zw324 · Accepted Answer · 2013-06-06 14:35:21Z

7

This is a type problem. If you use GHCi, load this function into it and use

:t lastButOne

to see its type, which is

lastButOne :: [[a]] -> [a]

this is because if need to have the same type on the then and else branches, and since you are returning a [] in the then branch, Haskell thinks you are trying to return a list always, and since you are returning head xs on the else branch, it thinks you are writing a function works on list of lists.

However, [1, 2] is not a list of lists, so GHC yelled at you about the type mismatch error.

Also note if you write out the type definition explicitly, it wouldn't compile:

lastButOne :: [a] -> a
lastButOne xs = if null xs || null (tail xs)
            then []
            else if null (tail (tail xs))
                 then head xs
                 else lastButOne (tail xs)

GHCi gets you an error:

Couldn't match type `a' with `[a0]'
  `a' is a rigid type variable bound by
      the type signature for lastButOne :: [a] -> a at k.hs:2:1
In the expression: []
In the expression:
  if null xs || null (tail xs) then
      []
  else
      if null (tail (tail xs)) then head xs else lastButOne (tail xs)
In an equation for `lastButOne':
    lastButOne xs
      = if null xs || null (tail xs) then
            []
        else
            if null (tail (tail xs)) then head xs else lastButOne (tail xs)

edited Jun 6, 2013 at 14:35

answered Jun 6, 2013 at 14:29

zw324

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1 Comment

kReoSo Over a year ago

Thank you. This answer is just what I wanted. So, [] is interpreted as a list and that was the problem...

Cubic · Accepted Answer · 2013-06-06 14:29:40Z

1

then [] here you return a list.

then head xs here you return what's in the list (a number in your case). I'm surprised this compiles at all. You should wrap the result with a Maybe, so the result of lastButOne [] and lastButOne [x] should be Nothing, and the result of lastButOne [...,x,_] should be Just x.

Or you could use the error pseudofunction in the error case.

answered Jun 6, 2013 at 14:29

Cubic

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kReoSo Over a year ago

I see. Following Ziyao's answer, the type of the lastButOne is [[a]]->[a], and it works well for such as lastButOne [[1,2], [1, 2, 3], [1, 2, 3, 4]]. I think, that's the reason why this is compiled successfully.

fortran · Accepted Answer · 2013-06-24 09:15:10Z

1

I think pattern matching is more elegant for this... Being myself a total newbie to Haskell (I just read a little bit about it long ago):

lastButOne ([]) = []
lastButOne (beforeLast:last:[]) = beforeLast
lastButOne (x:xs) = lastButOne xs

I know that it's not an explanation to your error, but sometimes the best solution is to avoid the problem at all!

edited Jun 24, 2013 at 9:15

answered Jun 6, 2013 at 14:34

fortran

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2 Comments

AndrewC Over a year ago

This is definitely a cleaner solution. @Jong-BeomKim - it would be worth looking at how much nicer this is than all the ifs you have at the moment.

kReoSo Over a year ago

Thank you, I know the pattern matching now, but it is introduced in chapter 3 and the exercise is at the end of the chapter 2. It seems like the exercises and some examples in Real World Haskell demand more knowledge than that is already taught. As a completely beginner, I should find more organized tutorial...

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