Hello, I'm currently going through the exercises in chapter 3 of Real World Haskell. One of the exercises challenges me to create a function which takes a list and makes a palindrome of it. I tried to solve it this way: palindrome :: [t] -> [t] palindrome xs = xs ++ (reverse xs) where reverse [] = [] reverse (x:xs) = (reverse xs) ++ x But when I try to compile this I get this error: Kapitel3.hs:14:32-33: error: … • Couldn't match type ‘t’ with ‘[t]’ ‘t’ is a rigid type variable bound by the type signature for: palindrome :: forall t. [t] -> [t] at /Users/jona/programmering/haskell/boken/Kapitel3.hs:13:15 Expected type: [[t]] Actual type: [t] • In the first argument of ‘reverse’, namely ‘xs’ In the second argument of ‘(++)’, namely ‘(reverse xs)’ In the expression: xs ++ (reverse xs) • Relevant bindings include xs :: [t] (bound at /Users/jona/programmering/haskell/boken/Kapitel3.hs:14:12) palindrome :: [t] -> [t] (bound at /Users/jona/programmering/haskell/boken/Kapitel3.hs:14:1) Compilation failed. It looks like I have used a function that want a list of lists, but I don't understand where. Also, is there some way to declare the type of my reverse-function in this case? Kind regards, Jona Ekenberg
Looks like you're missing the square brackets around your x in the
definition of reverse (the ++ takes two lists, hence the error message)
Try this
palindrome :: [t] -> [t]
palindrome xs = xs ++ (reverse xs)
where
reverse [] = []
reverse (x:xs) = (reverse xs) ++ *[*x*]*
to declare the type you can define it outside or annotate it within you're
code
palindrome :: [t] -> [t]
palindrome xs = xs ++ (reverse xs)
where
* reverse :: [t] -> [t]*
reverse [] = []
reverse (x:xs) = (reverse xs) ++ *[*x*]*
On Tue, Jul 4, 2017 at 12:07 PM Jona Ekenberg
Hello,
I'm currently going through the exercises in chapter 3 of Real World Haskell. One of the exercises challenges me to create a function which takes a list and makes a palindrome of it.
I tried to solve it this way: palindrome :: [t] -> [t] palindrome xs = xs ++ (reverse xs) where reverse [] = [] reverse (x:xs) = (reverse xs) ++ x
But when I try to compile this I get this error: Kapitel3.hs:14:32-33: error: … • Couldn't match type ‘t’ with ‘[t]’ ‘t’ is a rigid type variable bound by the type signature for: palindrome :: forall t. [t] -> [t] at /Users/jona/programmering/haskell/boken/Kapitel3.hs:13:15 Expected type: [[t]] Actual type: [t] • In the first argument of ‘reverse’, namely ‘xs’ In the second argument of ‘(++)’, namely ‘(reverse xs)’ In the expression: xs ++ (reverse xs) • Relevant bindings include xs :: [t] (bound at /Users/jona/programmering/haskell/boken/Kapitel3.hs:14:12) palindrome :: [t] -> [t] (bound at /Users/jona/programmering/haskell/boken/Kapitel3.hs:14:1) Compilation failed.
It looks like I have used a function that want a list of lists, but I don't understand where. Also, is there some way to declare the type of my reverse-function in this case?
Kind regards, Jona Ekenberg _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
-- Frederic Cogny +33 7 83 12 61 69
On Tue, Jul 04, 2017 at 12:06:39PM +0200, Jona Ekenberg wrote:
Hello,
I'm currently going through the exercises in chapter 3 of Real World Haskell. One of the exercises challenges me to create a function which takes a list and makes a palindrome of it.
I tried to solve it this way: palindrome :: [t] -> [t] palindrome xs = xs ++ (reverse xs) where reverse [] = [] reverse (x:xs) = (reverse xs) ++ x
Hello Jona, let's analyse the error. It points to this bit: palindrome xs = xs ++ (reverse xs) And it says: I expected [[t]], but you gave me [t]. Whenever I encounter such an error I try to write explicit type signatures so to make diagnosing easier. In your example palindrome :: [t] -> [t] palindrome xs = xs ++ (reverse xs) where reverse :: [t] -> [t] -- explicit type signature reverse [] = [] reverse (x:xs) = (reverse xs) ++ x If we :reload ghci complains again, the offending bit being reverse (x:xs) = (reverse xs) ++ x ^ Expected type is [t1] but we passed t. Not it is clear! The type of `++` is: λ> :t (++) (++) :: [a] -> [a] -> [a] and `x` is a single element. When we replace `x` with `[x]` everything works. Does that help? -F P.S.: Real World Haskell is an excellent book but sometimes can be a tad difficult to follow. If you want to integrate with another source, CIS194 [1] is an excellent choice: free, thorough, full of home-works and interactive. [1] http://www.cis.upenn.edu/~cis194/fall16/
Thank you both for your answers, I somehow thought ++ acted as both append
and concat, since I've mostly used it on strings where I haven't had to
think about it.
And thank you for the tips regarding adding explicit types and the reading
material. So far I feel that I'm able to follow along quite well, but it's
nice to have a second source!
Grateful regards,
Jona
Den 4 juli 2017 12:21 em skrev "Francesco Ariis"
Hello,
I'm currently going through the exercises in chapter 3 of Real World Haskell. One of the exercises challenges me to create a function which takes a list and makes a palindrome of it.
I tried to solve it this way: palindrome :: [t] -> [t] palindrome xs = xs ++ (reverse xs) where reverse [] = [] reverse (x:xs) = (reverse xs) ++ x
Hello Jona, let's analyse the error. It points to this bit: palindrome xs = xs ++ (reverse xs) And it says: I expected [[t]], but you gave me [t]. Whenever I encounter such an error I try to write explicit type signatures so to make diagnosing easier. In your example palindrome :: [t] -> [t] palindrome xs = xs ++ (reverse xs) where reverse :: [t] -> [t] -- explicit type signature reverse [] = [] reverse (x:xs) = (reverse xs) ++ x If we :reload ghci complains again, the offending bit being reverse (x:xs) = (reverse xs) ++ x ^ Expected type is [t1] but we passed t. Not it is clear! The type of `++` is: λ> :t (++) (++) :: [a] -> [a] -> [a] and `x` is a single element. When we replace `x` with `[x]` everything works. Does that help? -F P.S.: Real World Haskell is an excellent book but sometimes can be a tad difficult to follow. If you want to integrate with another source, CIS194 [1] is an excellent choice: free, thorough, full of home-works and interactive. [1] http://www.cis.upenn.edu/~cis194/fall16/ _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
On 2017-07-04 12:06, Jona Ekenberg wrote:
Hello,
I'm currently going through the exercises in chapter 3 of Real World Haskell. One of the exercises challenges me to create a function which takes a list and makes a palindrome of it.
I tried to solve it this way:
palindrome :: [t] -> [t] palindrome xs = xs ++ (reverse xs) where reverse [] = [] reverse (x:xs) = (reverse xs) ++ x
But when I try to compile this I get this error:
[..] This is caused by Haskell trying hard to make your code type check but the noticing that it just cannot make the parts fit together. I suspect the cause of this is the definition reverse (x:xs) = (reverse xs) ++ x What's noteworthy about this is: 1. The type of the ':' constructor you used in the pattern match is 'a -> [a] -> [a]', i.e. the first argument is of some type 'a' (in Haskell nomenclature, 'x :: a') and the second argument is a list (i.e. 'xs :: [a]'). So your compiler knows that no matter what type 'x' is, 'xs' will be a list of those things. 2. The type of the '++' function you used on the right-hand side is '[a] -> [a] -> [a]', i.e. the first and the second argument must be of the same type (a list of things). Since the second argument must be a list, this means that 'x' must be a list and hence (considering 1. above) 'xs' must be a list of lists. This means that your 'reverse' definition must be of type '[[a]] -> [a]': it takes a list of list of things and yields a list of things. I.e. it takes and returns different type sof things. This however is in conflict with your first definition: palindrome xs = xs ⧺ (reverse xs) For 'xs ⧺ (reverse xs)' to be sound (to 'type-check'), the expressions 'xs' and '(reverse xs)' have to be of the same type. And since 'xs' is also an argument to 'reverse' it means that 'reverse' has to take and yield values of the same type. You can resolve this conflict by changing reverse (x:xs) = (reverse xs) ++ x to reverse (x:xs) = (reverse xs) ++ [x] -- Frerich Raabe - raabe@froglogic.com www.froglogic.com - Multi-Platform GUI Testing
participants (4)
-
Francesco Ariis -
Frederic Cogny -
Frerich Raabe -
Jona Ekenberg