data Task = Task
{ id :: String
, description :: String
, dependsOn :: [String]
, dependentTasks :: [String]
} deriving (Eq, Show, Generic, ToJSON, FromJSON)
type Storage = Map String Task
s :: Storage
s = empty
addTask :: Task -> Storage -> Storage
addTask (Task id desc dep dept) = insert id (Task id desc dep dept)
removeTask :: String -> Storage -> Storage
removeTask tid = delete tid
changes = [addTask (Task "1" "Description" [] []), removeTask "1"]
main = putStrLn . show $ foldl (\s c -> c s) s changes
Suppose I have the following code. I want to store changes list in a json file. But I don't know how to do that with Aeson, aside probably from writing a custom parser and there must be a better way to do that obviously. Like maybe using language extension to derive (Generic, ToJSON, FromJSON) for addTask and removeTask etc...
EDIT. For all people that say "You can't serialize function".
Read the comments to an answer to this question.
Instance Show for function
That said, it's not possible to define Show to actually give you more
? detail about the function. – Louis Wasserman May 12 '12 at 14:51
Sure it is. It can show the type (given via Typeable); or it can show some of the inputs and outputs (as is done in QuickCheck).
EDIT2. Okay, I got that I can't have function name in serialization. But can this be done via template Haskell? I see that aeson supports serialization via template Haskell, but as newcomer to Haskell can't figure out how to do that.
Reading between the lines a bit, a recurring question here is, "Why can't I serialize a function (easily)?" The answer -- which several people have mentioned, but not explained clearly -- is that Haskell is dedicated to referential transparency. Referential transparency says that you can replace a definition with its defined value (and vice versa) without changing the meaning of the program.
So now, let's suppose we had a hypothetical serializeFunction, which in the presence of this code:
foo x y = x + y + 3
Would have this behavior:
> serializeFunction (foo 5)
"foo 5"
I guess you wouldn't object too strenuously if I also claimed that in the presence of
bar x y = x + y + 3
we would "want" this behavior:
> serializeFunction (bar 5)
"bar 5"
And now we have a problem, because by referential transparency
serializeFunction (foo 5)
= { definition of foo }
serializeFunction (\y -> 5 + y + 3)
= { definition of bar }
serializeFunction (bar 5)
but "foo 5" does not equal "bar 5".
The obvious followup question is: why do we demand referential transparency? There are at least two good reasons: first, it allows equational reasoning like above, hence eases the burden of refactoring; and second, it reduces the amount of runtime information that's needed, hence improving performance.
Of course, if you can come up with a representation of functions that respects referential transparency, that poses no problems. Here are some ideas in that direction:
printing the type of the function
instance (Typeable a, Typeable b) => Show (a -> b) where
show = show . typeOf
-- can only write a Read instance for trivial functions
printing the input-output behavior of the function (which can also be read back in)
creating a data type that combines a function with its name, and then printing that name
data Named a = Named String a
instance Show (Named a) where
show (Named n _) = n
-- perhaps you could write an instance Read (Map String a -> Named a)
(and see also cloud haskell for a more complete working of this idea)
constructing an algebraic data type that can represent all the expressions you care about but contains only basic types that already have a Show instance and serializing that (e.g. as described in the other answer)
But printing a bare function's name is in conflict with referential transparency.
Make a data type for your functions and an evaluation function:
data TaskFunction = AddTask Task | RemoveTask String
deriving (Eq, Show, Generic, ToJSON, FromJSON)
eval :: TaskFunction -> Storage -> Storage
eval (AddTask t) = addTask t
eval (RemoveTask t) = removeTask t
changes = [AddTask (Task "1" "Description" [] []), RemoveTask "1"]
main = putStrLn . show $ foldl (\s c -> c s) s (eval <$> changes)
Related
I am trying to define a parser in Haskell. I am a total beginner and somehow didn't manage to find any solution to my problem at all.
For the first steps I tried to follow the instructions on the slides of a powerpoint presentation. But I constantly get the error "Not in scope: type variable ‘a’":
type Parser b = a -> [(b,a)]
item :: Parser Char
item = \inp -> case inp of
[] -> []
(x:xs) -> [(x:xs)]
error: Not in scope: type variable ‘a’
|
11 | type Parser b = a -> [(b,a)]
| ^
I don't understand the error but moreover I don't understand the first line of the code as well:
type Parser b = a -> [(b,a)]
What is this supposed to do? On the slide it just tells me that in Haskell, Parsers can be defined as functions. But that doesn't look like a function definition to me. What is "type" doing here? If it s used to specify the type, why not use "::" like in second line above? And "Parser" seems to be a data type (because we can use it in the type definition of "item"). But that doesn't make sense either.
The line derives from:
type Parser = String -> (String, Tree)
The line I used in my code snippet above is supposed to be a generalization of that.
Your help would be much appreciated. And please bear in mind that I hardly know anything about Haskell, when you write an answer :D
There is a significant difference between the type alias type T = SomeType and the type annotation t :: SomeType.
type T = Int simply states that T is just another name for the type Int. From now on, every time we use T, it will be replaced with Int by the compiler.
By contrast, t :: Int indicates that t is some value of type Int. The exact value is to be specified by an equation like t = 42.
These two concepts are very different. On one hand we have equations like T = Int and t = 42, and we can replace either side with the other side, replacing type with types and values with values. On the other hand, the annotation t :: Int states that a value has a given type, not that the value and the type are the same thing (which is nonsensical: 42 and Int have a completely different nature, a value and a type).
type Parser = String -> (String, Tree)
This correctly defines a type alias. We can make it parametric by adding a parameter:
type Parser a = String -> (String, a)
In doing so, we can not use variables in the right hand side that are not parameters, for the same reason we can not allow code like
f x = x + y -- error: y is not in scope
Hence you need to use the above Parser type, or some variation like
type Parser a = String -> [(String, a)]
By contrast, writing
type Parser a = b -> [(b, a)] -- error
would use an undeclared type b, and is an error. At best, we could have
type Parser a b = b -> [(b, a)]
which compiles. I wonder, though, is you really need to make the String type even more general than it is.
So, going back to the previous case, a possible way to make your code run is:
type Parser a = String -> [(a, String)]
item :: Parser Char
item = \inp -> case inp of
[] -> []
(x:xs) -> [(x, xs)]
Note how [(x, xs)] is indeed of type [(Char, String)], as needed.
If you really want to generalize String as well, you need to write:
type Parser a b = b -> [(b, a)]
item :: Parser Char String
item = \inp -> case inp of
[] -> []
(x:xs) -> [(xs, x)]
I am trying to write a simple tool in Haskell as a learning exercise, and have encountered something that I cannot figure out. Here is a simple sample illustrating it.
idMap :: a -> a
idMap x = map id x
main = do
print $ idMap [1, 2]
According to my understanding, this example should compile and print [1, 2] when run. However, if fails to compile with the following message:
source_file.hs:2:18: error:
• Couldn't match expected type ‘[b0]’ with actual type ‘a’
‘a’ is a rigid type variable bound by
the type signature for:
idMap :: forall a. a -> a
at source_file.hs:1:10
• In the second argument of ‘map’, namely ‘x’
In the expression: map id x
In an equation for ‘idMap’: idMap x = map id x
• Relevant bindings include
x :: a
(bound at source_file.hs:2:7)
idMap :: a -> a
(bound at source_file.hs:2:1)
It kind of makes sense, given that the signature of map is (a -> b) -> [a] -> [b] so the input type is not necessarily the same as the output type, but the signature of id is a -> a so surely it follows that map id would have a signature of (a -> a) -> [a] -> [a]?
The second part I don't really understand is why this is an exception given that all the types (a and b as above) are Integer. It would make sense to me that since the signature of idMap is a -> a, there should only be a compile exception if it is used in a situation where the output type is expected to be different from input type.
Finally, how would I make this code actually work? My real code is a little more complicated and I am relying on the output type matching the input type elsewhere in the code so I don't want to change the signature of idMap, I want to know what I would need to do to write a function with that signature.
You are applying idMap to a list. Therefore, we know the argument type should be some list type. Further, you expect the return type to be a list ([1,2]), so the return type should be a list as well. Since the id function is fully polymorphic (a -> a), we can map it over a list of any type a and return a list of items of the same type a. Thus, your final type signature should be [a] -> [a].
Regarding your second question, while it's true that the argument type and return type are both the same, the type as a -> a is not true for all types a. Based on the type signature of map, idMap must accept a list argument. We can declare type signatures for functions that are more constrained than necessary, but not less.
Your implementation of idMap...
idMap x = map id x
... involves applying map id on x. As a consequence, x must be a list.
GHCi> :t map id
map id :: [b] -> [b]
This will work just fine:
idMap :: [a] -> [a]
idMap x = map id x
Note that, thanks to the use of id, x and idMap x do have the same type (as do their elements), as you expected.
[...] a pair of functions tofun : int -> ('a -> 'a) and fromfun : ('a -> 'a) ->
int such that (fromfun o tofun) n evaluates to n for every n : int.
Anyone able to explain to me what this is actually asking for? I'm looking for more of an explanation of that than an actual solution to this.
What this is asking for is:
1) A higher-order function tofun which when given an integer returns a polymorphic function, one which has type 'a->'a, meaning that it can be applied to values of any type, returning a value of the same type. An example of such a function is:
- fun id x = x;
val id = fn : 'a -> 'a
for example, id "cat" = "cat" and id () = (). The later value is of type unit, which is a type with only 1 value. Note that there is only 1 total function from unit to unit (namely, id or something equivalent). This underscores the difficulty with coming up with defining tofun: it returns a function of type 'a -> 'a, and other than the identity function it is hard to think of other functions. On the other hand -- such functions can fail to terminate or can raise an error and still have type 'a -> 'a.
2) fromfun is supposed to take a function of type 'a ->'a and return an integer. So e.g. fromfun id might evaluate to 0 (or if you want to get tricky it might never terminate or it might raise an error)
3) These are supposed to be inverses of each other so that, e.g. fromfun (tofun 5) needs to evaluate to 5.
Intuitively, this should be impossible in a sufficiently pure functional language. If it is possible in SML, my guess is that it would be by using some of the impure features of SML (which allow for side effects) to violate referential transparency. Or, the trick might involve raising and handling errors (which is also an impure feature of SML). If you find an answer which works in SML it would be interesting to see if it could be translated to the annoyingly pure functional language Haskell. My guess is that it wouldn't translate.
You can devise the following property:
fun prop_inverse f g n = (f o g) n = n
And with definitions for tofun and fromfun,
fun tofun n = ...
fun fromfun f = ...
You can test that they uphold the property:
val prop_test_1 =
List.all
(fn i => prop_inverse fromfun tofun i handle _ => false)
[0, ~1, 1, valOf Int.maxInt, valOf Int.minInt]
And as John suggests, those functions must be impure. I'd also go with exceptions.
I am new to Erlang and noticed that there is no native function to create json string from lists (Or is there?). I use this method to create json string in Erlang but do not know if it will not malfunction.
Here is an example of my method:
-module(index).
-export([test/0]).
test() ->
Ma = "Hello World", Mb = "Hello Erlang",
A = "{\"Messages\" : [\"" ++ Ma ++ "\", \""++Mb++"\"], \"Usernames\" : [\"Username1\", \"Username2\"]}",
A.
The result is:
388> test().
"{\"Messages\" : [\"Hello World\", \"Hello Erlang\"], \"Usernames\" : [\"Username1\", \"Username2\"]}"
389>
I think this is the expected result but is there any chance that this method may malfunction when included special characters, such as: <, >, & / \ " ??
What precautions should I take to make this method stronger?
If Ma or Mb contains double quotes or whatever control characters, the parsing from string to JSON will fail. This parsing may never occur in Erlang, as Erlang does not have string to JSON conversion built-in.
It's a good idea to use binaries (<<"I am a binary string">>), as lists consume a lot more resources.
We're using jiffy, which is implemented as a NIF and hence is reasonably fast and it allows for document construction like so:
jiffy:decode(<<"{\"foo\": \"bar\"}">>).
{[{<<"foo">>,<<"bar">>}]}
Doc = {[{foo, [<<"bing">>, 2.3, true]}]}.
{[{foo,[<<"bing">>,2.3,true]}]}
jiffy:encode(Doc).
<<"{\"foo\":[\"bing\",2.3,true]}">>
I had this very same problem, searched high and low and in the end came up with my own method. This is purely just pointing people in the right directions to finding a solution for themselves. Note: I tried jiffy but as I'm using rebar3 it's not currently compatible.
Im using MS sql server so i use the Erlang odbc module: http://erlang.org/doc/man/odbc.html
The odbc:sql_query/2 gives me back {selected, Columns, Results}
From here i can take the Columns which is a list of strings & the Results, a list of rows represented each as a tuple, then create a few functions to output valid Erlang code to be able to serialize correctly to Json based on a number of factors. Here's the full code:
make the initial query:
Sql = "SELECT * FROM alloys;",
Ref = connect(),
case odbc:sql_query(Ref, Sql) of
{selected, Columns, Results} ->
set_json_from_sql(Columns, Results, []);
{error, Reason} ->
{error, Reason}
end.
Then the input function is set_json_from_sql/3 that calls the below functions:
format_by_type(Item) ->
if
is_list(Item) -> list_to_binary(io_lib:format("~s", [Item]));
is_integer(Item) -> Item;
is_boolean(Item) -> io_lib:format("~a", [Item]);
is_atom(Item) -> Item
end.
json_by_type([H], [Hc], Data) ->
NewH = format_by_type(H),
set_json_flatten(Data, Hc, NewH);
json_by_type([H|T], [Hc|Tc], Data) ->
NewH = format_by_type(H),
NewData = set_json_flatten(Data, Hc, NewH),
json_by_type(T, Tc, NewData).
set_json_flatten(Data, Column, Row) ->
ResTuple = {list_to_binary(Column), Row},
lists:flatten(Data, [ResTuple]).
set_json_from_sql([], [], Data) -> jsone:encode([{<<"data">>, lists:reverse(Data)}]);
set_json_from_sql(Columns, [H], Data) ->
NewData = set_json_merge(H, Columns, Data),
set_json_from_sql([], [], NewData);
set_json_from_sql(Columns, [H|T], Data) ->
NewData = set_json_merge(H, Columns, Data),
set_json_from_sql(Columns, T, NewData).
set_json_merge(Row, Columns, Data) ->
TupleRow = json_by_type(tuple_to_list(Row), Columns, []),
lists:append([TupleRow], Data).
So set_json_from_sql/3 gives you your Json output after matching set_json_from_sql([], [], Data).
The key points here are that you need to call list_to_binary/1 for strings & atoms. Use jsone to encode Erlang objects to Json: https://github.com/sile/jsone
And, notice format_by_type/1 is used to match against Erlang object types, yes not ideal but works as long as you are aware of your DB's types or you can increase the extra guards to accommodate this.
This works for me
test()->
Ma = "Hello World", Mb = "Hello Erlang",
A = "{\"Messages\" : {{\"Ma\":\"" ++ Ma ++ "\"}, {\"Mb\":\""++Mb++"\"}}, {\"Usernames\" : {\"Username1\":\"usrname1\"}, {\"Username2\":\"usrname2\"}}",
io:format("~s~n",[A]).
Output
10> io:format("~s~n",[A]).
{"Messages" : {{"Ma":Hello World"}, {"Mb":Hello Erlang"}}, {"Usernames" : {"Username1":"usrname1"}, {"Username2":"usrname2"}}
ok
or use one of many libraries on github to convert erlang terms to json. My Tuple to JSON module is simple but effective.
Do it like a pro
-define(JSON_WRAPPER(Proplist), {Proplist}).
-spec from_list(json_proplist()) -> object().
from_list([]) -> new();
from_list(L) when is_list(L) -> ?JSON_WRAPPER(L).
-spec to_binary(atom() | string() | binary() | integer() | float() | pid() | iolist()) -> binary().
to_binary(X) when is_float(X) -> to_binary(mochinum:digits(X));
to_binary(X) when is_integer(X) -> list_to_binary(integer_to_list(X));
to_binary(X) when is_atom(X) -> list_to_binary(atom_to_list(X));
to_binary(X) when is_list(X) -> iolist_to_binary(X);
to_binary(X) when is_pid(X) -> to_binary(pid_to_list(X));
to_binary(X) when is_binary(X) -> X.
-spec recursive_from_proplist(any()) -> object().
recursive_from_proplist([]) -> new();
recursive_from_proplist(List) when is_list(List) ->
case lists:all(fun is_integer/1, List) of
'true' -> List;
'false' ->
from_list([{to_binary(K) ,recursive_from_proplist(V)}
|| {K,V} <- List
])
end;
recursive_from_proplist(Other) -> Other.
This question here is related to
Haskell Input Return Tuple
I wonder how we can passes the input from monad IO to another function in order to do some computation.
Actually what i want is something like
-- First Example
test = savefile investinput
-- Second Example
maxinvest :: a
maxinvest = liftM maximuminvest maxinvestinput
maxinvestinput :: IO()
maxinvestinput = do
str <- readFile "C:\\Invest.txt"
let cont = words str
let mytuple = converttuple cont
let myint = getint mytuple
putStrLn ""
-- Convert to Tuple
converttuple :: [String] -> [(String, Integer)]
converttuple [] = []
converttuple (x:y:z) = (x, read y):converttuple z
-- Get Integer
getint :: [(String, Integer)] -> [Integer]
getint [] = []
getint (x:xs) = snd (x) : getint xs
-- Search Maximum Invest
maximuminvest :: (Ord a) => [a] -> a
maximuminvest [] = error "Empty Invest Amount List"
maximuminvest [x] = x
maximuminvest (x:xs)
| x > maxTail = x
| otherwise = maxTail
where maxTail = maximuminvest xs
In the second example, the maxinvestinput is read from file and convert the data to the type maximuminvest expected.
Please help.
Thanks.
First, I think you're having some basic issues with understanding Haskell, so let's go through building this step by step. Hopefully you'll find this helpful. Some of it will just arrive at the code you have, and some of it will not, but it is a slowed-down version of what I'd be thinking about as I wrote this code. After that, I'll try to answer your one particular question.
I'm not quite sure what you want your program to do. I understand that you want a program which reads as input a file containing a list of people and their investments. However, I'm not sure what you want to do with it. You seem to (a) want a sensible data structure ([(String,Integer)]), but then (b) only use the integers, so I'll suppose that you want to do something with the strings too. Let's go through this. First, you want a function that can, given a list of integers, return the maximum. You call this maximuminvest, but this function is more general that just investments, so why not call it maximum? As it turns out, this function already exists. How could you know this? I recommend Hoogle—it's a Haskell search engine which lets you search both function names and types. You want a function from lists of integers to a single integer, so let's search for that. As it turns out, the first result is maximum, which is the more general version of what you want. But for learning purposes, let's suppose you want to write it yourself; in that case, your implementation is just fine.
Alright, now we can compute the maximum. But first, we need to construct our list. We're going to need a function of type [String] -> [(String,Integer)] to convert our formattingless list into a sensible one. Well, to get an integer from a string, we'll need to use read. Long story short, your current implementation of this is also fine, though I would (a) add an error case for the one-item list (or, if I were feeling nice, just have it return an empty list to ignore the final item of odd-length lists), and (b) use a name with a capital letter, so I could tell the words apart (and probably a different name):
tupledInvestors :: [String] -> [(String, Integer)]
tupledInvestors [] = []
tupledInvestors [_] = error "tupledInvestors: Odd-length list"
tupledInvestors (name:amt:rest) = (name, read amt) : tupledInvestors rest
Now that we have these, we can provide ourselves with a convenience function, maxInvestment :: [String] -> Integer. The only thing missing is the ability to go from the tupled list to a list of integers. There are several ways to solve this. One is the one you have, though that would be unusual in Haskell. A second would be to use map :: (a -> b) -> [a] -> [b]. This is a function which applies a function to every element of a list. Thus, your getint is equivalent to the simpler map snd. The nicest way would probably be to use Data.List.maximumBy :: :: (a -> a -> Ordering) -> [a] -> a. This is like maximum, but it allows you to use a comparison function of your own. And using Data.Ord.comparing :: Ord a => (b -> a) -> b -> b -> Ordering, things become nice. This function allows you to compare two arbitrary objects by converting them to something which can be compared. Thus, I would write
maxInvestment :: [String] -> Integer
maxInvestment = maximumBy (comparing snd) . tupledInvestors
Though you could also write maxInvestment = maximum . map snd . tupledInvestors.
Alright, now on to the IO. Your main function, then, wants to read from a specific file, compute the maximum investment, and print that out. One way to represent that is as a series of three distinct steps:
main :: IO ()
main = do dataStr <- readFile "C:\\Invest.txt"
let maxInv = maxInvestment $ words dataStr
print maxInv
(The $ operator, if you haven't seen it, is just function application, but with more convenient precedence; it has type (a -> b) -> a -> b, which should make sense.) But that let maxInv seems pretty pointless, so we can get rid of that:
main :: IO ()
main = do dataStr <- readFile "C:\\Invest.txt"
print . maxInvestment $ words dataStr
The ., if you haven't seen it yet, is function composition; f . g is the same as \x -> f (g x). (It has type (b -> c) -> (a -> b) -> a -> c, which should, with some thought, make sense.) Thus, f . g $ h x is the same as f (g (h x)), only easier to read.
Now, we were able to get rid of the let. What about the <-? For that, we can use the =<< :: Monad m => (a -> m b) -> m a -> m b operator. Note that it's almost like $, but with an m tainting almost everything. This allows us to take a monadic value (here, the readFile "C:\\Invest.txt" :: IO String), pass it to a function which turns a plain value into a monadic value, and get that monadic value. Thus, we have
main :: IO ()
main = print . maxInvestment . words =<< readFile "C:\\Invest.txt"
That should be clear, I hope, especially if you think of =<< as a monadic $.
I'm not sure what's happening with testfile; if you edit your question to reflect that, I'll try to update my answer.
One more thing. You said
I wonder how we can passes the input from monad IO to another function in order to do some computation.
As with everything in Haskell, this is a question of types. So let's puzzle through the types here. You have some function f :: a -> b and some monadic value m :: IO a. You want to use f to get a value of type b. This is impossible, as I explained in my answer to your other question; however, you can get something of type IO b. Thus, you need a function which takes your f and gives you a monadic version. In other words, something with type Monad m => (a -> b) -> (m a -> m b). If we plug that into Hoogle, the first result is Control.Monad.liftM, which has precisely that type signature. Thus, you can treat liftM as a slightly different "monadic $" than =<<: f `liftM` m applies f to the pure result of m (in accordance with whichever monad you're using) and returns the monadic result. The difference is that liftM takes a pure function on the left, and =<< takes a partially-monadic one.
Another way to write the same thing is with do-notation:
do x <- m
return $ f x
This says "get the x out of m, apply f to it, and lift the result back into the monad." This is the same as the statement return . f =<< m, which is precisely liftM again. First f performs a pure computation; its result is passed into return (via .), which lifts the pure value into the monad; and then this partially-monadic function is applied, via =<,, to m.
It's late, so I'm not sure how much sense that made. Let me try to sum it up. In short, there is no general way to leave a monad. When you want to perform computation on monadic values, you lift pure values (including functions) into the monad, and not the other way around; that could violate purity, which would be Very Bad™.
I hope that actually answered your question. Let me know if it didn't, so I can try to make it more helpful!
I'm not sure I understand your question, but I'll answer as best I can. I've simplified things a bit to get at the "meat" of the question, if I understand it correctly.
maxInvestInput :: IO [Integer]
maxInvestInput = liftM convertToIntegers (readFile "foo")
maximumInvest :: Ord a => [a] -> a
maximumInvest = blah blah blah
main = do
values <- maxInvestInput
print $ maximumInvest values
OR
main = liftM maximumInvest maxInvestInput >>= print