So basically I've past learning this part way back a month ago, and I can do more complicated stuff but I still don't understand when I need "Ord" or "Eq" etc in my type definitions. When I look it up online its just so confusing to me for some reason.
E.g.
my_min :: Ord a => a -> a -> a
my_min n1 n2 = if n1<n2 then n1 else n2
Why does this need Ord? Can you give an example of when you need Eq as well (a very simple one)? I need a very clear, basic explanation of when you need to put these, what to look out for to do that, and how to avoid using these at all if possible.
Usually I just need something like "[Int] -> Int -> [Int]" so i know ok this function takes an integer list and an integer, and returns an integer list. But when it includes Eq or Ord I have no idea.
What about this Eq type in this example I found in my lecture notes about finding two lists is identical, how does it apply?
identical :: Eq a =>[a]->[a]->Bool
identical [] [] = True
identical [] _ = False
identical _ [] = False
identical (h1:t1) (h2:t2) =
if h1==h2
then (identical t1 t2)
else False
Thank you.
Ord implies that the thing can be ordered, which means that you can say a is smaller (or greater) than b. Using only Eq is like saying: I know that these two items are not the same, but I cannot say which one is greater or smaller. For example if you take a traffic light as a data type:
data TLight = Red | Yellow | Green deriving (Show, Eq)
instance Eq TLight where
Green == Green = True
Yellow == Yellow = True
Red == Red = True
_ == _ = False
Now we can say: Red is unequal to Yellow but we cannot say what is greater. This is the reason why you could not use TLight in your my_min. You cannot say which one is greater.
To your second question: "Is there any case where you have to use Eq and Ord?":
Ord implies Eq. This means that if a type can be ordered, you can also check it for equality.
You said you have mostly dealt with [Int] -> Int -> [Int] and you then knew it takes a list of integer and an integer and returns an integer. Now if you want to generalise your function you have to ask yourself: Do the possible types I want to use in my function need any special functionality? like if they have to be able to be ordered or equated.
Lets do a few examples: Say we want to write a function which takes a list of type a and an element of type a and returns the lisy with the element consed onto it. How would it's type signature look like? Lets start with simply this:
consfunc :: [a] -> a -> [a]
Do we need any more functionality? No! Our type a can be anything because we do not need it to be able to be ordered simple because that is mot what our function should do.
Now what if we want to take a list and an element and check if the element is in the list already? The beginning type signature is:
elemfunc :: [a] -> a -> Bool
Now does our element have to be able to do something special? Yes it does, we have to be able to check if it is equal to any element in the list, which says that our type a has to be equatable, so our type signature looks like this:
elemfunc :: (Eq a) => [a] -> a -> Bool
Now what if we want to take a list and a element and insert it if it is smaller than the first element? Can you guess how the type signature would look like?
Lets begin with the standard again and ask ourselves: Do we need more than just knowing that the element and the list have to be of the same type: Yes, becuase our condition needs to perform a test that requires our type to be ordered, we have to include Ord in our type signature:
conditionalconsfunc :: (Ord a) => [a] -> a -> [a]
Edit:
Well you want to see if two lists are identical, so there are two things you have to look out for:
Your lists have to contain the same type and the things inside the list have to be equatable, hence the Eq.
If you are working with fixed types like Int, you never need class constraints. These only arise when working with polymorphic code.
You need Eq if you ever use the == or /= functions, or if you call any other functions that do. (I.e., if you call a function that has Eq in its type, then your type needs to have Eq as well.)
You need Ord if you ever use <, >, compare or similar functions. (Again, or if you call something that does.)
Note that you do not need Eq if you only do pattern matching. Thus, the following are different:
factorial 1 = 1
factorial n = n * factorial (n-1)
-- Only needs Num.
factorial n = if n == 1 then 1 else n * factorial (n-1)
-- Needs Num and Eq.
factorial n = if n < 2 then 1 else n * factorial (n-1)
-- Needs Num, Eq and Ord. (But Ord implies Eq automatically.)
Related
I am trying to learn haskell and saw a exercise which says
Write two different Haskell functions having the same type:
[a] -> [b] -> Int -> (a,b)
So from my understanding the expressions should take in two lists, an int and return a tuple of the type of the lists.
What i tried so far was
together :: [a] -> [b] -> Int -> (a,b)
together [] [] 0 = (0,0)
together [b] [a] x = if x == a | b then (b,a) else (0,0)
I know I am way off but any help is appreciated!
First you need to make your mind up what the function should return. That is partly determined by the signature. But still you can come up with a lot of functions that return different things, but have the same signature.
Here one of the most straightforward functions is probably to return the elements that are placed on the index determined by the third parameter.
It makes no sense to return (0,0), since a and b are not per se numerical types. Furthermore if x == a | b is not semantically valid. You can write this as x == a || x == b, but this will not work, since a and b are not per se Ints.
We can implement a function that returns the heads of the two lists in case the index is 0. In case the index is negative, or at least one of the two lists is exhausted, then we can raise an error. I leave it as an exercise what to do in case the index is greater than 0:
together :: [a] -> [b] -> Int -> (a,b)
together [] _ = error "List exhausted"
together _ [] = error "List exhausted"
together (a:_) (b:_) 0 = (a, b)
together (a:_) (b:_) n | n < 0 = error "Negative index!"
| …
you thus still need to fill in the ….
I generally dislike those "write any function with this signature"-type excercises precisely because of how arbitrary they are. You're supposed to figure out a definition that would make sense for that particular signature and implement it. In a lot of cases, you can wing it by ignoring as many arguments as possible:
fa :: [a] -> [b] -> Int -> (a,b)
fa (a:_) (b:_) _ = (a,b)
fa _ _ _ = error "Unfortunately, this function can't be made total because lists can be empty"
The error here is the important bit to note. You attempted to go around that problem by returning 0s, but this will only work when 0 is a valid value for types of a and b. Your next idea could be some sort of a "Default" value, but not every type has such a concept. The key observation is that without any knowledge about a type, in order to produce a value from a function, you need to get this value from somewhere else first*.
If you actually wanted a more sensible definition, you'd need to think up a use for that Int parameter; maybe it's the nth element from each
list? With the help of take :: Int -> [a] -> [a] and head :: [a] -> a this should be doable as an excercise.
Again, your idea of comparing x with a won't work for all types; not every type is comparable with an Int. You might think that this would make generic functions awfully limited; that's the point where you typically learn about how to express certain expectations about the types you get, which will allow you to operate only on certain subsets of all possible types.
* That's also the reason why id :: a -> a has only one possible implementation.
Write two different Haskell functions having the same type:
[a] -> [b] -> Int -> (a,b)
As Willem and Bartek have pointed out, there's a lot of gibberish functions that have this type.
Bartek took the approach of picking two based on what the simplest functions with that type could look like. One was a function that did nothing but throw an error. And one was picking the first element of each list, hoping they were not empty and failing otherwise. This is a somewhat theoretical approach, since you probably don't ever want to use those functions in practice.
Willem took the approach of suggesting an actually useful function with that type and proceeded to explore how to exhaust the possible patterns of such a function: For lists, match the empty list [] and the non-empty list a:_, and for integers, match some stopping point, 0 and some categories n < 0 and ….
A question that arises to me is if there is any other equally useful function with this type signature, or if a second function would necessarily have to be hypothetically constructed. It would seem natural that the Int argument has some relation to the positions of elements in [a] and [b], since they are also integers, especially because a pair of single (a,b) is returned.
But the only remotely useful functions (in the sense of not being completely silly) that I can think of are small variations of this: For example, the Int could be the position from the end rather than from the beginning, or if there's not enough elements in one of the lists, it could default to the last element of a list rather than an error. Neither of these are very pleasing to make ("from the end" conflicts with the list being potentially infinite, and having a fall-back to the last element of a list conflicts with the fact that lists don't necessarily have a last element), so it is tempting to go with Bartek's approach of writing the simplest useless function as the second one.
If I do the following
functionS (x,y) = y
:t functionS
functionS :: (a, b) -> b
Now with this function:
functionC x y = if (x > y) then True else False
:t function
I would expect to get:
functionC :: (Ord a, Ord b) => a -> b -> Bool
But I get:
functionC :: Ord a => a -> a -> Bool
GHCI seems to be ok with the 2 previous results, but why does it give me the second? Why the type variable a AND b aren't defined?
I think you might be misreading type signatures. Through no fault of your own––the examples you using to inform your thinking are kind of confusing. In particular, in your tuple example
functionS :: (a,b) -> b
functionS (x,y) = y
The notation (_,_) means two different things. In the first line, (a,b) refers to a type, the type of pairs whose first element has type a and second has type b. In the second line, (x,y) refers to a specfiic pair, where x has type a and y has type b. While this "pun" provides a useful mnemonic, it can be confusing as you are first getting the hang of it. I would rather that the type of pairs be a regular type constructor:
functionS :: Pair a b -> b
functionS (x,y) = y
So, moving on to your question. In the signature you are given
functionC :: Ord a => a -> a -> Bool
a is a type. Ord a says that elements of the type a are orderable with respect to each other. The function takes two arguments of the same type. Some types that are orderable are Integer (numerically), String (lexicographically), and a bunch of others. That means that you can tell which of two Integers is the smaller, or which of two Strings are the smaller. However we don't necessarily know how to tell whether an Integer is smaller than a String (and this is good! Have you seen what kinds of shenanigans javascript has to do to support untyped equality? Haskell doesn't have to solve this problem at all!). So that's what this signature is saying –– there is only one single orderable type, a, and the function takes two elements of this same type.
You might still be wondering why functionS's signature has two different type variables. It's because there is no constraint confining them to be the same, such as having to order them against each other. functionS works equally well with a pair where both components are integers as when one is an integer and the other is a string. It doesn't matter. And Haskell always picks the most general type that works. So if they are not forced to be the same, they will be different.
There are more technical ways to explain all this, but I felt an intuitive explanation was in order. I hope it's helpful!
The assignment I have: A function numOccurences that takes a value and a list, returning the number of times that value appears in the list. I am learning haskell and am getting frustrated, this is my code for this:
numOccurences:: b -> [a] -> Int
numOccurences n [ls]
|([ls] !! n==True) = (numOccurences(n (tail [ls])))+1
|otherwise = 0
The errors I am getting are as follows:
https://imgur.com/a/0lTBn
A few pointers:
First, in your type signature, using different type variables (i.e. b and a) creates the possibility that you could look for occurrences of a value of one type, in a list with another type, which in this case is not what you want. So instead of two type variables, you just want to use one.
Second, whatever the concrete type of your list is, whether it's [Char], [Int], etc., it needs to be equatable (i.e. it needs to derive the Eq typeclass), so it makes sense to use the class constraint (Eq a) => in your type signature.
Third, since we're traversing a list, let's use pattern matching to safely break off the first element of the list for comparison, and let's also add a base case (i.e. what we do with an empty list), since we're using recursion, and we only want the recursive pattern to match as long as there are elements in our list.
Lastly, try to avoid using indexing (i.e. !!), where you can avoid it, and use pattern matching instead, as it's safer and easier to reason about.
Here's how your modified function might look, based on the above pointers:
numOccurences :: (Eq a) => a -> [a] -> Int
numOccurences _ [] = 0
numOccurences n (x:xs)
| n == x = 1 + numOccurences n xs
| otherwise = numOccurences n xs
This question already has answers here:
Understanding Haskell's Bool Deriving an Ord
(3 answers)
Closed 7 years ago.
Can someone please explain the following output?
Prelude> compare True False
GT
it :: Ordering
Prelude> compare False True
LT
it :: Ordering
Why are Bool type values ordered in Haskell - especially, since we can demonstrate that values of True and False are not exactly 1 and 0 (unlike many other languages)?
This is how the derived instance of Ord works:
data D = A | B | C deriving Ord
Given that datatype, we get C > B > A. Bool is defined as False | True, and it kind of makes sense when you look at other examples such as:
Maybe a = Nothing | Just a
Either a b = Left a | Right b
In each of the case having "some" ("truthy") value is greater than having no values at all (or having "left" or "bad" or "falsy" value).
While Bool is not Int, it can be converted to the 0,1 fragment of Int since it is an Enum type.
fromEnum False = 0
fromEnum True = 1
Now, the Enum could have been different, reversing 0 and 1, but that would probably be surprising to most programmers thinking about bits.
Since it has an Enum type, everything else being equal, it's better to define an Ord instance which follows the same order, satisfying
compare x y = compare (fromEnum x) (fromEnum y)
In fact, each instance generated from deriving (Eq, Ord, Enum) follows such property.
On a more theoretical note, logicians tend to order propositions from the strongest to the weakest (forming a lattice). In this structure, False (as a proposition) is the bottom, i.e. the least element, while True is the top. While this is only a convention (theory would be just as nice if we picked the opposite ordering), it's a good thing to be consistent.
Minor downside: the implication boolean connective is actually p <= q expressing that p implies q, instead of the converse as the "arrow" seems to indicate.
Let me answer your question with a question: Why is there an Ord instance for ()?
Unlike Bool, () has only one possible value: (). So why the hell would you ever want to compare it? There is only one value possible!
Basically, it's useful if all or most of the standard basic types have instances for common classes. It makes it easier to derive instances for your own types. If Foo doesn't have an Ord instance, and your new type has a single Foo field, then you can't auto-derive an Ord instance.
You might, for example, have some kind of tree type where we can attach several items of information to the leaves. Something like Tree x y z. And you might want to have an Eq instance to compare trees. It would be annoying if Tree () Int String didn't have an Eq instance just because () doesn't. So that's why () has Eq (and Ord and a few others).
Similar remarks apply to Bool. It might not sound particularly useful to compare two bool values, but it would be irritating if your Ord instance vanishes as soon as you put a bool in there.
(One other complicating factor is that sometimes we want Ord because there's a logically meaningful ordering for things, and sometimes we just want some arbitrary order, typically so we can use something as a key for Data.Map or similar. Arguably there ought to be two separate classes for that… but there isn't.)
Basically, it comes from math. In set theory or category theory boolean functions are usually thought of as classifiers of subsets/subobjects. In plain terms, function f :: a -> Bool is identified with filter f :: [a] -> [a]. So, if we change one value from False to True, the resulting filtered list (subset, subobject, whatever) is going to have more elements. Therefore, True is considered "bigger" than False.
I need to create a function of two parameters, an Int and a [Int], that returns a new [Int] with all occurrences of the first parameter removed.
I can create the function easily enough, both with list comprehension and list recursion. However, I do it with these parameters:
deleteAll_list_comp :: Integer -> [Integer] -> [Integer]
deleteAll_list_rec :: (Integer -> Bool) -> [Integer] -> [Integer]
For my assignment, however, my required parameters are
deleteAll_list_comp :: (Eq a) => a -> [a] -> [a]
deleteAll_list_rec :: (Eq a) => a -> [a] -> [a]
I don't know how to read this syntax. As Google has told me, (Eq a) merely explains to Haskell that a is a type that is comparable. However, I don't understand the point of this as all Ints are naturally comparable. How do I go about interpreting and implementing the methods using these parameters? What I mean is, what exactly are the parameters to begin with?
#groovy #pelotom
Thanks, this makes it very clear. I understand now that really it is only asking for two parameters as opposed to three. However, I still am running into a problem with this code.
deleteAll_list_rec :: (Eq a) => a -> [a] -> [a]
delete_list_rec toDelete [] = []
delete_list_rec toDelete (a:as) =
if(toDelete == a) then delete_list_rec toDelete as
else a:(delete_list_rec toDelete as)
This gives me a "The type signature for deleteAll_list_rec
lacks an accompanying binding" which makes no sense to me seeing as how I did bind the requirements properly, didn't I? From my small experience, (a:as) counts as a list while extracting the first element from it. Why does this generate an error but
deleteAll_list_comp :: (Eq a) => a -> [a] -> [a]
deleteAll_list_comp toDelete ls = [x | x <- ls, toDelete==x]
does not?
2/7/13 Update: For all those who might stumble upon this post in the future with the same question, I've found some good information about Haskell in general, and my question specifically, at this link : http://learnyouahaskell.com/types-and-typeclasses
"Interesting. We see a new thing here, the => symbol. Everything before the => symbol is >called a class constraint. We can read the previous type declaration like this: the >equality function takes any two values that are of the same type and returns a Bool. The >type of those two values must be a member of the Eq class (this was the class constraint).
The Eq typeclass provides an interface for testing for equality. Any type where it makes >sense to test for equality between two values of that type should be a member of the Eq >class. All standard Haskell types except for IO (the type for dealing with input and >output) and functions are a part of the Eq typeclass."
One way to think of the parameters could be:
(Eq a) => a -> [a] -> [a]
(Eq a) => means any a's in the function parameters should be members of the
class Eq, which can be evaluated as equal or unequal.*
a -> [a] means the function will have two parameters: (1) an element of
type a, and (2) a list of elements of the same type a (we know that
type a in this case should be a member of class Eq, such as Num or
String).
-> [a] means the function will return a list of elements of the same
type a; and the assignment states that this returned list should
exclude any elements that equal the first function parameter,
toDelete.
(* edited based on pelotom's comment)
What you implemented (rather, what you think you implemented) is a function that works only on lists of Integers, what the assignment wants you to do is create one that works on lists of all types provided they are equality-comparable (so that your function will also work on lists of booleans or strings). You probably don't have to change a lot: Try removing the explicit type signatures from your code and ask ghci about the type that it would infer from your code (:l yourfile.hs and then :t deleteAll_list_comp). Unless you use arithmetic operations or similar things, you will most likely find that your functions already work for all Eq a.
As a simpler example that may explain the concept: Let's say we want to write a function isequal that checks for equality (slightly useless, but hey):
isequal :: Integer -> Integer -> Bool
isequal a b = (a == b)
This is a perfectly fine definition of isequal, but the type constraints that I have manually put on it are way stronger than they have to. In fact, in the absence of the manual type signature, ghci infers:
Prelude> :t isequal
isequal :: Eq a => a -> a -> Bool
which tells us that the function will work for all input types, as long as they are deriving Eq, which means nothing more than having a proper == relation defined on them.
There is still a problem with your _rec function though, since it should do the same thing as your _comp function, the type signatures should match.