Suppose I'm writing a function that takes a list of integers and returns only those integers in the list that are less than 5.2. I might do something like this:
belowThreshold = filter (< 5.2)
Easy enough, right? But now I want to constrain this function to only work with input lists of type [Int] for design reasons of my own. This seems like a reasonable request. Alas, no. A declaration that constraints the types as so:
belowThreshold :: [Integer] -> [Integer]
belowThreshold = filter (< 5.2)
Causes a type error. So what's the story here? Why does doing filter (< 5.2) seem to convert my input list into Doubles? How can I make a version of this function that only accepts integer lists and only returns integer lists? Why does the type system hate me?
Check the inferred type of belowThreshold in ghci before adding your annoatation:
> :t belowThreshold
belowThreshold :: [Double] -> [Double]
It sounds like you expected Num a => [a] -> [a] when you said "constrain this function". You are actually changing the type of the function when you add the [Integer] -> [Integer] annotation.
To make this work, use an explicit conversion:
belowThreshold = filter ((< 5.2) . fromIntegral)
Now belowThreshold :: [Integer] -> [Integer] like you wanted. But the integers are converted to doubles before comparison to 5.2.
So why do you need the conversion? The type error probably misled you: the list of Integers wasn't being converted to Doubles by comparison to 5.2, the real problem is that only Doubles can be compared to Doubles, so you must pass a list of Doubles to belowThreshold. Haskell has no implicit conversions, not even between numbers. If you want conversions, you have to write them yourself.
I want to constrain this function to only work with input lists of type [Int] for design reasons of my own. This seems like a reasonable request.
Well, from the perspective of the type system, no. Is this reasonable code?
'c' < "foo"
What about this?
12 < "bar"
All of these values are instances of Ord, but you can't use them together with (<). Haskell has no implicit conversions. So even if two values are both instances of Num as well as Ord, you won't be able to compare them with (<) if they are of different types.
You are trying to compare an Integer to a double (5.2). Haskell doesn't like that. Try using
filter (< 6)
If you must use a double (let's say it is an argument), I would use ceiling:
filter (< (ceiling 5.2))
Now if you want a function that takes in the bounding value as 'any' (relevant) numeric value, you can make your own type class to ceiling the number for you.
class Ceilingable a where
ceil :: (Integral b) => a -> b
instance (RealFrac a) => Ceilingable a where
ceil = ceiling
instance (Integral a) => Ceilingable a where
ceil = fromIntegral
belowThreshold :: (Ceilingable a) => a -> [Integer] -> [Integer]
belowThreshold threshold = filter (< ceil threshold)
The syntax 5.2 is valid for any Fractional. Int is not an instance of Fractional, nor can or should it be. As what to do when converting an arbitrary Rational to an Int is underspecified.
The conversion to a Double from an arbitrary fraction, however makes perfectly good sense (within the range of the type).
Your expectation is driven by the presence of implicit coercions in many languages.
However, those come with a cost. You have to manually ensure that the entire system of coercions is confluent. Haskell does not do this, choosing instead to let numeric literal syntax leverage the type system. To convert between them you need to use fromIntegral to make explicit the need for coercion, this avoids relying on confluence and allows programmers to define new numeric types.
belowThreshold = filter (\x -> fromIntegral x < 5.2)
This is analogous to using an explicit conversion in C++, like ((double)x < 5.2). Although, this statement only works because of defaulting, because 5.2 could be used as a member of any Fractional, and the result of 'fromIntegral x' is any Num, a superclass of Fractional, so fromIntegral x < 5.2 is underspecified, it merely knows that it needs to compare two Fractional values of the same type and it chooses Double as a reasonable default, based on a 'default' statement.
Also note that Int is not the only Integral type, so the above method works on any list of Integral values:
belowThreshold :: Integral a => [a] -> [a]
Related
I'm working with ByteStrings, and by extension a lot of Word8.
When i write sum word8List, the returned value is also Word8, but i know my data, and that sum is too large for a Word8. Should i avoid sum, or does it work differently than i think?
Is there some general function for summing numbers without overflowing, like Num a => a -> Integer?
The sum will use the (+) defined for that type, so for a Word8, it will make a wraparound.
You can however first perform a mapping, for example with fromIntegral :: (Integral a, Num b) => a -> b, and thus convert the numbers to another numeric type (like Integer, Word16, Word32, etc.). For example:
sum (map fromIntegral word8List) :: Integer
I'm just starting Learn You a Haskell for Great Good, and I'm having a bit of trouble with type classes. I would like to create a function that takes any number type and forces it to be a double.
My first thought was to define
numToDouble :: Num -> Double
But I don't think that worked because Num isn't a type, it's a typeclass (which seems to me to be a set of types). So looking at read, shows (Read a) => String -> a. I'm reading that as "read takes a string, and returns a thing of type a which is specified by the user". So I wrote the following
numToDouble :: (Num n) => n -> Double
numToDouble i = ((i) :: Double)
Which looks to me like "take thing of type n (must be in the Num typeclass, and convert it to a Double". This seems reasonable becuase I can do 20::Double
This produces the following output
Could not deduce (n ~ Double)
from the context (Num n)
bound by the type signature for numToDouble :: Num n => n -> Double
I have no idea what I'm reading. Based on what I can find, it seems like this has something to do with polymorphism?
Edit:
To be clear, my question is: Why isn't this working?
The reason you can say "20::Double" is that in Haskell an integer literal has type "Num a => a", meaning it can be any numeric type you like.
You are correct that a typeclass is a set of types. To be precise, it is the set of types that implement the functions in the "where" clause of the typeclass. Your type signature for your numToDouble correctly expresses what you want to do.
All you know about a value of type "n" in your function is that it implements the Num interface. This consists of +, -, *, negate, abs, signum and fromInteger. The last is the only one that does type conversion, but its not any use for what you want.
Bear in mind that Complex is also an instance of Num. What should numToDouble do with that? The Right Thing is not obvious, which is part of the reason you are having problems.
However lower down the type hierarchy you have the Real typeclass, which has instances for all the more straightforward numerical types you probably want to work with, like floats, doubles and the various types of integers. That includes a function "toRational" which converts any real value into a ratio, from which you can convert it to a Double using "fromRational", which is a function of the "Fractional" typeclass.
So try:
toDouble :: (Real n) => n -> Double
toDouble = fromRational . toRational
But of course this is actually too specific. GHCI says:
Prelude> :type fromRational . toRational
fromRational . toRational :: (Fractional c, Real a) => a -> c
So it converts any real type to any Fractional type (the latter covers anything that can do division, including things that are not instances of Real, like Complex) When messing around with numeric types I keep finding myself using it as a kind of generic numerical coercion.
Edit: as leftaroundabout says,
realToFrac = fromRational . toRational
You can't "convert" anything per se in Haskell. Between specific types, there may be the possibility to convert – with dedicated functions.
In your particular example, it certainly shouldn't work. Num is the class1 of all types that can be treated as numerical types, and that have numerical values in them (at least integer ones, so here's one such conversion function fromInteger).
But these types can apart from that have any other stuff in them, which oftentimes is not in the reals and can thus not be approximated by Double. The most obvious example is Complex.
The particular class that has only real numbers in it is, suprise, called Real. What is indeed a bit strange is that its method is a conversion toRational, since the rationals don't quite cover the reals... but they're dense within them, so it's kind of ok. At any rate, you can use that function to implement your desired conversion:
realToDouble :: Real n => n -> Double
realToDouble i = fromRational $ toRational i
Incidentally, that combination fromRational . toRational is already a standard function: realToFrac, a bit more general.
Calling type classes "sets of types" is kind of ok, much like you can often get away without calling any kind of collection in maths a set – but it's not really correct. The most problematic thing is, you can't really say some type is not in a particular class: type classes are open, so at any place in a project you could declare an instance for some type to a given class.
Just to be 100% clear, the problem is
(i) :: Double
This does not convert i to a Double, it demands that i already is a Double. That isn't what you mean at all.
The type signature for your function is correct. (Or at least, it means exactly what you think it means.) But your function's implementation is wrong.
If you want to convert one type of data to another, you have to actually call a function of some sort.
Unfortunately, Num itself only allows you to convert an Integer to any Num instance. You're trying to convert something that isn't necessarily an Integer, so this doesn't help. As others have said, you probably want fromRational or similar...
There is no such thing as numeric casts in Haskell. When you write i :: Double, what that means isn't "cast i to Double"; it's just an assertion that i's type is Double. In your case, however, your function's signature also asserts that i's type is Num n => n, i.e., any type n (chosen by the caller) that implements Num; so for example, n could be Integer. Those two assertions cannot be simultaneously true, hence you get an error.
The confusing thing is that you can say 1 :: Double. But that's because in Haskell, a numeric literal like 1 has the same meaning as fromInteger one, where one :: Integer is the Integer whose value is one.
But that only works for numeric literals. This is one of the surprising things if you come to Haskell from almost any other language. In most languages you can use expressions of mixed numeric types rather freely and rely on implicit coercions to "do what I mean"; in Haskell, on the other hand you have to use functions like fromIntegral or fromRational all the time. And while most statically typed languages have a syntax for casting from one numeric type to another, in Haskell you just use a function.
I'm playing around with beginner Haskell, and I wanted to write an average function. It seemed like the simplest thing in the world, right?
Wrong.
It seems like Haskell's type system forbids average from working on a generic numeric type - I can get it to work on a list of Integrals, or an list of Fractionals, but not both.
I want:
average :: (Num a, Fractional b) => [a] -> b
average xs = ...
But I can only get:
averageInt :: (Integral a, Fractional b) => [a] -> b
averageInt xs = fromIntegral (sum xs) / fromIntegral (length xs)
or
averageFrac :: (Fractional a) => [a] -> a
averageFrac xs = sum xs / fromIntegral (length xs)
and the second one seems to work. Until I try to pass a variable.
*Main> averageFrac [1,2,3]
2.0
*Main> let x = [1,2,3]
*Main> :t x
x :: [Integer]
*Main> averageFrac x
<interactive>:1:0:
No instance for (Fractional Integer)
arising from a use of `averageFrac ' at <interactive>:1:0-8
Possible fix: add an instance declaration for (Fractional Integer)
In the expression: average x
In the definition of `it': it = averageFrac x
Apparently, Haskell is really picky about its types. That makes sense. But not when they could both be [Num]
Am I missing an obvious application of RealFrac?
Is there way to coerce Integrals into Fractionals that doesn't choke when it gets a Fractional input?
Is there some way to use Either and either to make some sort of polymorphic average function that would work on any sort of numeric array?
Does Haskell's type system outright forbid this function from ever existing?
Learning Haskell is like learning Calculus. It's really complicated and based on mountains of theory, and sometimes the problem is so mindbogglingly complex that I don't even know enough to phrase the question correctly, so any insight will be warmly accepted.
(Also, footnote: this is based off a homework problem. Everybody agrees that averageFrac, above, gets full points, but I have a sneaking suspicion that there is a way to make it work on both Integral AND Fractional arrays)
So fundamentally, you're constrained by the type of (/):
(/) :: (Fractional a) => a -> a -> a
BTW, you also want Data.List.genericLength
genericLength :: (Num i) => [b] -> i
So how about removing the fromIntegral for something more general:
import Data.List
average xs = realToFrac (sum xs) / genericLength xs
which has only a Real constraint (Int, Integer, Float, Double)...
average :: (Real a, Fractional b) => [a] -> b
So that'll take any Real into any Fractional.
And note all the posters getting caught by the polymorphic numeric literals in Haskell. 1 is not an integer, it is any number.
The Real class provides only one method: the ability to turn a value in class Num to a rational. Which is exactly what we need here.
And thus,
Prelude> average ([1 .. 10] :: [Double])
5.5
Prelude> average ([1 .. 10] :: [Int])
5.5
Prelude> average ([1 .. 10] :: [Float])
5.5
Prelude> average ([1 .. 10] :: [Data.Word.Word8])
5.5
The question has been very well answered by Dons, I thought I might add something.
When calculating the average this way :
average xs = realToFrac (sum xs) / genericLength xs
What your code will do is to traverse the list twice, once to calculate the sum of its elements, and once to get its length.
As far as I know, GHC isn't able yet to optimize this and compute both the sum and length in a single pass.
It doesn't hurt even as a beginner to think about it and about possible solutions, for example the average function might be written using a fold that computes both the sum and length; on ghci :
:set -XBangPatterns
import Data.List
let avg l=let (t,n) = foldl' (\(!b,!c) a -> (a+b,c+1)) (0,0) l in realToFrac(t)/realToFrac(n)
avg ([1,2,3,4]::[Int])
2.5
avg ([1,2,3,4]::[Double])
2.5
The function doesn't look as elegant, but the performance is better.
More information on Dons blog:
http://donsbot.wordpress.com/2008/06/04/haskell-as-fast-as-c-working-at-a-high-altitude-for-low-level-performance/
Since dons has done such a good job at answering your question, I'll work on questioning your question....
For example, in your question, where you first run an average on a given list, getting a good answer. Then, you take what looks like the exact same list, assign it to a variable, then use the function the variable...which then blows up.
What you've run into here is a set-up in the compiler, called the DMR: the D readed M onomorphic R estriction. When you passed the list straight into the function, the compiler made no assumption about which type the numbers were, it just inferred what types it could be based on usage, and then picked one once it couldn't narrow the field down any more. It's kind of like the direct opposite of duck-typing, there.
Anyway, when you assigned the list to a variable, the DMR kicked in. Since you've put the list in a variable, but given no hints on how you want to use it, the DMR made the compiler pick a type, in this case, it picked one that matched the form and seemed to fit: Integer. Since your function couldn't use an Integer in its / operation (it needs a type in the Fractional class), it makes that very complaint: there's no instance of Integer in the Fractional class. There are options you can set in GHC so that it doesn't force your values into a single form ("mono-morphic", get it?) until it needs to, but it makes any error messages slightly tougher to figure out.
Now, on another note, you had a reply to dons' answer that caught my eye:
I was mislead by the chart on the last page of cs.ut.ee/~varmo/MFP2004/PreludeTour.pdf
that shows Floating NOT inheriting properties from Real, and I then assumed that
they would share no types in common.
Haskell does types differently from what you're used to. Real and Floating are type classes, which work more like interfaces than object classes. They tell you what you can do with a type that's in that class, but it doesn't mean that some type can't do other things, any more than having one interface means that a(n OO-style) class can't have any others.
Learning Haskell is like learning Calculus
I'd say learning Haskell is like learning Swedish - there are lots of little, simple things (letters, numbers) that look and work the same, but there are also words that look like they should mean one thing, when they actually mean something else. But once you get fluent in it, your regular friends will be amazed at how you can spout off this oddball stuff that makes gorgeous beauties do amazing tricks. Curiously, there are many folks involved in Haskell from the beginnings, who also know Swedish. Maybe that metaphor is more than just a metaphor...
:m Data.List
let list = [1..10]
let average = div (sum list) (genericLength list)
average
I'm amazed that after all of these years, no one has pointed out that Don Stewart's average doesn't work with complex numbers, while OP's averageFrac does work with complex numbers. Neither one is unambiguously superior to the other.
The fundamental reason why you can't write
average :: (Num a, Fractional b) => [a] -> b
is that it can be instantiated at a type like
average :: [Complex Double] -> Double
Haskell's numeric classes support conversions that are a little bit lossy, like Rational to Double, Double to Float, and Integer to Int, but don't support extremely lossy conversions like complex to real, or fractional to integral. You can't convert Complex Double to Double without explicitly taking (e.g.) the real part of it, which is not something that average should be doing. Therefore, you can't write average :: [Complex Double] -> Double. Therefore, you can't write average with any type that can be specialized to [Complex Double] -> Double.
The most Haskellish type for average is probably OP's averageFrac. Generally, functions that aren't dedicated to type conversion should be leaving the type conversion to the caller as much as possible. averageFrac will work with practically any numeric type, either directly or after coercion of the input list. The caller, being closer to the source of the data, is more likely to know whether it needs to be coerced or not (and if it doesn't know, it can leave the decision to its caller). In contrast, Don Stewart's average just doesn't support complex numbers, even with coercion. You'd either have to rewrite it from scratch or else call it twice with the real and imaginary projections of the list (and then write another wrapper for quaternions that calls it four times, etc.).
Yeah, Haskell's type system is very picky. The problem here is the type of fromIntegral:
Prelude> :t fromIntegral
fromIntegral :: (Integral a, Num b) => a -> b
fromIntegral will only accept an Integral as a, not any other kind of Num. (/), on the other hand only accepts fractional. How do you go about making the two work together?
Well, the sum function is a good start:
Prelude> :t sum
sum :: (Num a) => [a] -> a
Sum takes a list of any Num and returns a Num.
Your next problem is the length of the list. The length is an Int:
Prelude> :t length
length :: [a] -> Int
You need to convert that Int into a Num as well. That's what fromIntegral does.
So now you've got a function that returns a Num and another function that returns a Num. There are some rules for type promotion of numbers you can look up, but basically at this point you're good to go:
Prelude> let average xs = (sum xs) / (fromIntegral (length xs))
Prelude> :t average
average :: (Fractional a) => [a] -> a
Let's give it a trial run:
Prelude> average [1,2,3,4,5]
3.0
Prelude> average [1.2,3.4,5.6,7.8,9.0]
5.4
Prelude> average [1.2,3,4.5,6,7.8,9]
5.25
I have a haskell function that that calculates the size of the list of finite Ints. I need the output type to be an Integer because the value will actually be larger than the maximum bound of Int (the result will be -1 to be exact if the output type is an Int)
size :: a -> Integer
size a = (maxBound::Int) - (minBound::Int)
I understand the difference between Ints (bounded) and Integers (unbounded) but I'd like to make an Integer from an Int. I was wondering if there was a function like fromInteger, that will allow me to convert an Int to an Integer type.
You'll need to convert the values to Integers, which can be done by the fromIntegral function (numeric casting for Haskell):
fromIntegral :: (Integral a, Num b) => a -> b
It converts any type in the Integral class to any type in the (larger) Num class. E.g.
fromIntegral (maxBound::Int) - fromIntegral (minBound::Int)
However, I would not really trust the approach you're taking -- it seems very fragile. The behaviour in the presence of types that admit wraparound is pretty suspect.
What do you really mean by: "the size of the list of finite Ints". What is the size in this sense, if it isn't the length of the list?
I believe you are looking for:
fromIntegral :: (Integral a, Num b) => a -> b
which will convert an Integer to an Int
Perhaps you were assuming that Haskell, like many main-stream languages like C and (to a certain extent) Java, has implicit numeric coercions. It doesn't: Int and Integer are totally unrelated types and there is a special function for converting between them: fromIntegral. It belongs to the Num typeclass. Look at the documentation: essentially, fromIntegral does more than that: it is a generic "construct the representation of an arbitrary integral number", t.i. if you're implementing some kind of numbers and instantiating Num, you must provide a way to construct integral numbers of your type. For example, in the Num instance for complex numbers, fromIntegral creates a complex number with a zero imaginary part and an integral real part.
The only sense in which Haskell has implicit numeric coercions is that integer literals are overloaded, and when you write 42, the compiler implicitly interprets it as "fromIntegral (42::Integer)", so you can use integers in whatever contexts where a Num type is required.
I am trying to revise for my functional programming exam, and am stumped on the first questions on past papers, and yes, we arent allowed solution sheets, here is an example of the 1st question on a past paper.
For each of the following expressions give its type in Haskell (for an expression that has many types, just give one type).
(True, "hello", 42)
[42, 4, 2]
length [True]
filter even
I think personally that the answer for one and two would be a tuple of bool, String and int and a list of ints respectively, is this correct to assume? and secondly how would you answer 3 and 4, i am sure length True just outputs a list of all elements that are of that length, and that filter even just alters a list of ints to a list that are of all even numbers, though how could i show this as an answer?
If you want to get types of variables offline with ghci you have to type
:t expression
if you want to create variables in ghci, you have to use let without using 'in' (as in do notation for monads, I don't know if you have seen them yet) :
let var = expr
If you check it all by yourself, you should be able to remember it more easily for your exams. (good luck for it ;))
length [True] will be Int, and it would return 1. You can check that with ghci or lambdabot.
filter even will be (Integral a) => [a] -> [a]
for example, [Int] -> [Int]
And I think this is kind of pointless because lambdabot can tell all those things to you.
To be precise, the type of [42, 4, 2] is going to be
Num a => [a]
This is because an integer literal in Haskell is treated as having an implicit "fromIntegral" in front of it, so the real expression is [fromIntegral 42, fromIntegral 4, fromIntegral 2].
"fromIntegral" is part of the Num class, and has the type
fromIntegral :: (Integral a, Num b) => a -> b
This says that it converts an instance of some Integral type (i.e. Int or Integer) into an arbitrary other numeric type (Int, Float, Double, Complex ...). This is why you can say something like "43.2 + 1" without getting a type error.
"length [True]" is going to have type Int, because "length" has type "[a] -> Int", and the argument (a list of Bool) is provided.
"filter even" is a little bit more complicated. Start with the type of "filter":
filter :: (a -> Bool) -> [a] -> [a]
The first parameter (the bit in brackets) is itself a function that takes a list item and returns a Bool. Remember that the "->" operator in Haskell types is right associative, so if you put in the implied brackets you see that the type is:
filter :: (a -> Bool) -> ([a] -> [a])
In other words if you give it the first argument, you get back a new function that expects the second argument. In this case the first argument is:
even :: (Integral a) => a -> Bool
This introduces a slight wrinkle: "even" requires its argument to be an Integral type (i.e. Int or Integer, as above), so this constraint has to be propagated to the result. If it were not then you could write this:
filter even "foo"
Hence the answer is:
filter even :: (Integral a) => [a] -> [a]
You can see that the Integral constraint comes from the type of "even", while the rest of the type comes from "filter".