I have just typed in the RandomState example from real world haskell. It looks like this:
import System.Random
import Control.Monad.State
type RandomState a = State StdGen a
getRandom :: Random a => RandomState a
getRandom =
get >>= \gen ->
let (val, gen') = random gen in
put gen' >>
return val
getTwoRandoms :: Random a => RandomState (a, a)
getTwoRandoms = liftM2 (,) getRandom getRandom
It works, but the result doesn't get displayed. I get the error message:
No instance for (Show (RandomState (Int, Int)))
arising from a use of `print' at <interactive>:1:0-38
Possible fix:
add an instance declaration for (Show (RandomState (Int, Int)))
In a stmt of a 'do' expression: print it
I am having some trouble adding an instance for Show RandomState. Can anyone show me how this is done?
Thanks.
For the sake of being explicit, as jberryman and the comments on the question imply: Something of type RandomState (a, a) is a function, not a value. To do anything with it, you want to run it with an initial state.
I'm guessing you want something like this:
> fmap (runState getTwoRandoms) getStdGen
((809219598,1361755735),767966517 1872071452)
This is essentially what the runTwoRandoms function a bit further in RWH is doing.
Since RandomState is a synonym for State and there isn't an instance of show defined for State, you won't be able to show it.
You would also not be able to derive show because State is just a wrapper for a function and Haskell has no way to define a show for functions that would be useful:
Prelude> show (+)
<interactive>:1:0:
No instance for (Show (a -> a -> a))
arising from a use of `show' at <interactive>:1:0-7
Possible fix: add an instance declaration for (Show (a -> a -> a))
In the expression: show (+)
In the definition of `it': it = show (+)
EDIT: Forgot to add the other piece: GHCi is giving you that error because it uses show behind the scenes on the expressions you enter... REPL and all that.
Related
I'm learning Haskell and would like to write functions to recursively process lists nested arbitrarily deep.
For example I'd like to write recurReverse, which in the basis case, acts just like the built in reverse, but when passed a nested list, would reverse all the elements of the sublists recursively as well:
recurReverse [1,2]
>> [2,1]
recurReverse [[1],[2],[3,4]]
>> [[4,3],[2],[1]]
recurReverse [[[1,2]]]
>> [[[2,1]]]
Currently I have the basic reverse down:
rev [] = []
rev (h:t) = rev t ++ [h]
But I need more than this -- in the case when the head h is also a list (as opposed to an atom in the LISP sense), I'd like to be able to reverse h as well and return something like rev t ++ [rev h]. When I try that, I get a compiler error saying something like I can't rev h because rev is of type [t] -> [t] but I'm trying to call it on type t, which makes sense. How can I get around this?
as opposed to an atom in the LISP sense
Well, there is no such thing in Haskell. Any type which you don't know a priori (which you can't, if you're doing recursion over types) could possibly be a list itself. There's no notion of atomicity, of “not-list-being” which you could use as a base case for this recursion.
That is, unless you make the distinction explicit. This can be done quite nicely in Haskell, with GADTs:
data Nest t where
Egg :: t -> Nest t
Nest :: [Nest t] -> Nest [t]
nestReverse :: Nest t -> Nest t
nestReverse (Egg x) = Egg x
nestReverse (Nest xs) = Nest . reverse $ map nestReverse xs
If you don't like this... well, there is another way, but it's considered ugly / un–Haskell‑ish.
class Atomeous l where
recurReverse :: l -> l
instance {-# OVERLAPPABLE #-} Atomeous l where
recurReverse = id
instance Atomeous l => Atomeous [l] where
recurReverse = reverse . map recurReverse
Now, recurReverse has your intended behaviour. The first instance is for “atomic” types; because it is marked OVERLAPPABLE the compiler will only use this instance if it can't find “a better one” – which it does precisely for lists; these get the recursive call over all elements.
Could someone explain what the |> operator does? This code was taken from the reference here:
let m = PairsMap.(empty |> add (0,1) "hello" |> add (1,0) "world")
I can see what it does, but I wouldn't know how to apply the |> operator otherwise.
For that matter, I have no idea what the Module.() syntax is doing either. An explanation on that would be nice too.
Module.(e) is equivalent to let open Module in e. It is a shorthand syntax to introduce things in scope.
The operator |> is defined in module Pervasives as let (|>) x f = f x. (In fact, it is defined as an external primitive, easier to compile. This is unimportant here.) It is the reverse application function, that makes it easier to chain successive calls. Without it, you would need to write
let m = PairsMap.(add (1,0) "world" (add (0,1) "hello" empty))
that requires more parentheses.
The |> operator looks like the | in bash.
The basic idea is that
e |> f = f e
It is a way to write your applications in the order of execution.
As an exemple you could use it (I don't particularly think you should though) to avoid lets:
12 |> fun x -> e
instead of
let x = 12 in e
For the Module.() thing, it is to use a specific function of a given module.
You probably have seen List.map before.
You could of course use open List and then only refer to the function with map. But if you also open Array afterwards, map is now referring to Array.map so you need to use List.map.
The |> operator represents reverse function application. It sounds complicated but it just means you can put the function (and maybe a few extra parameters) after the value you want to apply it to. This lets you build up something that looks like a Unix pipeline:
# let ( |> ) x f = f x;;
val ( |> ) : 'a -> ('a -> 'b) -> 'b = <fun>
# 0.0 |> sin |> exp;;
- : float = 1.
The notation Module.(expr) is used to open the module temporarily for the one expression. In other words, you can use names from the module directly in the expression, without having to prefix the module name.
I'm teaching myself OCaml and I sometimes need to create a function where I'm not really sure what the proper solution should be. Here's one that I'm a little confused about.
I need a function that will prompt the user for individual float values and return everything entered in a float list. I can create this function but I'm not sure if its the proper/best way to do it in Ocaml.
Here's my attempt.
let rec get_floats() =
match
(
try Some(read_float())
with
| float_of_string -> None
)
with
| None -> []
| Some s -> s :: get_floats();;
This code works buts I'm at a loss deciding if its a 'proper OCaml' solution. Note, to exit the function and return the float list just enter a non-integer value.
(I hope that) this is a simple peephole rewrite involving no thought whatsoever of the function in your question:
let rec get_floats() =
try
let f = read_float() in (* as suggested by Martin Jambon *)
f :: (get_floats())
with
| float_of_string -> []
The idea I tried to apply here is that you do not need to convert the success/failure of read_float into an option that you immediately match: just do what you have to do with the value read, and let the with handle the failure case.
Now that I think of it, I should point out that in both your question and my rewrite, float_of_string is a fresh variable. If you meant to match a specific exception, you failed at it: all exception constructors, like datatype constructors, are Capitalized. You might as well have written with _ -> instead of with float_of_string ->, and a recent version of OCaml with all warnings active should tell you that your function (or mine) binds a variable float_of_string without ever using it.
Thanks everyone for the help. This works.
let rec get_floats() =
try
let x = read_float() in
x :: get_floats()
with
| _ -> [];;
List.iter (fun x -> print_endline(string_of_float x)) (get_floats());;
I have a question about tuples and lists in Haskell. I know how to add input into a tuple a specific number of times. Now I want to add tuples into a list an unknown number of times; it's up to the user to decide how many tuples they want to add.
How do I add tuples into a list x number of times when I don't know X beforehand?
There's a lot of things you could possibly mean. For example, if you want a few copies of a single value, you can use replicate, defined in the Prelude:
replicate :: Int -> a -> [a]
replicate 0 x = []
replicate n | n < 0 = undefined
| otherwise = x : replicate (n-1) x
In ghci:
Prelude> replicate 4 ("Haskell", 2)
[("Haskell",2),("Haskell",2),("Haskell",2),("Haskell",2)]
Alternately, perhaps you actually want to do some IO to determine the list. Then a simple loop will do:
getListFromUser = do
putStrLn "keep going?"
s <- getLine
case s of
'y':_ -> do
putStrLn "enter a value"
v <- readLn
vs <- getListFromUser
return (v:vs)
_ -> return []
In ghci:
*Main> getListFromUser :: IO [(String, Int)]
keep going?
y
enter a value
("Haskell",2)
keep going?
y
enter a value
("Prolog",4)
keep going?
n
[("Haskell",2),("Prolog",4)]
Of course, this is a particularly crappy user interface -- I'm sure you can come up with a dozen ways to improve it! But the pattern, at least, should shine through: you can use values like [] and functions like : to construct lists. There are many, many other higher-level functions for constructing and manipulating lists, as well.
P.S. There's nothing particularly special about lists of tuples (as compared to lists of other things); the above functions display that by never mentioning them. =)
Sorry, you can't1. There are fundamental differences between tuples and lists:
A tuple always have a finite amount of elements, that is known at compile time. Tuples with different amounts of elements are actually different types.
List an have as many elements as they want. The amount of elements in a list doesn't need to be known at compile time.
A tuple can have elements of arbitrary types. Since the way you can use tuples always ensures that there is no type mismatch, this is safe.
On the other hand, all elements of a list have to have the same type. Haskell is a statically-typed language; that basically means that all types are known at compile time.
Because of these reasons, you can't. If it's not known, how many elements will fit into the tuple, you can't give it a type.
I guess that the input you get from your user is actually a string like "(1,2,3)". Try to make this directly a list, whithout making it a tuple before. You can use pattern matching for this, but here is a slightly sneaky approach. I just remove the opening and closing paranthesis from the string and replace them with brackets -- and voila it becomes a list.
tuplishToList :: String -> [Int]
tuplishToList str = read ('[' : tail (init str) ++ "]")
Edit
Sorry, I did not see your latest comment. What you try to do is not that difficult. I use these simple functions for my task:
words str splits str into a list of words that where separated by whitespace before. The output is a list of Strings. Caution: This only works if the string inside your tuple contains no whitespace. Implementing a better solution is left as an excercise to the reader.
map f lst applies f to each element of lst
read is a magic function that makes a a data type from a String. It only works if you know before, what the output is supposed to be. If you really want to understand how that works, consider implementing read for your specific usecase.
And here you go:
tuplish2List :: String -> [(String,Int)]
tuplish2List str = map read (words str)
1 As some others may point out, it may be possible using templates and other hacks, but I don't consider that a real solution.
When doing functional programming, it is often better to think about composition of operations instead of individual steps. So instead of thinking about it like adding tuples one at a time to a list, we can approach it by first dividing the input into a list of strings, and then converting each string into a tuple.
Assuming the tuples are written each on one line, we can split the input using lines, and then use read to parse each tuple. To make it work on the entire list, we use map.
main = do input <- getContents
let tuples = map read (lines input) :: [(String, Integer)]
print tuples
Let's try it.
$ runghc Tuples.hs
("Hello", 2)
("Haskell", 4)
Here, I press Ctrl+D to send EOF to the program, (or Ctrl+Z on Windows) and it prints the result.
[("Hello",2),("Haskell",4)]
If you want something more interactive, you will probably have to do your own recursion. See Daniel Wagner's answer for an example of that.
One simple solution to this would be to use a list comprehension, as so (done in GHCi):
Prelude> let fstMap tuplist = [fst x | x <- tuplist]
Prelude> fstMap [("String1",1),("String2",2),("String3",3)]
["String1","String2","String3"]
Prelude> :t fstMap
fstMap :: [(t, b)] -> [t]
This will work for an arbitrary number of tuples - as many as the user wants to use.
To use this in your code, you would just write:
fstMap :: Eq a => [(a,b)] -> [a]
fstMap tuplist = [fst x | x <- tuplist]
The example I gave is just one possible solution. As the name implies, of course, you can just write:
fstMap' :: Eq a => [(a,b)] -> [a]
fstMap' = map fst
This is an even simpler solution.
I'm guessing that, since this is for a class, and you've been studying Haskell for < 1 week, you don't actually need to do any input/output. That's a bit more advanced than you probably are, yet. So:
As others have said, map fst will take a list of tuples, of arbitrary length, and return the first elements. You say you know how to do that. Fine.
But how do the tuples get into the list in the first place? Well, if you have a list of tuples and want to add another, (:) does the trick. Like so:
oldList = [("first", 1), ("second", 2)]
newList = ("third", 2) : oldList
You can do that as many times as you like. And if you don't have a list of tuples yet, your list is [].
Does that do everything that you need? If not, what specifically is it missing?
Edit: With the corrected type:
Eq a => [(a, b)]
That's not the type of a function. It's the type of a list of tuples. Just have the user type yourFunctionName followed by [ ("String1", val1), ("String2", val2), ... ("LastString", lastVal)] at the prompt.
I have a question about tuples and lists in Haskell. I know how to add input into a tuple a specific number of times. Now I want to add tuples into a list an unknown number of times; it's up to the user to decide how many tuples they want to add.
How do I add tuples into a list x number of times when I don't know X beforehand?
There's a lot of things you could possibly mean. For example, if you want a few copies of a single value, you can use replicate, defined in the Prelude:
replicate :: Int -> a -> [a]
replicate 0 x = []
replicate n | n < 0 = undefined
| otherwise = x : replicate (n-1) x
In ghci:
Prelude> replicate 4 ("Haskell", 2)
[("Haskell",2),("Haskell",2),("Haskell",2),("Haskell",2)]
Alternately, perhaps you actually want to do some IO to determine the list. Then a simple loop will do:
getListFromUser = do
putStrLn "keep going?"
s <- getLine
case s of
'y':_ -> do
putStrLn "enter a value"
v <- readLn
vs <- getListFromUser
return (v:vs)
_ -> return []
In ghci:
*Main> getListFromUser :: IO [(String, Int)]
keep going?
y
enter a value
("Haskell",2)
keep going?
y
enter a value
("Prolog",4)
keep going?
n
[("Haskell",2),("Prolog",4)]
Of course, this is a particularly crappy user interface -- I'm sure you can come up with a dozen ways to improve it! But the pattern, at least, should shine through: you can use values like [] and functions like : to construct lists. There are many, many other higher-level functions for constructing and manipulating lists, as well.
P.S. There's nothing particularly special about lists of tuples (as compared to lists of other things); the above functions display that by never mentioning them. =)
Sorry, you can't1. There are fundamental differences between tuples and lists:
A tuple always have a finite amount of elements, that is known at compile time. Tuples with different amounts of elements are actually different types.
List an have as many elements as they want. The amount of elements in a list doesn't need to be known at compile time.
A tuple can have elements of arbitrary types. Since the way you can use tuples always ensures that there is no type mismatch, this is safe.
On the other hand, all elements of a list have to have the same type. Haskell is a statically-typed language; that basically means that all types are known at compile time.
Because of these reasons, you can't. If it's not known, how many elements will fit into the tuple, you can't give it a type.
I guess that the input you get from your user is actually a string like "(1,2,3)". Try to make this directly a list, whithout making it a tuple before. You can use pattern matching for this, but here is a slightly sneaky approach. I just remove the opening and closing paranthesis from the string and replace them with brackets -- and voila it becomes a list.
tuplishToList :: String -> [Int]
tuplishToList str = read ('[' : tail (init str) ++ "]")
Edit
Sorry, I did not see your latest comment. What you try to do is not that difficult. I use these simple functions for my task:
words str splits str into a list of words that where separated by whitespace before. The output is a list of Strings. Caution: This only works if the string inside your tuple contains no whitespace. Implementing a better solution is left as an excercise to the reader.
map f lst applies f to each element of lst
read is a magic function that makes a a data type from a String. It only works if you know before, what the output is supposed to be. If you really want to understand how that works, consider implementing read for your specific usecase.
And here you go:
tuplish2List :: String -> [(String,Int)]
tuplish2List str = map read (words str)
1 As some others may point out, it may be possible using templates and other hacks, but I don't consider that a real solution.
When doing functional programming, it is often better to think about composition of operations instead of individual steps. So instead of thinking about it like adding tuples one at a time to a list, we can approach it by first dividing the input into a list of strings, and then converting each string into a tuple.
Assuming the tuples are written each on one line, we can split the input using lines, and then use read to parse each tuple. To make it work on the entire list, we use map.
main = do input <- getContents
let tuples = map read (lines input) :: [(String, Integer)]
print tuples
Let's try it.
$ runghc Tuples.hs
("Hello", 2)
("Haskell", 4)
Here, I press Ctrl+D to send EOF to the program, (or Ctrl+Z on Windows) and it prints the result.
[("Hello",2),("Haskell",4)]
If you want something more interactive, you will probably have to do your own recursion. See Daniel Wagner's answer for an example of that.
One simple solution to this would be to use a list comprehension, as so (done in GHCi):
Prelude> let fstMap tuplist = [fst x | x <- tuplist]
Prelude> fstMap [("String1",1),("String2",2),("String3",3)]
["String1","String2","String3"]
Prelude> :t fstMap
fstMap :: [(t, b)] -> [t]
This will work for an arbitrary number of tuples - as many as the user wants to use.
To use this in your code, you would just write:
fstMap :: Eq a => [(a,b)] -> [a]
fstMap tuplist = [fst x | x <- tuplist]
The example I gave is just one possible solution. As the name implies, of course, you can just write:
fstMap' :: Eq a => [(a,b)] -> [a]
fstMap' = map fst
This is an even simpler solution.
I'm guessing that, since this is for a class, and you've been studying Haskell for < 1 week, you don't actually need to do any input/output. That's a bit more advanced than you probably are, yet. So:
As others have said, map fst will take a list of tuples, of arbitrary length, and return the first elements. You say you know how to do that. Fine.
But how do the tuples get into the list in the first place? Well, if you have a list of tuples and want to add another, (:) does the trick. Like so:
oldList = [("first", 1), ("second", 2)]
newList = ("third", 2) : oldList
You can do that as many times as you like. And if you don't have a list of tuples yet, your list is [].
Does that do everything that you need? If not, what specifically is it missing?
Edit: With the corrected type:
Eq a => [(a, b)]
That's not the type of a function. It's the type of a list of tuples. Just have the user type yourFunctionName followed by [ ("String1", val1), ("String2", val2), ... ("LastString", lastVal)] at the prompt.