Haskell Recursion and Type Error

Question

I'm teaching myself Haskell and the best way to learn any programming language is to use it. My current "exercise" is an implementation of take. The pseudo-code is:

take(0, list) = [] --empty list
take(n, list) = const(head(list), take(n-1, tail(list))

What I've worked out in Haskell is:

myTake :: (Num a) => a -> [b] -> [b]
myTake 0 l = []
myTake n (l:ls) = l :  myTake n-1 ls

This doesn't compile when I load the file in GHCi. This is the error message I get:

Couldn't match expected type `[b]'
       against inferred type `[b1] -> [b1]'
In the second argument of `(:)', namely `myTake n - 1 ls'
In the expression: l : myTake n - 1 ls
In the definition of `myTake':
    myTake n (l : ls) = l : myTake n - 1 ls

My current Haskell resource is "Learn You a Haskell for Great Good!" and I've read the section on types several times trying to figure this out. Google has been unusually unhelpful too. I think that I don't entirely understand typing yet. Can anyone explain what's going wrong?

Josh Lee · Accepted Answer · 2011-02-24 04:35:52Z

7

myTake n - 1 ls

is parsed like

(myTake n) - (1 ls)

because function application binds higher than any infix operator. Parenthesize it:

myTake (n - 1) ls

answered Feb 24, 2011 at 4:35

Josh Lee

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4 Comments

T Suds Over a year ago

Thanks, that solved it. I'm glad it was just something simple like that.

fuz Over a year ago

As a rule of thumb, always parenthesize expression which are an argument to a prefix function, if they are more than just a variable name.

Dan Burton Over a year ago

Note this is why x : f y works, because the function application f y binds more tightly than :, so it's as if you had parenthesized x : (f y)

porges Over a year ago

Actually the error message is this way because : binds more tightly than -, so it's parsed as (l : (myTake n)) - (1 ls). This is why it complains that the second argument of : is a function - myTake is only partially applied.

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