Bug in hindent (5.3.4) with Haskell choice operator? - haskell

Working through Graham Hutton's Programming in Haskell, second edition, I have the following piece of code:
int :: Parser Int
int =
do char '-'
n <- nat
return (-n)
<|> nat
The formatting above is the result of applying hindent 5.3.4 to the piece of code.
As far as I know, that code is equivalent to:
int :: Parser Int
int =
(do char '-'
n <- nat
return (-n)) <|>
nat
In fact, the latter formatting is the result of the same hindent, and I think it is correct.
Isn't the former formatting, if not incorrect, at least confusing? Is that a bug in hindent? Worth reporting at https://github.com/mihaimaruseac/hindent/issues, maybe?
By the way, the formatting in the book is:
int :: Parser Int
int = do char '-'
n <- nat
return (-n)
<|> nat

Since it looks like a bug, I have posted an issue.

Related

"string" Implementation in Text.Parser.Char

First, just some quick context. I'm going through the Haskell Programming From First Principles book, and ran into the following exercise.
Try writing a Parser that does what string does, but using char.
I couldn't figure it out, so I checked out the source for the implementation. I'm currently trying to wrap my head around it. Here it is:
class Parsing m => CharParsing m where
-- etc.
string :: CharParsing m => String -> m String
string s = s <$ try (traverse_ char s) <?> show s
My questions are as follows, from most to least specific.
Why is show necessary?
Why is s <$ necessary? Doesn't traverse char s <?> s work the same? In other words, why do we throw away the results of the traversal?
What is going on with the traversal? I get what a list traversal does, so I guess I'm confused about the Applicative/Monad instances for Parser. On a high level, I get that the traversal applies char, which has type CharParsing m => Char -> m Char, to every character in string s, and then collects all the results into something of type Parser [Char]. So the types make sense, but I have no idea what's going on in the background.
Thanks in advance!
1) Why is show necessary?
Because showing a string (or a Text, etc.) escapes special characters, which makes sense for error messages:
GHCi> import Text.Parsec -- Simulating your scenario with Parsec.
GHCi> runParser ((\s -> s <$ try (traverse_ char s) <?> s) "foo\nbar") () "" "foo"
Left (line 1, column 4):
unexpected end of input
expecting foo
bar
GHCi> runParser ((\s -> s <$ try (traverse_ char s) <?> show s) "foo\nbar") () "" "foo"
Left (line 1, column 4):
unexpected end of input
expecting "foo\nbar"
2) Why is s <$ necessary? Doesn't traverse char s <?> s work the same? In other words, why do we throw away the results of the traversal?
The result of the parse is unnecessary because we know in advance that it would be s (if the parse were successful). traverse would needlessly reconstruct s from the results of parsing each individual character. In general, if the results are not needed it is a good idea to use traverse_ (which just combines the effects, discarding the results without trying to rebuild the data structure) rather than traverse, so that is likely why the function is written the way it is.
3) What is going on with the traversal?
traverse_ char s (traverse_, and not traverse, as explained above) is a parser. It tries to parse, in order, each character in s, while discarding the results, and it is built by sequencing parsers for each character in s. It may be helpful to remind that traverse_ is just a fold which uses (*>):
-- Slightly paraphrasing the definition in Data.Foldable:
traverse_ :: (Foldable t, Applicative f) => (a -> f b) -> t a -> f ()
traverse_ f = foldr (\x u -> f x *> u) (pure ())

Parsing to Free Monads

Say I have the following free monad:
data ExampleF a
= Foo Int a
| Bar String (Int -> a)
deriving Functor
type Example = Free ExampleF -- this is the free monad want to discuss
I know how I can work with this monad, eg. I could write some nice helpers:
foo :: Int -> Example ()
foo i = liftF $ Foo i ()
bar :: String -> Example Int
bar s = liftF $ Bar s id
So I can write programs in haskell like:
fooThenBar :: Example Int
fooThenBar =
do
foo 10
bar "nice"
I know how to print it, interpret it, etc. But what about parsing it?
Would it be possible to write a parser that could parse arbitrary
programs like:
foo 12
bar nice
foo 11
foo 42
So I can store them, serialize them, use them in cli programs etc.
The problem I keep running into is that the type of the program depends on which program is being parsed. If the program ends with a foo it's of
type Example () if it ends with a bar it's of type Example Int.
I do not feel like writing parsers for every possible permutation (it's simple here because there are only two possibilities, but imagine we add
Baz Int (String -> a), Doo (Int -> a), Moz Int a, Foz String a, .... This get's tedious and error-prone).
Perhaps I'm solving the wrong problem?
Boilerplate
To run the above examples, you need to add this to the beginning of the file:
{-# LANGUAGE DeriveFunctor #-}
import Control.Monad.Free
import Text.ParserCombinators.Parsec
Note: I put up a gist containing this code.
Not every Example value can be represented on the page without reimplementing some portion of Haskell. For example, return putStrLn has a type of Example (String -> IO ()), but I don't think it makes sense to attempt to parse that sort of Example value out of a file.
So let's restrict ourselves to parsing the examples you've given, which consist only of calls to foo and bar sequenced with >> (that is, no variable bindings and no arbitrary computations)*. The Backus-Naur form for our grammar looks approximately like this:
<program> ::= "" | <expr> "\n" <program>
<expr> ::= "foo " <integer> | "bar " <string>
It's straightforward enough to parse our two types of expression...
type Parser = Parsec String ()
int :: Parser Int
int = fmap read (many1 digit)
parseFoo :: Parser (Example ())
parseFoo = string "foo " *> fmap foo int
parseBar :: Parser (Example Int)
parseBar = string "bar " *> fmap bar (many1 alphaNum)
... but how can we give a type to the composition of these two parsers?
parseExpr :: Parser (Example ???)
parseExpr = parseFoo <|> parseBar
parseFoo and parseBar have different types, so we can't compose them with <|> :: Alternative f => f a -> f a -> f a. Moreover, there's no way to know ahead of time which type the program we're given will be: as you point out, the type of the parsed program depends on the value of the input string. "Types depending on values" is called dependent types; Haskell doesn't feature a proper dependent type system, but it comes close enough for us to have a stab at making this example work.
Let's start by forcing the expressions on either side of <|> to have the same type. This involves erasing Example's type parameter using existential quantification.†
data Ex a = forall i. Wrap (a i)
parseExpr :: Parser (Ex Example)
parseExpr = fmap Wrap parseFoo <|> fmap Wrap parseBar
This typechecks, but the parser now returns an Example containing a value of an unknown type. A value of unknown type is of course useless - but we do know something about Example's parameter: it must be either () or Int because those are the return types of parseFoo and parseBar. Programming is about getting knowledge out of your brain and onto the page, so we're going to wrap up the Example value with a bit of GADT evidence which, when unwrapped, will tell you whether a was Int or ().
data Ty a where
IntTy :: Ty Int
UnitTy :: Ty ()
data (a :*: b) i = a i :&: b i
type Sig a b = Ex (a :*: b)
pattern Sig x y = Wrap (x :&: y)
parseExpr :: Parser (Sig Ty Example)
parseExpr = fmap (\x -> Sig UnitTy x) parseFoo <|>
fmap (\x -> Sig IntTy x) parseBar
Ty is (something like) a runtime "singleton" representative of Example's type parameter. When you pattern match on IntTy, you learn that a ~ Int; when you pattern match on UnitTy you learn that a ~ (). (Information can be made to flow the other way, from types to values, using classes.) :*:, the functor product, pairs up two type constructors ensuring that their parameters are equal; thus, pattern matching on the Ty tells you about its accompanying Example.
Sig is therefore called a dependent pair or sigma type - the type of the second component of the pair depends on the value of the first. This is a common technique: when you erase a type parameter by existential quantification, it usually pays to make it recoverable by bundling up a runtime representative of that parameter.
Note that this use of Sig is equivalent to Either (Example Int) (Example ()) - a sigma type is a sum, after all - but this version scales better when you're summing over a large (or possibly infinite) set.
Now it's easy to build our expression parser into a program parser. We just have to repeatedly apply the expression parser, and then manipulate the dependent pairs in the list.
parseProgram :: Parser (Sig Ty Example)
parseProgram = fmap (foldr1 combine) $ parseExpr `sepBy1` (char '\n')
where combine (Sig _ val) (Sig ty acc) = Sig ty (val >> acc)
The code I've shown you is not exemplary. It doesn't separate the concerns of parsing and typechecking. In production code I would modularise this design by first parsing the data into an untyped syntax tree - a separate data type which doesn't enforce the typing invariant - then transform that into a typed version by type-checking it. The dependent pair technique would still be necessary to give a type to the output of the type-checker, but it wouldn't be tangled up in the parser.
*If binding is not a requirement, have you thought about using a free applicative to represent your data?
†Ex and :*: are reusable bits of machinery which I lifted from the Hasochism paper
So, I worry that this is the same sort of premature abstraction that you see in object-oriented languages, getting in the way of things. For example, I am not 100% sure that you are using the structure of the free monad -- your helpers for example simply seem to use id and () in a rather boring way, in fact I'm not sure if your Int -> x is ever anything other than either Pure :: Int -> Free ExampleF Int or const (something :: Free ExampleF Int).
The free monad for a functor F can basically be described as a tree whose data is stored in leaves and whose branching factor is controlled by the recursion in each constructor of the functor F. So for example Free Identity has no branching, hence only one leaf, and thus has the same structure as the monad:
data MonoidalFree m x = MF m x deriving (Functor)
instance Monoid m => Monad (MonoidalFree m) where
return x = MF mempty x
MF m x >>= my_x = case my_x x of MF n y -> MF (mappend m n) y
In fact Free Identity is isomorphic to MonoidalFree (Sum Integer), the difference is just that instead of MF (Sum 3) "Hello" you see Free . Identity . Free . Identity . Free . Identity $ Pure "Hello" as the means of tracking this integer. On the other hand if you have data E x = L x | R x deriving (Functor) then you get a sort of "path" of Ls and Rs before you hit this one leaf, Free E is going to be isomorphic to MonoidalFree [Bool].
The reason I'm going through this is that when you combine Free with an Integer -> x functor, you get an infinitely branching tree, and when I'm looking through your code to figure out how you're actually using this tree, all I see is that you use the id function with it. As far as I can tell, that restricts the recursion to either have the form Free (Bar "string" Pure) or else Free (Bar "string" (const subExpression)), in which case the system would seem to reduce completely to the MonoidalFree [Either Int String] monad.
(At this point I should pause to ask: Is that correct as far as you know? Was this what was intended?)
Anyway. Aside from my problems with your premature abstraction, the specific problem that you're citing with your monad (you can't tell the difference between () and Int has a bunch of really complicated solutions, but one really easy one. The really easy solution is to yield a value of type Example (Either () Int) and if you have a () you can fmap Left onto it and if you have an Int you can fmap Right onto it.
Without a much better understanding of how you're using this thing over TCP/IP we can't recommend a better structure for you than the generic free monads that you seem to be finding -- in particular we'd need to know how you're planning on using the infinite-branching of Int -> x options in practice.

Haskell: Function that converts Char to Word8

I'm trying to find a function that converts a Char to a Word8, and another that converts it back. I've looked on Hoogle and there are no functions of type Char -> Word8. I want to do this because Word8 has an instance of the Num class, and I would be able to add numbers to them to change them to another number. For example, my function would look something like:
shift :: Char -> Int -> Char
shift c x = toChar $ (toWord8 c) + x
Any ideas?
You can use fromEnum and toEnum to go via Int instead. This has the added benefit of also supporting Unicode.
> toEnum (fromEnum 'a' + 3) :: Char
'd'
I found these using FP Complete's Hoogle: http://haddocks.fpcomplete.com/fp/7.7/20131212-1/utf8-string/Codec-Binary-UTF8-String.html#v:encodeChar
A Char doesn't necessarily contain only one byte, so the conversions you're describing could lose information. Would these functions work instead?

How to override Show instance of some basic types in Haskell?

I'm writting some programs in Haskell, dealing with a lot of basic types like Word32/Word64 etc..
I use ghci to test the functions frequently, see the results in terminal.
To be convenient and fast, I always show data in hexadecimal e.g.
data Human = M Int | F Int
instance Show Human where
show M x = printf "man, age %d" x
show F x = printf "woman, age %d" x
but I want basic types to be showed in hexadecimal (espacially in ghci).
I found instance declaration cannot be overridden.
and I don't want to warp all of them up like:
newtype MyInt = MyInt Int
instance Show MyInt where
...
It looks a little bit stupid.
Can I modify some code in the package base for ghc?
I just want everything becoming "hex". I just want ghci showing "hex". how could I achieve it?
EDIT
Since all of us agree that override Show is not proper and impractical,
Any answer of "better ways to show Numeric in hexadecimal in ghci" is welcomed.
No, there is no way to achieve this without newtypes; instances cannot be overriden.
If you really want this, I would suggest defining your own typeclass, ShowHex, like Show but with all the instances printing in hex. However, I would consider your Show instance incorrect; Show instances are designed for debugging and serialisation, and should output syntactically valid code.1 Yours doesn't, so I would suggest either defining your own typeclass for displaying these values, or simply using a function.
Modifying the code to base for this is impractical; not only would this change in semantics for the instances break a lot of packages, but it'd be a huge pain to get GHC to actually use your modified version.
1 Ideally, the code they produce should be semantically valid Haskell that produces a value comparing equal to show's input, but this is not strictly necessary.
That would be abusing the Show instance. It's not really meant for formatting. If you want to show something in hexadecimal, just use a function to do the conversion. For example, you can use showHex from Numeric to make a small helper like this:
> import Numeric
Numeric> let hex x = showHex x ""
Numeric> hex 123456
"1e240"
One extreme solution would be to use {-# LANGUAGE NoImplicitPrelude #-}, and import your own "Prelude" instead. That would probably be a lot more work than it's worth for your case, though.
Agreeing with #ehird and #hammar that this could be abused. In the case of wanting some numbers to always show as hex, I think it's reasonable because "0xff" is a legitimate representation of a number. So this:
{-# LANGUAGE GeneralizedNewtypeDeriving #-}
module HexNumber where
import Numeric
import Text.Read
import qualified Text.Read.Lex as L
newtype HexInt a = HexInt { int :: a }
deriving (Eq, Ord, Num, Enum)
instance (Show a, Integral a) => Show (HexInt a) where
show hi = "0x" ++ showHex (int hi) ""
instance (Num a) => Read (HexInt a) where
-- Couldn't figure out how to write this instance so just copy/paste from Text.Read
readPrec = readNumber convertInt
readListPrec = readListPrecDefault
readList = readListDefault
readNumber :: Num a => (L.Lexeme -> ReadPrec a) -> ReadPrec a
readNumber convert =
parens
( do x <- lexP
case x of
L.Symbol "-" -> do y <- lexP
n <- convert y
return (negate n)
_ -> convert x
)
convertInt :: Num a => L.Lexeme -> ReadPrec a
convertInt (L.Number n)
| Just i <- L.numberToInteger n = return (fromInteger i)
convertInt _ = pfail
Now I can:
> let x = 10 :: HexInt Int
> x
0xa
> x * 2
0x14
> let x = 10 :: HexInt Integer
> x
0xa
> x * 2
0x14
> read "0xa" :: HexInt Int
0xa
> read "10" :: HexInt Int
0xa
This seems very useful to me working with low-level stuff a lot. Maybe I'll put it on Hackage.

Haskell int to float and char to float

Is there a function in haskell which converts from int to float, and from char to float?
I know that there is a function that converts from char to int and int to char.
fromIntegral will convert from Int to Float.
For Char to Float, it depends. If you want to get the ASCII value of a Char (ignoring Unicode for now), use Data.Char.ord:
Prelude Data.Char> fromIntegral (ord '2') :: Float
50.0
If you want to read the digit of a Char, i.e. '2' becomes the value 2, you can do this:
char2float :: Char -> Float
char2float n = fromInteger (read [n])
Prelude Data.Char> char2float '2'
2.0
If you're going to do a lot of this, you might consider using an actual parsing library to get actual error handling.
Questions like this can be answered with hoogle.
For example, Hoogle for "Char -> Int" and the first function listed will do it (ord, mentioned in other answers, is the second result):
digitToInt :: Char -> Int
Though your need for a function :: Char -> Float does mandate using read (third result down) or a combination of digitToInt and a function :: Int -> Float:
digitToFloat = toEnum . digitToInt
did you try:
intToFloat :: Int -> Float
intToFloat n = fromInteger (toInteger n)
Additionally see here
If I get it right, you can use Read typeclass for these purposes. And it means you can convert some basic types to some other basic types.
Converting from Int to a Float:
Prelude> 1::Float
1.0
BUT I don't know which means converting from char to Float - maybe [Char] to Float or String to Float? Anyway:
Prelude> read("1")::Float
1.0
Check out this link: http://book.realworldhaskell.org/read/using-typeclasses.html

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