In Haskell we have the function (==) :: Eq a => a->a->Bool which is fine for the large number of datatypes for which an instance of Eq can be defined. However there are some types for which there is no reasonable way to define an instance of Eq. one example is the simple function type (a->b)
I am looking for a function that will tell me if two values are actually the same -- not equal but identical.
For example f(id,id) ==> True f((+),(-)) = False
Clarification:
I don't want to know if two function do the same thing. It is not possible in the general case to do so. I want to know if I've gotten back the same thing I started with. Let me give a stripped down example:
data Foo = Foo (Foo->Foo) --contains a function so no Eq instance
x :: Foo
x = Foo id -- or for that matter Foo undefined
y :: Foo
y = Foo (const x)
a :: Foo
a = let (Foo fy) = y
in fy x
It is clear that by inspection once evaluated, a will be x. But let's assume I don't know the function in y but I want to test if the Foo I put in is the same one I got back - that is does fy x give me x. How do I do this?
One way that wasn't mentioned in Pointer equality in Haskell? is reallyUnsafePtrEquality#. As the name suggests it can be unpredictable and probably should't be used but it can be interesting to see how ghc works. Here's how you can use it in ghci:
> :set -package ghc-prim
> import GHC.Prim
> import GHC.Types
> isTrue# (reallyUnsafePtrEquality# id id)
True
> let a x = x + 2 :: Int
> isTrue# (reallyUnsafePtrEquality# a a)
True
> let b x = x + 2 :: Int
> isTrue# (reallyUnsafePtrEquality# a b)
False
If the function isn't monomorphic it doesn't work:
> let c x = x + 2
> isTrue# (reallyUnsafePtrEquality# c c)
False
Some more examples:
> let d = c in isTrue# (reallyUnsafePtrEquality# d d)
False
> :set -XMonomorphismRestriction
> let d = c in isTrue# (reallyUnsafePtrEquality# d d)
True
You can compare polymorphic types if you wrap them in a newtype:
> :set -XRankNTypes
> newtype X = X (forall a. Num a => a -> a)
> let d = X c
> isTrue# (reallyUnsafePtrEquality# d d)
True
Applying anything makes them not equal
> isTrue# (reallyUnsafePtrEquality# (id ()) (id ()))
False
But when compiling with optimisations this is True.
Hopefully this is enough to convince you that what you want is a bad idea. One of the solutions in Pointer equality in Haskell? would be a better.
Related
How can I define a function with the following signature,
f :: [Int???] -> [Int]
f xs = _ -- what I do with xs doesn't matter for the question
where a is a List of Int's
such that the first argument's inputs, i.e. list elements, must be >= 0, but <= 5 at compile-time?
In other words,
f [6] would fail to compile.
How about:
f :: [Int] -> [Int]
f = filter (\x -> x >= 0 && x <= 5)
Or do you want to enforce the bounds on the type (dependent types)?
If you want to restrict the range of the Int that is allowed you are probably better of using a smart constructor. Have a look here. The idea is that you create your own datatype and your own custom constructor:
newtype Range0_5 = Range0_5 { unRange :: Int }
makeRange0_5 :: Int -> Maybe Range0_5
makeRange0_5 x
| x >= 0 && x <= 5 = Just $ Range0_5 x
| otherwise = Nothing
If you make a smart constructor, it is important to not expose it to the user of the module. This can be done by simply not exporting the Range0_5 constructor.
However this is not a compile time check. Other languages than Haskell might be more appropriate if you really need such a feature.
Since the range is fairly small, you could also make a sum type to represent it:
data Range0_5 = Int0 | Int1 | Int2 | Int3 | Int4 | Int5
If the signature is
f :: [Int] -> [Int]
(which was the original form of the question), then it is impossible to enforce your constraint at compile time. This follows from the standard diagonalization argument of the Halting problem.
Suppose the compiler could detect that
f[g x]
should not compile. By incorporating the source code of the compiler into g, it could choose the opposite of the compiler's decision.
Following your comment on Liquid Haskell (which seems like a very interesting project), note the following:
{-# type Even = {v:Int | v mod 2 = 0} #-}
{-# foo :: n:Even -> {v:Bool | (v <=> (n mod 2 == 0))} #-}
foo :: Int -> Bool
foo n = if n^2 - 1 == (n + 1) * (n - 1) then True else foo (n - 1)
LiquidHaskell claims this function is unsafe, because, potentially foo n calls foo (n - 1). Note, however, that this will never happen: it will only be called if the relationship n2 - 1 ≠ (n + 1) (n - 1), which can never happen.
Again, this is not a criticism of the quality of LiquidHaskell, but rather just pointing out that it, too, cannot solve Halting Problem like issues.
When I have the two functions:
a)
three :: Int -> Maybe Int
three a
| a == 3 = Just 3
| otherwise = Nothing
b)
takeOne :: Int -> Int
takeOne a = (a - 1)
how do I call function a as a parameter to function b? i.e How do I let function b accept a 'Maybe Int' in place of an 'Int'?
At the minute when I try
takeOne (three 3)
I get the error:
ERROR - Type error in application
*** Expression : takeThree (three 3)
*** Term : three 3
*** Type : Maybe Int
*** Does not match : Int
Thanks.
You've got a few options, but I'd say the easiest is fmap:
fmap :: Functor f => (a -> b) -> f a -> f b
Example:
> fmap takeOne $ three 3
Just 2
> fmap takeOne $ three 2
Nothing
Another option would be to use the function maybe, which takes a default value, a function to apply to any value inside the Just, and then the Maybe a to apply this to. An example should make it clear
> maybe 0 takeOne $ three 3
2
> maybe 0 takeOne $ three 2
0
Another alternative if you just want to give a default value is to use the function fromMaybe from Data.Maybe:
> import Data.Maybe
> fromMaybe 0 $ three 3
3
> fromMaybe 0 $ three 2
0
In Haskell, there is a typeclass called Functor defined as
class Functor f where
fmap :: (a -> b) -> f a -> f b
There are many, many types that are instances of Functor. In fact, all parametrized data structures are Functors, as are all Applicatives and Monads. The easiest mental model of a Functor is that it's just a fancy name for a container. For lists, fmap = map, for example. All it does is map a function over the elements inside a container.
Some more examples are:
> fmap (+1) (Left "error")
Left "error"
> fmap (+1) (Right 1)
Right 2
> x <- fmap (++", world") getLine
Hello
> x
Hello, world
> fmap (+1) [1..5]
[2,3,4,5,6]
> fmap (+1) ("fst", 2)
("fst", 3)
Even functions are Functors! Here fmap = (.), it's just normal function composition:
> let lengthPlusOne = fmap (+1) length
> lengthPlusOne "Hello"
6
One other option of course is to write your own.
data IAcceptedAMaybeInt = GotAnswer Int | NothingAtTheMoment deriving Show
pleaseAcceptAMaybeInt f g a = case g a of
Just b -> GotAnswer (f b)
otherwise -> NothingAtTheMoment
Output:
*Main> pleaseAcceptAMaybeInt takeOne three 3
GotAnswer 2
*Main> pleaseAcceptAMaybeInt takeOne three 2
NothingAtTheMoment
I'm reading about lazy evaluations in haskell and have a question. For example we have following computations:
Prelude> let x = 1 + 1 :: Int
Prelude> let y = (x,x)
And after getting value of x:
Prelude> :sprint x
x = _
It's unevaluated. Ok, now let's get value of y:
Prelude> :sprint y
y = (_,_)
It is unevaluated too, because y depends on x and it's unevaluated. Now let's try the same example but without ::Int:
Prelude> let x = 1 + 1
Prelude> let y = (x, x)
Prelude> :sprint y
y = _
Why y value is _ instead (_, _) when we're trying without ::Int?
I see that they have different types:
Prelude> let x = 1 + 1
Prelude> :t x
x :: Num a => a
Prelude> let x = 1 + 1 :: Int
Prelude> :t x
x :: Int
But why values of y depends on it?
Thank you.
What is happening is that when you've specified x to have the type Num a => a, the compiler can't possibly know which instance of Num to use when performing 1 + 1. What it does instead is use defaulting. GHC defines default types for certain typeclasses so that when there's no possible way to determine what concrete type to use it can still give meaningful results without raising errors. So when you see
> let x :: Num a => a
| x = 1 + 1
> x
2
> :sprint x
x = _
This is because GHCi chooses Integer as its default type for Num, but when it performs this operation it doesn't store the result in x's memory location, since there isn't a way to know if that is even the correct answer. This is why you see x = _ from :sprint, it hasn't actually evaluated x :: Num a => a, it's evaluated x :: Integer. You can even mess with this default yourself:
> newtype MyInt = MyInt Int deriving (Eq)
>
> instance Show MyInt where
| show (MyInt i) = show i
> instance Num MyInt where
| (MyInt x) + (MyInt y) = MyInt (x - y)
| fromInteger = MyInt . fromInteger
>
> default (MyInt)
> x
0
So now we've said that 1 + 1 = 0! Keep in mind that you will probably never have a use for this functionality of GHC, but it's good to know about.
import Data.Function (on)
import Data.List (sort)
data Monomial = Monomial
{ m_coeff :: Coefficient
, m_powers :: [(Variable, Power)]
}
deriving ()
instance Ord Monomial where
(>=) = on (>=) m_powers
instance Eq Monomial where
(==) = on (==) m_powers
That's an excerpt from my code, cut down to principal size. Let's try comparing:
*Main> (Monomial 1 [("x",2)]) > (Monomial (-1) [])
/* Computation hangs here */
*Main> (Monomial 1 [("x",2)]) < (Monomial (-1) [])
/* Computation hangs here */
On a side note, it's interesting that if I replace s/(>=)/(>)/g in instance declaration, it will not hang on the fist pair, but still will on the second:
*Main> (Monomial 1 [("x",2)]) > (Monomial (-1) [])
True
*Main> (Monomial 1 [("x",2)]) < (Monomial (-1) [])
/* Computation hangs here */
Although the standard states minimal declaration of Eq instance to be either$compare$ or $(>=)$.
What might be the problem here? (>=) on lists seems to work just fine.
Short answer:
You need to provide either (<=) or compare to have a complete definition for Ord, not (>=).
Longer explanation:
It is common for type classes in Haskell to have default implementations of some methods implemented in terms of other methods. You can then choose which ones you want to implement. For example, Eq looks like this:
class Eq a where
(==), (/=) :: a -> a -> Bool
x /= y = not (x == y)
x == y = not (x /= y)
Here, you must either implement (==) or (/=), otherwise trying to use either of them will cause an infinite loop. Which methods you need to provide is usually listed as the minimal complete definition in the documentation.
The minimal complete definition for Ord instances, as listed in the documentation, is either (<=) or compare. Since you've only provided (>=), you have not provided a complete definition, and therefore some of the methods will loop. You can fix it by e.g. changing your instance to provide compare instead.
instance Ord Monomial where
compare = compare `on` m_powers
Let's look at the default instance for Ord:
class (Eq a) => Ord a where
compare :: a -> a -> Ordering
(<), (<=), (>), (>=) :: a -> a -> Bool
max, min :: a -> a -> a
compare x y = if x == y then EQ
-- NB: must be '<=' not '<' to validate the
-- above claim about the minimal things that
-- can be defined for an instance of Ord:
else if x <= y then LT
else GT
x < y = case compare x y of { LT -> True; _ -> False }
x <= y = case compare x y of { GT -> False; _ -> True }
x > y = case compare x y of { GT -> True; _ -> False }
x >= y = case compare x y of { LT -> False; _ -> True }
-- These two default methods use '<=' rather than 'compare'
-- because the latter is often more expensive
max x y = if x <= y then y else x
min x y = if x <= y then x else y
So, if you supply >= and == as above, only, then you are in trouble, since:
> is defined in terms of compare
But
compare is defined in terms of <=
<= is defined in terms of compare
So you have an infinite loop!
A minimum definition must defined <= or compare, not '>=`.
So I'm writing a program which returns a procedure for some given arithmetic problem, so I wanted to instance a couple of functions to Show so that I can print the same expression I evaluate when I test. The trouble is that the given code matches (-) to the first line when it should fall to the second.
{-# OPTIONS_GHC -XFlexibleInstances #-}
instance Show (t -> t-> t) where
show (+) = "plus"
show (-) = "minus"
main = print [(+),(-)]
returns
[plus,plus]
Am I just committing a mortal sin printing functions in the first place or is there some way I can get it to match properly?
edit:I realise I am getting the following warning:
Warning: Pattern match(es) are overlapped
In the definition of `show': show - = ...
I still don't know why it overlaps, or how to stop it.
As sepp2k and MtnViewMark said, you can't pattern match on the value of identifiers, only on constructors and, in some cases, implicit equality checks. So, your instance is binding any argument to the identifier, in the process shadowing the external definition of (+). Unfortunately, this means that what you're trying to do won't and can't ever work.
A typical solution to what you want to accomplish is to define an "arithmetic expression" algebraic data type, with an appropriate show instance. Note that you can make your expression type itself an instance of Num, with numeric literals wrapped in a "Literal" constructor, and operations like (+) returning their arguments combined with a constructor for the operation. Here's a quick, incomplete example:
data Expression a = Literal a
| Sum (Expression a) (Expression a)
| Product (Expression a) (Expression a)
deriving (Eq, Ord, Show)
instance (Num a) => Num (Expression a) where
x + y = Sum x y
x * y = Product x y
fromInteger x = Literal (fromInteger x)
evaluate (Literal x) = x
evaluate (Sum x y) = evaluate x + evaluate y
evaluate (Product x y) = evaluate x * evaluate y
integer :: Integer
integer = (1 + 2) * 3 + 4
expr :: Expression Integer
expr = (1 + 2) * 3 + 4
Trying it out in GHCi:
> integer
13
> evaluate expr
13
> expr
Sum (Product (Sum (Literal 1) (Literal 2)) (Literal 3)) (Literal 4)
Here's a way to think about this. Consider:
answer = 42
magic = 3
specialName :: Int -> String
specialName answer = "the answer to the ultimate question"
specialName magic = "the magic number"
specialName x = "just plain ol' " ++ show x
Can you see why this won't work? answer in the pattern match is a variable, distinct from answer at the outer scope. So instead, you'd have to write this like:
answer = 42
magic = 3
specialName :: Int -> String
specialName x | x == answer = "the answer to the ultimate question"
specialName x | x == magic = "the magic number"
specialName x = "just plain ol' " ++ show x
In fact, this is just what is going on when you write constants in a pattern. That is:
digitName :: Bool -> String
digitName 0 = "zero"
digitName 1 = "one"
digitName _ = "math is hard"
gets converted by the compiler to something equivalent to:
digitName :: Bool -> String
digitName x | x == 0 = "zero"
digitName x | x == 1 = "one"
digitName _ = "math is hard"
Since you want to match against the function bound to (+) rather than just bind anything to the symbol (+), you'd need to write your code as:
instance Show (t -> t-> t) where
show f | f == (+) = "plus"
show f | f == (-) = "minus"
But, this would require that functions were comparable for equality. And that is an undecidable problem in general.
You might counter that you are just asking the run-time system to compare function pointers, but at the language level, the Haskell programmer doesn't have access to pointers. In other words, you can't manipulate references to values in Haskell(*), only values themselves. This is the purity of Haskell, and gains referential transparency.
(*) MVars and other such objects in the IO monad are another matter, but their existence doesn't invalidate the point.
It overlaps because it treats (+) simply as a variable, meaning on the RHS the identifier + will be bound to the function you called show on.
There is no way to pattern match on functions the way you want.
Solved it myself with a mega hack.
instance (Num t) => Show (t -> t-> t) where
show op =
case (op 6 2) of
8 -> "plus"
4 -> "minus"
12 -> "times"
3 -> "divided"