With dependent types, you can either capture function properties in the type signature, like in concatenation with length-indexed lists
(++) : Vect m a -> Vect n a -> Vect (m + n) a
or you can not use dependent types in your signature, like concatenation with standard lists
(++) : List a -> List a -> List a
and write proofs about (++)
appendAddsLength : (xs, ys : List a) -> length (xs ++ ys) = length xs + length ys
lengthNil : length [] = 0
lengthCons : (x : a) -> (xs : List a) -> length (x :: xs) = length xs + 1
Is there any difference between these approaches beyond ergonomics?
The most obvious difference is that, with (++) on Vects, the length is statically known: you can operate on it at compile time. Moreover, you don't need to write any additional proofs in order to ensure that (++) has the expected behavior on Vects, whereas you need to do it for Lists.
That is, (++) on Vect is correct by construction. The compiler will always enforce the desired properties, whether you like it or not, and without the user taking any additional action.
It's important to note that length xs is not really interchangeable with the statically-known size in general. On these data types, length is a function which actually re-computes the length of a List or Vect by walking through it and incrementing a counter:
libs/prelude/Prelude/Types.idr:L403-L407
namespace List
||| Returns the length of the list.
public export
length : List a -> Nat
length [] = Z
length (x :: xs) = S (length xs)
libs/base/Data/Vect.idr:25-28
public export
length : (xs : Vect len elem) -> Nat
length [] = 0
length (_::xs) = 1 + length xs
Even with Vect, the length built into to the type by construction, but the result of applying the length function to a List or Vect is not fundamental at all. In fact, Data.Vect contains a proof that Data.Vect.length that applying length to a Vect n t always returns n:
libs/base/Data/Vect.idr:30-34
||| Show that the length function on vectors in fact calculates the length
export
lengthCorrect : (xs : Vect len elem) -> length xs = len
lengthCorrect [] = Refl
lengthCorrect (_ :: xs) = rewrite lengthCorrect xs in Refl
Using the above proof, we can assert statically, without actually executing length, that the result of length is propositionally equal to the statically-known length of the Vect. But this assurance is not available for List. And it's much more cumbersome to work with in general, likely requiring the use of with ... proof and rewrite a lot more than just using the correct-by-construction type.
Related
I'm learning Haskell and have some problems with list comprehension.
If I define a function to get a list of the divisors of a given number, I get an error.
check n = [x | x <- [1..(floor (n/2))], mod n x == 0]
I don't get why it's causing an error. If I want to generate a list from 1 to n/2 I can do it with [1..(floor (n/2))], but not if I do it in the list comprehension.
I tried another way but I get also an error (in this code I want to get all so called "perfect numbers")
f n = [1..(floor (n/2))]
main = print $ filter (\t -> foldr (+) 0 (f t) == t) [2..100]
Usually it is better to start writing a signature. While signatures are often not required, it makes it easier to debug a single function.
The signature of your check function is:
check :: (RealFrac a, Integral a) => a -> [a]
The type of input (and output) a thus needs to be both a RealFrac and an Integral. While technically speaking we can make such type, it does not make much sense.
The reason this happens is because of the use of mod :: Integral a => a -> a -> a this requires x and n to be both of the same type, and a should be a member of the Integral typeclass.
Another problem is the use of n/2, since (/) :: Fractional a => a -> a -> a requires that n and 2 have the same type as n / 2, and n should also be of a type that is a member of Fractional. To make matters even worse, we use floor :: (RealFrac a, Integral b) => a -> b which enforces that n (and thus x as well) have a type that is a member of the RealFrac typeclass.
We can prevent the Fractional and RealFrac type constaints by making use of div :: Integral a => a -> a -> a instead. Since mod already required n to have a type that is a member of the Integral typeclass, this thus will not restrict the types further:
check n = [x | x <- [1 .. div n 2], mod n x == 0]
This for example prints:
Prelude> print (check 5)
[1]
Prelude> print (check 17)
[1]
Prelude> print (check 18)
[1,2,3,6,9]
I'm new to haskell and nothing I've read online really helped me with this. My assignment is to implement a typeclass VEC that has a function factor which either calculates the multiplication of two Integers or performs the dot product of two matrices if the user input wasn't two numbers but instead two lists of integers of type vector. My code looks as follows:
data Vector = VECTOR [Int] deriving (Show, Eq)
class VEC a where
factor :: a -> a -> a
instance VEC Int where
factor a b = a*b
instance VEC Vector where
factor xs [] = 0
factor xs ys = (head xs) * (head ys) + factor (tail xs) (tail ys)
I assumed the Vectors were of type [Int] but now I'm not so sure, since I get the following error message when trying to compile the code using Hugs:
Hugs> :l kdp
ERROR file:kdp.hs:7 - Type error in instance member binding
*** Term : []
*** Type : [a]
*** Does not match : Vector
So my questions would be: What does the first line actually mean? It was given along with the assignment and I've seen many similar definitions of data types but none with this particular pattern. And then how do I fix the problem of not being able to use predefined functions such as tail or in the above error case comparing a list of Integers with my custom data type? My guess is I have to define the operations on my own but I couldn't figure out how to do that.
When you write an instance for a class like
class C a where
method :: a -> a
you replace all of the parameter appearances (a) with the type you are writing instance for. For example in your case:
{-# LANGUAGE InstanceSigs #-}
instance VEC Vector where
factor :: Vector -> Vector -> Vector
factor _ _ = undefined
So you cannot match arguments of type Vector with pattern [] nor use head nor tail functions on it as they are working on lists. However, your Vector consists of lists, so you can simply unpack it:
instance VEC Vector where
factor _ (Vector []) = Vector [0] -- you need to return Vector as well
factor (Vector xs) (Vector ys) =
let Vector [x] = factor (Vector $ tail xs) (Vector $ tail ys)
in Vector [(head xs) * (head ys) + x]
This is very ugly and partial tho, you can use some builtin Data.List machinery to make it more sexy:
instance VEC Vector where
factor (Vector xs) (Vector ys) =
Vector [sum (zipWith (*) xs ys)]
As you are using deriving (Show, Eq), the == operator should just work.
You can generalize your type class to take two type variables to accommodate the two different operations of integer multiplication and vector dot products. This requires two GHC-specific extensions, though, precluding your use of Hugs.
{-# LANGUAGE MultiParamTypeClasses, FunctionalDependencies #-}
-- a -> b means that the second type is uniquely determined by the
-- first. Without it, you would need explicitly type an expression
-- involving factor.
class VEC a b | a -> b where
factor :: a -> a -> b
instance VEC Int Int where
-- a ~ Int, b ~ Int
factor x y = x * y
-- newtype instead of data, for efficiency if nothing else
newtype Vector = VECTOR [Int] deriving (Show, Eq)
instance VEC Vector Int where
-- a ~ Vector, b ~ Int
factor (VECTOR xs) (VECTOR ys) = sum $ zipWith (*) xs ys
main = do
print $ factor (3 :: Int) 3
print $ factor (VECTOR [1,2,3]) (VECTOR [4,5,6])
Without the functional dependency, you would have to write
main = do
print (factor (3 :: Int) 3 :: Int)
print (factor (VECTOR [1,2,3]) (VECTOR [4,5,6]) :: Int)
I want to write a function that takes a mathematical function (/,x,+,-), a number to start with and a list of numbers. Then, it's supposed to give back a list.
The first element is the starting number, the second element the value of the starting number plus/minus/times/divided by the first number of the given list. The third element is the result of the previous result plus/minus/times/divided by the second result of the given list, and so on.
I've gotten everything to work if I tell the code which function to use but if I want to let the user input the mathematical function he wants, there are problems with the types. Trying :t (/) for example gives out Fractional a => a -> a -> a, but if you put that at the start of your types, it fails.
Is there a specific type to distinguish these functions (/,x,+,-)? Or is there another way to write this function succesfully?
prefix :: (Fractional a, Num a) => a -> a -> a -> a -> [a] -> [a]
prefix (f) a b = [a] ++ prefix' (f) a b
prefix' :: (Fractional a, Num a) => a -> a -> a -> a -> [a] -> [a]
prefix' (z) x [] = []
prefix' (z) x y = [x z (head y)] ++ prefix' (z) (head (prefix' (z) x y)) (tail y)
A right solution would be something like this:
prefix (-) 0 [1..5]
[0,-1,-3,-6,-10,-15]
Is there a specific type to distinguish these functions (/,*,+,-)?
I don't see a reason to do this. Why is \x y -> x+y considered "better" than \x y -> x + y + 1. Sure adding two numbers is something that most will consider more "pure". But it is strange to restrict yourself to a specific subset of functions. It is also possible that for some function \x y -> f x y - 1 "happens" to be equal to (+), except that the compiler can not determine that.
The type checking will make sure that one can not pass functions that operate on numbers, given the list contains strings, etc. But deliberately restricting this further is not very useful. Why would you prevent programmers to use your function for different purposes?
Or is there another way to write this function succesfully?
What you here describe is the scanl :: (b -> a -> b) -> b -> [a] -> [b] function. If we call scanl with scanl f z [x1, x2, ..., xn], then we obtain a list [z, f z x1, f (f z x1) x2, ...]. scanl can be defined as:
scanl :: (b -> a -> b) -> b -> [a] -> [b]
scanl f = go
where go z [] = [z]
go z (x:xs) = z : go (f z x) xs
We thus first emit the accumulator (that starts with the initial value), and then "update" the accumulator to f z x with z the old accumulator, and x the head of the list, and recurse on the tail of the list.
If you want to restrict to these four operations, just define the type yourself:
data ArithOp = Plus | Minus | Times | Div
as_fun Plus = (+)
as_fun Minus = (-)
as_fun Times = (*)
as_fun Div = (/)
This question already has answers here:
Can you overload + in haskell?
(5 answers)
Closed 4 years ago.
I am trying to override the + symbol in an effort to learn how to define my own types. I am very new to Haskell, and I cannot seem to get past this error.
Here is my simple new type:
newtype Matrix x = Matrix x
(+):: (Num a, Num b, Num c) => Matrix [[a]] -> Matrix [[b]] -> Matrix [[c]]
x + y = Matrix zipWith (\ a b -> zipWith (+) a b) x y
When I try to load this into ghci, I get the error
linear_algebra.hs:9:42:
Ambiguous occurrence ‘+’
It could refer to either ‘Main.+’, defined at linear_algebra.hs:9:3
or ‘Prelude.+’,
imported from ‘Prelude’ at linear_algebra.hs:1:1
(and originally defined in ‘GHC.Num’)
Failed, modules loaded: none.
Replacing my last line of code with
x + y = Matrix zipWith (\ a b -> zipWith (Prelude.+) a b) x y
gives me the error
Couldn't match expected type ‘([Integer] -> [Integer] -> [Integer])
-> Matrix [[a]] -> Matrix [[b]] -> Matrix [[c]]’
with actual type ‘Matrix
((a0 -> b0 -> c0) -> [a0] -> [b0] -> [c0])’
Relevant bindings include
y :: Matrix [[b]] (bound at linear_algebra.hs:9:5)
x :: Matrix [[a]] (bound at linear_algebra.hs:9:1)
(+) :: Matrix [[a]] -> Matrix [[b]] -> Matrix [[c]]
(bound at linear_algebra.hs:9:1)
The function ‘Matrix’ is applied to four arguments,
but its type ‘((a0 -> b0 -> c0) -> [a0] -> [b0] -> [c0])
-> Matrix ((a0 -> b0 -> c0) -> [a0] -> [b0] -> [c0])’
has only one
In the expression:
Matrix zipWith (\ a b -> zipWith (Prelude.+) a b) x y
In an equation for ‘+’:
x + y = Matrix zipWith (\ a b -> zipWith (Prelude.+) a b) x y
Failed, modules loaded: none.
Can you please help me understand what the error is? I would really appreciate it. Thanks!
First of all, the type
(+) :: (Num a, Num b, Num c) => Matrix [[a]] -> Matrix [[b]] -> Matrix [[c]]
is too much general. It states that you can sum any numeric matrix with any other numeric matrix, even if the element types are different, to produce a matrix of a third numeric type (potentially distinct from the first two). That is, in particular a matrix of floats can be summed to a matrix of doubles to produce a matrix of ints.
You want instead
(+) :: Num a => Matrix [[a]] -> Matrix [[a]] -> Matrix [[a]]
I would recommend to move the "list of list" type inside the newtype
newtype Matrix a = Matrix [[a]]
reflecting that the list of lists implements the Matrix concept. That gives the type signature
(+) :: Num a => Matrix a -> Matrix a -> Matrix a
To "override" (+): there's no overriding/overloading in Haskell. The closest options are:
define a module-local function (+). This will clash with Prelude.(+), so that every + will now need to be qualified to remove the ambiguity. We can not write x + y, but we need x Prelude.+ y or x MyModuleName.+ y.
implement a Num instance for Matrix a. This is not a great idea since a matrix is not exactly a number, but we can try anyway.
instance Num a => Num (Matrix a) where
Matrix xs + Matrix ys = Matrix (zipWith (zipWith (+)) xs ys)
-- other Num operators here
(*) = error "not implemented" -- We can't match dimension
negate (Matrix xs) = Matrix (map (map negate) xs)
abs = error "not implemented"
signum = error "not implemented"
fromInteger = error "not implemented"
This is very similar to your code, which lacks some parentheses. Not all the other methods can be implemented in a completely meaningful way, since Num is for numbers, not matrices.
use a different operator, e.g. (^+) or whatever
I have below code to take the args to set some offset time.
setOffsetTime :: (Ord a, Num b)=>[a] -> b
setOffsetTime [] = 200
setOffsetTime (x:xs) = read x::Int
But compiler says "Could not deduce (b ~ Int) from the context (Ord a, Num b) bound by the type signature for setOffsetTime :: (Ord a, Num b) => [a] -> b
Also I found I could not use 200.0 if I want float as the default value. The compilers says "Could not deduce (Fractional b) arising from the literal `200.0'"
Could any one show me some code as a function (not in the prelude) that takes an arg to store some variable so I can use in other function? I can do this in the main = do, but hope
to use an elegant function to achieve this.
Is there any global constant stuff in Hasekll? I googled it, but seems not.
I wanna use Haskell to replace some of my python script although it is not easy.
I think this type signature doesn't quite mean what you think it does:
setOffsetTime :: (Ord a, Num b)=>[a] -> b
What that says is "if you give me a value of type [a], for any type a you choose that is a member of the Ord type class, I will give you a value of type b, for any type b that you choose that is a member of the Num type class". The caller gets to pick the particular types a and b that are used each time setOffsetTime is called.
So trying to return a value of type Int (or Float, or any particular type) doesn't make sense. Int is indeed a member of the type class Num, but it's not any member of the type class Num. According to that type signature, I should be able to make a brand new instance of Num that you've never seen before, import setOffsetTime from your module, and call it to get a value of my new type.
To come up with an acceptable return value, you can only use functions that likewise return an arbitrary Num. You can't use any functions of particular concrete types.
Existential types are essentially a mechanism for allowing the callee to choose the value for a type variable (and then the caller has to be written to work regardless of what that type is), but that's not really something you want to be getting into while you're still learning.
If you are convinced that the implementation of your function is correct, i.e., that it should interpret the first element in its input list as the number to return and return 200 if there is no such argument, then you only need to make sure that the type signature matches that implementation (which it does not do, right now).
To do so, you could, for example, remove the type signature and ask ghci to infer the type:
$ ghci
GHCi, version 7.6.2: http://www.haskell.org/ghc/ :? for help
Loading package ghc-prim ... linking ... done.
Loading package integer-gmp ... linking ... done.
Loading package base ... linking ... done.
Prelude> :{
Prelude| let setOffsetTime [] = 200
Prelude| setOffsetTime (x : xs) = read x :: Int
Prelude| :}
Prelude> :t setOffsetTime
setOffsetTime :: [String] -> Int
Prelude> :q
Leaving GHCi.
$
And indeed,
setOffsetTime :: [String] -> Int
setOffsetTime [] = 200
setOffsetTime (x : xs) = read x :: Int
compiles fine.
If you want a slightly more general type, you can drop the ascription :: Int from the second case. The above method then tells you that you can write
setOffsetTime :: (Num a, Read a) => [String] -> a
setOffsetTime [] = 200
setOffsetTime (x : xs) = read x
From the comment that you added to your question, I understand that you want your function to return a floating-point number. In that case, you can write
setOffsetTime :: [String] -> Float
setOffsetTime [] = 200.0
setOffsetTime (x : xs) = read x
or, more general:
setOffsetTime :: (Fractional a, Read a) => [String] -> a
setOffsetTime [] = 200.0
setOffsetTime (x : xs) = read x