I have the following functions:
which (x:xs) = worker x xs
worker x [] = x
worker x (y:ys)
| x > y = worker y ys
| otherwise = worker x ys
and am wondering how I should define the types signatures of these above functions which and worker?
For Example, which of the following ways would be best as a type signature for worker?
worker :: Num a => a -> [a] -> a,
or
worker :: Ord a => a -> [a] -> a?
I'm just really confused and don't get which these three I should choose. I'd appreciate your thoughts. Thanks.
If you define the function without an explicit type signature, Haskell will infer the most general one. If you’re unsure, this is the easiest way to figure out how your definition will be read; you can then copy it into your source code. A common mistake is incorrectly typing a function and then getting a confusing type error somewhere else.
Anyway, you can get info on the Num class by typing :i Num into ghci, or by reading the documentation. The Num class gives you +, *, -, negate, abs, signum, fromInteger, as well as every function of Eq and Show. Notice that < and > aren’t there! Requiring values of Num and attempting to compare them will in fact produce a type error — not every kind of number can be compared.
So it should be Ord a => ..., as Num a => ... would produce a type error if you tried it.
If you think about what your functions do, you'll see that which xs returns the minimum value in xs. What can have a minimum value? A list of something Orderable!
Ask ghci and see what it says. I just copy-pasted your code as is into a file and loaded it into ghci. Then I used :t which is a special ghci command to determine the type of something.
ghci> :t which
which :: (Ord t) => [t] -> t
ghci> :t worker
worker :: (Ord a) => a -> [a] -> a
Haskell's type inference is pretty smart in most cases; learn to trust it. Other answers sufficiently cover why Ord should be used in this case; I just wanted to make sure ghci was clearly mentioned as a technique for determining the type of something.
I would always go with the Ord type constraint. It is the most general, so it can be reused more often.
There is no advantage to using Num over Ord.
Int may have a small advantage as it is not polymorphic and would not require a dictionary lookup. I would stil use Ord and use the specialize pragma if I needed to for performance.
Edit: Altered my answer after comments.
It depends on what you want to being able to compare. If you want to being able to compare Double, Float, Int, Integer and Char then use Ord. If you only want to being able to compare Int then just use Int.
If you have another problem like this, just look at the instances of the type class to tell which types you want to be able to use in the function.
Ord documentation
Related
If I want to run a simulation of the abstract machine on the code below, how do I know what would be in the workspace, stack, and heap?
let rec map (f: 'a -> 'b) (y: 'a list): 'b list =
begin match y with
| [] -> []
| h :: t -> (f h) :: (map f t)
end in
let x = map (fun t -> t + 1) [0; 1; 2] in
0 :: x
You're using terminology that we (people reading StackOverflow) don't share. In particular, we don't know what abstract machine you're talking about.
Speaking very generally about computer systems, the workspace usually contains the current definitions. At the beginning it might contain predefined functions and so on. Your code uses some predefined type names (like 'a list) and constructors (like ::). I don't know if they need to be in your workspace specifically.
Very possibly your definitions of map and x would need to be loaded into the workspace as the first step.
The stack contains a record of the functions that have been called but haven't returned. In general it will be empty when you start an evaluation.
The heap is a general term for the set of values that exist at the moment but might disappear later (when no longer needed). Unless you count your values named map and x, the heap could also be empty at the beginning.
Sorry I can't be more specific. It sounds like you're taking a class, and you might want to consult some of the class resources (including the professor or TA :-)
I'm pretty new to haskell and don't get how to work with Maybe [a]. Normally I'm coding OOP (VB.NET) and in my free time I want to learn haskell (don't ask why ;) ).
So, what do I want to do? I want to read two files with numerical IDs and find out only the IDs that match in both files. Reading the files is not a big thing, it works wonderfully easy. Now, I get two lists of Maybe[Ids] (for a simple example, just think the IDs are Int). So the function I need looks like this
playWithMaybe :: (Maybe [a]) -> (Maybe [Int]) -> [Int]
Now I want to access list members as I used to like this
playWithMaybe (x:xs) (y:ys) = undefined
But unfortunately it's not allowed, GHC says for both lists
Couldn't match expected type ‘Maybe [Int]’ with actual type ‘[t0]’
So I played around a little bit but didn't find a way to access list members. Can someone help me out? A little bit of explanation would be a great thing!
To approach your problem from a different direction, I would argue you don't want a function that processes two Maybe [a]. Bear with me:
Fundamentally, the operation you want to do operates on two lists to give you a new list, eg,
yourFunction :: [a] -> [a] -> [a]
yourFunction a b = ...
That's fine, and you can and should write yourFunction as such. The fact that the data you have is Maybe [a] captures some additional, auxiliary information: the operation that created your input lists may have failed. The next step is to chain together yourFunction with the auxiliary information. This is exactly the purpose of do notation, to mix pure operations (like yourFunction) with contexts (the fact that the creation of one of your input lists may have failed):
playWithMaybe :: Maybe [a] -> Maybe [a] -> Maybe [a]
playWithMaybe maybeA maybeB =
do a <- maybeA -- if A is something, get it; otherwise, pass through Nothing
b <- maybeB -- if B is something, get it; otherwise, pass through Nothing
Just (yourFunction a b) -- both inputs succeeded! do the operation, and return the result
But then it turns out there are other kinds of contexts you might want to work with (a simple one, instead of Maybe that just captures "something bad happened", we can use Either to capture "something bad happened, and here is a description of what happened). Looking back at playWithMaybe, the "Maybe-ness" only shows up in one place, the Just in the last line. It turns out Haskell offers a generic function pure to wrap a pure value, like what we get from yourFunction, in a minimal context:
playWithMaybe' :: Maybe [a] -> Maybe [a] -> Maybe [a]
playWithMaybe' maybeA maybeB =
do a <- maybeA
b <- maybeB
pure (yourFunction a b)
But then Haskell also has a generic type to abstract the idea of a context, the Monad. This lets us make our function even more generic:
playWithMonad :: Monad m => m [a] -> m [a] -> m [a]
playWithMonad mA mB =
do a <- mA
b <- mB
pure (yourFunction a b)
Now we have something very generic, and it turns out it is so generic, it's already in the standard library! (This is getting quite subtle, so don't worry if it doesn't all make sense yet.)
import Control.Applicative
play :: Monad m => m [a] -> m [a] -> m [a]
play mA mB = liftA2 yourFunction mA mB
or even
import Control.Applicative
play' :: Monad m => m [a] -> m [a] -> m [a]
play' = liftA2 yourFunction
Why did I switch from Monad to Applicative suddenly? Applicative is similar to Monad, but even more generic, so given the choice, it is generally better to use Applicative if you can (similar to my choice to use pure instead of return earlier). For a more complete explanation, I strongly recommend Learn You a Haskell (http://learnyouahaskell.com/chapters), in particular chapters 11 and 12. Note- definitely read chapter 11 first! Monads only makes sense after you have a grasp on Functor and Applicative.
In general:
yourFunction Nothing Nothing = ...
yourFunction (Just xs) Nothing =
case xs of
[] -> ...
x':xs' -> ...
-- or separately:
yourFunction (Just []) Nothing = ...
yourFunction (Just (x:xs)) Nothing = ...
et cetera. Which cases need to be treated separately depends on the specific function. More likely you would combine functions working on Maybe with functions working on [].
If you want to "Just return an list with no elements" for Nothing, then you can write
maybeToList1 :: Maybe [a] -> [a]
maybeToList1 Nothing = []
maybeToList1 (Just xs) = xs
A better way to write the same function is maybeToList1 = maybe [] id (docs for maybe) or maybeToList1 = fromMaybe [], but since you are just starting you may want to come back to this one later.
Like others have said, [Int] and Maybe [Int] are not the same thing. Maybe [Int] includes the extra information that the list may or may not be there. You said that you read the Ids from files. Perhaps, the Maybe signifies whether the file existed or not, while the empty list signifies that the file did exist but contained no Ids.
If you want to work with the list, you first need to define what to do if there is no list. You can use this function to extract the list:
fromMaybe :: a -> Maybe a -> a
Maybe you want to treat having no list the same as having an empty list:
fromMaybe [] :: Maybe [a] -> [a]
Maybe you want to crash the entire program:
fromMaybe (error "Missing list") :: Maybe a -> a
There is also the more general maybe function, which you can use if the default value is not of the same type as what's contained in the Maybe:
maybe :: b -> (a -> b) -> Maybe a -> b
Both these functions are defined in the module Data.Maybe.
You can also just work with the lists as if they existed and worry about their existence later by using Applicative. You said you wanted to find the Ids common to both lists. You can do it like this:
maybeCommonIds :: Maybe [Int] -> Maybe [Int] -> Maybe [Int]
maybeCommonIds xs ys = intersect <$> xs <*> ys
intersect is defined in Data.List. Using maybeCommonIds will result in Maybe [Int]. The list contained inside will hold the common ids, but if either of the two lists did not exist, there is not list of common ids. In your case, this might be the same as having no common ids. In that case, you might want to pass the result to fromMaybe [], to get back the list.
A past paper problem asked me; to define a function p :: [a] -> [a] that swaps every two items in a list. Your function should swap the first with the second item, the third
with the fourth, and so on. define one by list comprehension another by recursion.
Well this is what I came up with:
import Test.QuickCheck
p :: [a] -> [a]
p [] = []
p xs = concat [ [y,x] | ((x,y),i) <- zip (zip xs (tail xs)) [1..], odd i]
q :: [a] -> [a]
q [] = []
q (x:y:zs) | even (length zs) = [y,x] ++ q zs
| otherwise = error "The list provided is not of even length"
prop_2 xs = even (length xs) ==> p xs == q xs
check2 = quickCheck prop_2
The functions work fine, but I wanted to check if the two are identical, so I put the quickCheck below; but this gives me an error for some reason saying
"ambiguous type variable [a0] arising from the use of prop_2"
I just don't understand what's wrong here, I looks perfectly sensible to me...
what exactly is Haskell complaining??
Let's start by commenting out check2 and asking GHCi what the type of prop_2 is:
*Main> :t prop_2
prop_2 :: Eq a => [a] -> Property
Ok, so the way you've written prop_2, it works for lists of any element type that is in the equality class.
Now, you want to pass prop_2 to the quickCheck function. Let's look at the type of quickCheck next:
*Main> :t quickCheck
quickCheck :: Testable prop => prop -> IO ()
This function actually has an immensely general type. It works on anything that's in the Testable class. So how does this Testable class work? There are a couple of base instances here, for example:
instance Testable Bool
instance Testable Property -- this is a simplification, but it's more or less true
These instances are defined in the QuickCheck library and tell you that you can quick-check
constant Booleans as well as elements of type Property.
Now, testing properties that do not depend on any inputs isn't particularly interesting. The interesting instance is this one:
instance (Arbitrary a, Show a, Testable prop) => Testable (a -> prop)
What this says is that if you know how to generate random values of a particular type (Arbitrary a) and how to show values of that type (Show a), then you can also test functions from that type a to an already testable type prop.
Why? Because that's how QuickCheck operates. In such a situation, QuickCheck will consult the Arbitrary instance in order to come up with random test cases of type a, apply the function to each of them, and check if the outcome is positive. If any of the tests fails, it will print a message informing you about the test failure, and it will print the test case (that's why there's a Show requirement, too).
Now, in our situation this means that we should be able to quick-check prop_2: it is a function that results in a Property. The important thing is that the function argument (of type [a] as long as Eq a holds) is a member of class Arbitrary and a member of class Show.
Here we arrive at the source of the error. The information available is not sufficient to make that conclusion. As I said in the beginning, prop_2 works for lists of any element type that admits equality. There's no built-in rule that says that all these type are in Arbitrary and Show. But even if there was, what kind of lists should QuickCheck generate? Should it generate lists of Booleans, lists of unit type, lists of functions, lists of characters, lists of integers? So many options, and the choice of element type may well affect whether you find a bug or not. (Consider that GHC would pick the unit type () with just one element. Then your property would hold for any two functions p and q that preserve the input lists' length, regardless of whether they have your desired swapping property or not.)
This is why you need to provide extra type information to GHC so that it can resolve which element type to use for the list. Doing so is simple enough. You can either annotate prop_2 itself:
prop_2 :: [Integer] -> Property
Or, if you don't want that (because you might want to run the tests on different kinds of lists without reimplementing prop_2), you can add a type annotation when calling quickCheck:
check2 = quickCheck (prop_2 :: [Integer] -> Property)
Now the code compiles, and we can run check2:
Main*> check2
+++ OK, passed 100 tests.
Given a tuple of type (Int, a) such as (n,c), I wish to construct a list [a] where the element c is repeated n times, i.e., (4, 'b') becomes "bbbb". My current solution is the following:
decode :: (Int, a) -> [a]
decode (n, a) = map (\x -> a) [1..n]
As you can see, I'm mapping an anonymous function that always returns a over a list of n elements, the first n positive integers. Is there a more efficient way to do this? I feel bad about constructing a list of integers and never using it. Another solution would use a helper function and recurse down n, but that seems messy and overcomplicated. Is there perhaps something akin to the following python code?
'b'*4
uncurry replicate
Prelude> :t uncurry replicate
uncurry replicate :: (Int, b) -> [b]
Prelude> uncurry replicate (4, 'b')
"bbbb"
There is a builtin replicate for that.
Check out Hoogle for when you need to find if there is already a function that does what you want somewhere.
You want replicate.
A good way to find those things:
http://haskell.org/hoogle/?hoogle=Int+-%3E+a+-%3E+%5Ba%5D
map :: (a -> b) -> [a] -> [b]
fmap :: Functor f => (a -> b) -> f a -> f b
liftM :: Monad m => (a -> b) -> m a -> m b
Why do we have three different functions that do essentially the same thing?
map exists to simplify operations on lists and for historical reasons (see What's the point of map in Haskell, when there is fmap?).
You might ask why we need a separate map function. Why not just do away with the current
list-only map function, and rename fmap to map instead? Well, that’s a good question. The
usual argument is that someone just learning Haskell, when using map incorrectly, would much
rather see an error about lists than about Functors.
-- Typeclassopedia, page 20
fmap and liftM exist because monads were not automatically functors in Haskell:
The fact that we have both fmap and liftM is an
unfortunate consequence of the fact that the Monad type class does not require
a Functor instance, even though mathematically speaking, every monad is a
functor. However, fmap and liftM are essentially interchangeable, since it is
a bug (in a social rather than technical sense) for any type to be an instance
of Monad without also being an instance of Functor.
-- Typeclassopedia, page 33
Edit: agustuss's history of map and fmap:
That's not actually how it happens. What happened was that the type of map was generalized to cover Functor in Haskell 1.3. I.e., in Haskell 1.3 fmap was called map. This change was then reverted in Haskell 1.4 and fmap was introduced. The reason for this change was pedagogical; when teaching Haskell to beginners the very general type of map made error messages more difficult to understand. In my opinion this wasn't the right way to solve the problem.
-- What's the point of map in Haskell, when there is fmap?