other questions and problems, although similar, are not quite like this one. in this specific compiler error, the Haskell GHC won't compile the following code, for the following reason. I don't understand at all - the code is pretty straight forward.
--factorial
fact :: int -> int
fact 0 = 1
fact n | n > 0 = n * fact(n - 1)
main = print (fact 10)
(error:)
No instance for (Ord int) arising from a use of `>'
Possible fix:
add (Ord int) to the context of
the type signature for fact :: int -> int
In the expression: n > 0
In a stmt of a pattern guard for
an equation for `fact':
n > 0
In an equation for `fact': fact n | n > 0 = n * fact (n - 1)
Can you explain the problem to me?
Int is what you want:
fact :: int -> int
-->
fact :: Int -> Int
Since in Haskell, types need to begin with a cap.
Edit: Thank Yuras for commenting this:
Or if you want you could use a type class:
fact :: Integral a => a -> a
And you can name the type variable whichever you like, including int. Also, Num might fit your purpose better if you want to define factorial over general numbers.
Related
If I compile the following source file with ghc -Wall:
main = putStr . show $ squareOfSum 5
squareOfSum :: Integral a => a -> a
squareOfSum n = (^2) $ sum [1..n]
I get:
powerTypes.hs:4:18: warning: [-Wtype-defaults]
• Defaulting the following constraints to type ‘Integer’
(Integral b0) arising from a use of ‘^’ at powerTypes.hs:4:18-19
(Num b0) arising from the literal ‘2’ at powerTypes.hs:4:19
• In the expression: (^ 2)
In the expression: (^ 2) $ sum [1 .. n]
In an equation for ‘squareOfSum’:
squareOfSum n = (^ 2) $ sum [1 .. n]
|
4 | squareOfSum n = (^2) $ sum [1..n]
| ^^
I understand that the type of (^) is:
Prelude> :t (^)
(^) :: (Integral b, Num a) => a -> b -> a
which means it works for any a^b provided a is a Num and b is an Integral. I also understand the type hierarchy to be:
Num --> Integral --> Int or Integer
where --> denotes "includes" and the first two are typeclasses while the last two are types.
Why does ghc not conclusively infer that 2 is an Int, instead of "defaulting the constraints to Integer". Why is ghc defaulting anything? Is replacing 2 with 2 :: Int a good way to resolve this warning?
In Haskell, numeric literals have a polymorphic type
2 :: Num a => a
This means that the expression 2 can be used to generate a value in any numeric type. For instance, all these expression type-check:
2 :: Int
2 :: Integer
2 :: Float
2 :: Double
2 :: MyCustomTypeForWhichIDefinedANumInstance
Technically, each time we use 2 we would have to write 2 :: T to choose the actual numeric type T we want. Fortunately, this is often not needed since type inference can frequently deduce T from the context. E.g.,
foo :: Int -> Int
foo x = x + 2
Here, x is an Int because of the type annotation, and + requires both operands to have the same type, hence Haskell infers 2 :: Int. Technically, this is because (+) has type
(+) :: Num a => a -> a -> a
Sometimes, however, type inference can not deduce T from the context. Consider this example involving a custom type class:
class C a where bar :: a -> String
instance C Int where bar x = "Int: " ++ show x
instance C Integer where bar x = "Integer: " ++ show x
test :: String
test = bar 2
What is the value of test? Well, if 2 is an Int, then we have test = "Int: 2". If it is an Integer, then we have test = "Integer: 2". If it's another numeric type T, we can not find an instance for C T.
This code is inherently ambiguous. In such a case, Haskell mandates that numeric types that can not be deduced are defaulted to Integer (the programmer can change this default to another type, but it's not relevant now). Hence we have test = "Integer: 2".
While this mechanism makes our code type check, it might cause an unintended result: for all we know, the programmer might have wanted 2 :: Int instead. Because of this, GHC chooses the default, but warns about it.
In your code, (^) can work with any Integral type for the exponent. But, in principle, x ^ (2::Int) and x ^ (2::Integer) could lead to different results. We know this is not the case since we know the semantics of (^), but for the compiler (^) is only a random function with that type, which could behave differently on Int and Integer. Consider, e.g.,
a ^ n = if n + 3000000000 < 0 then 0 else 1
When n = 2, if we use n :: Int the if guard could be true on a 32 bit system. This is not the case when using n :: Integer which never overflows.
The standard solution, in these cases, is to resolve the warning using something like x ^ (2 :: Int).
This question already has answers here:
`String' is applied to too many type arguments
(2 answers)
Closed 2 years ago.
I'm starting to learn Haskell and I'm struggling with syntax in Functions. I'm trying to create a function that receives a Number 'e' and a Number 'n', returning a list with of 'n' times 'e'.
Example: repn 3 5 would return [3,3,3,3,3]:
repn :: Int a => a -> a -> [a]
repn e n
| n >= 1 = (take n (repeat e))
| otherwise = []
But I'm getting this error:
* Expected kind `* -> Constraint', but `Int' has kind `*'
* In the type signature: replica :: Int a => a -> a
As a rule of thumb, writing something of the form Foo a => ... only makes sense of Foo is a type class. Unlike in OO languages, a class and a type in Haskell are completely different things. Int is a type, therefore it cannot be used in this way; instead you should simply use
repn :: Int -> Int -> [Int]
...though actually there's no reason to restrict the list-element type at all, the signature could as well be
repn :: a -> Int -> [a]
Alternatively, you can use a type class: the class of “int-like types” is Integral. Then you can't use take directly, as that is restricted to Int for the counter argument; however you can convert any integral type to Int:
repn :: Integral n => a -> n -> [a]
repn e n = take (fromIntegral n) $ repeat e
-- Note no case distinction is needed, because `take` already yields
-- an empty list if `n<1`.
Your type definition is not correct. The double arrow is used to show required typeclasses (aka kind * -> Constraint), or a constraint on types. However, Int is itself a type (aka kind *).
You can't specify types in this way. In your implementation, n must be an Int, but e could be anything. Since you specify that it should be a number, though, you can constrain it with the Num typeclass. The correct version of your signature therefore is:
repn :: Num a => a -> Int -> [a]
If you want e to be constrained to an Int, then your signature should be
repn :: Int -> Int -> [Int]
If you don't need any constraints on e, your signature can be
repn :: a -> Int -> [a]
These are all valid type signatures for the function you have written. (Though in actual code, I would just point you to the builtin replicate function.)
other questions and problems, although similar, are not quite like this one. in this specific compiler error, the Haskell GHC won't compile the following code, for the following reason. I don't understand at all - the code is pretty straight forward.
--factorial
fact :: int -> int
fact 0 = 1
fact n | n > 0 = n * fact(n - 1)
main = print (fact 10)
(error:)
No instance for (Ord int) arising from a use of `>'
Possible fix:
add (Ord int) to the context of
the type signature for fact :: int -> int
In the expression: n > 0
In a stmt of a pattern guard for
an equation for `fact':
n > 0
In an equation for `fact': fact n | n > 0 = n * fact (n - 1)
Can you explain the problem to me?
Int is what you want:
fact :: int -> int
-->
fact :: Int -> Int
Since in Haskell, types need to begin with a cap.
Edit: Thank Yuras for commenting this:
Or if you want you could use a type class:
fact :: Integral a => a -> a
And you can name the type variable whichever you like, including int. Also, Num might fit your purpose better if you want to define factorial over general numbers.
In Haskell, if I create a dataype like this:
data MyT = MyT Int deriving (Show)
myValue = MyT 42
I can get the Int value passing 'myValue' to a function and doing pattern matching:
getInt :: MyT -> Int
getInt (MyT n) = n
It seems to me that something simpler should be possible. Is there another way?
Also, I tried a lambda function:
(\(MyT n) -> n) myValue
It doesn't work and I don't understand why not.
I get the error:
The function `\ (MyT n) -> n' is applied to two arguments,
but its type `MyT -> Int' has only one
EDIT:
Of course, sepp2k below, is right about my lambda function working OK. I was doing:
(\(MyT n) -> n) myT 42
instead of
(\(MyT n) -> n) (myT 42)
If you want to get at the value of MyT inside a larger function without defining a helper function, you could either use case of or pattern matching in local variable definitions. Here are examples of that (assuming that g produces a MyT and f takes an Int):
Using case:
myLargerFunction x = f (case g x of MyT n => n)
Or with local variables:
myLargerFunction x = f myInt
where MyT myInt = g x
Or using let instead of where:
myLargerFunction x =
let MyT myInt = g x in
f myInt
Your lambda function should (and in fact does) also work fine. Your error message suggests that in your real code you're really doing something like (\(MyT n) -> n) myValue somethingElse (presumably by accident).
You can use the record syntax
data MyT = MyT {unMyT :: Int} deriving (Show)
which gives you the projection for free
unMyT :: MyT -> Int
This is nice if your data type has only one constructor (including newtypes). For data types involving more than one constrctor, projection functions tend to be unsafe (e.g., head,tail), and pattern matching is usually preferred instead. GHC checks for non-exhaustive patterns if you enable warnings, and can help to spot errors.
NewTypes create a distinct type and do not have an extra level of indirection like algebraic datatypes. See the Haskell report for more information:
http://www.haskell.org/onlinereport/decls.html#sect4.2.3
Prelude> newtype Age = Age { unAge :: Int } deriving (Show)
Prelude> let personAge = Age 42
Prelude> personAge
Age {unAge = 42}
Prelude> (unAge personAge) + 1
43
Using a lambda function:
Prelude> (\(Age age) -> age * 2) personAge
84
I need a function which works like:
some :: (Int, Maybe Int) -> Int
some a b
| b == Nothing = 0
| otherwise = a + b
Use cases:
some (2,Just 1)
some (3,Nothing)
map some [(2, Just 1), (3,Nothing)]
But my code raise the error:
The equation(s) for `some' have two arguments,
but its type `(Int, Maybe Int) -> Int' has only one
I don't understand it.
Thanks in advance.
When you write
foo x y = ...
That is notation for a curried function, with a type like:
foo :: a -> b -> c
You have declared your function to expect a tuple, so you must write it:
some :: (Int, Maybe Int) -> Int
some (x, y) = ...
But Haskell convention is usually to take arguments in the former curried form. Seeing funcitons take tuples as arguments is very rare.
For the other part of your question, you probably want to express it with pattern matching. You could say:
foo :: Maybe Int -> Int
foo Nothing = 0
foo (Just x) = x + 1
Generalizing that to the OP's question is left as an exercise for the reader.
Your error doesn't come from a misunderstanding of Maybe: The type signature of some indicates that it takes a pair (Int, Maybe Int), while in your definition you provide it two arguments. The definition should thus begin with some (a,b) to match the type signature.
One way to fix the problem (which is also a bit more idiomatic and uses pattern matching) is:
some :: (Int, Maybe Int) -> Int
some (a, Nothing) = a
some (a, Just b) = a + b
It's also worth noting that unless you have a really good reason for using a tuple as input, you should probably not do so. If your signature were instead some :: Int -> Maybe Int -> Int, you'd have a function of two arguments, which can be curried. Then you'd write something like
some :: Int -> Maybe Int -> Int
some a Nothing = a
some a (Just b) = a + b
Also, you might want to add the following immediate generalization: All Num types are additive, so you might aswell do
some :: (Num n) => n -> Maybe n -> n
some a Nothing = a
some a (Just b) = a + b
(I've violated the common practice of using a, b, c... for type variables so as not to confuse the OP since he binds a and b to the arguments of some).