I have a lot of code of the style:
do
x <- getSomething
case x of
this -> ...
that -> ...
other -> ...
Any way of me combining the "x <- ..." and "case x of" lines to eliminate the need for a variable?
You could use the bind operator >>= to pipe the x.
import System.Environment (getArgs)
main :: IO ()
main = getArgs >>= process
where process ["xxx"] = putStrLn "You entered xxx"
process ["yyy"] = putStrLn "You entered yyy"
process _ = putStrLn "Error"
I do this as
foo "this" = return 2
foo "that" = return 3
main = foo =<< getSomething
The nice thing about this approach is that if foo is pure then this becomes
main = foo <$> getSomething
So the code retains the same shape for somewhat varying circumstances.
If you want something extremely close to:
getSomething >>= caseOf
this -> expr1
that -> expr2
other -> expr3
Then I think you're just out of luck - there is no such syntax in Haskell. But know you are not alone. Mark Jones defined the Habit language to include a sort of monadic case with the syntax:
case<- getSomething of
Nothing -> expr1
Just x -> expr2 x
He refers to this as a "Case-From" statement in the language definition.
Related
I'm trying to learn how to work with IO in Haskell by writing a function that, if there is a flag, will take a list of points from a file, and if there is no flag, it asks the user to enter them.
dispatch :: [String] -> IO ()
dispatch argList = do
if "file" `elem` argList
then do
let (path : otherArgs) = argList
points <- getPointsFile path
else
print "Enter a point in the format: x;y"
input <- getLine
if (input == "exit")
then do
print "The user inputted list:"
print $ reverse xs
else (inputStrings (input:xs))
if "help" `elem` argList
then help
else return ()
dispatch [] = return ()
dispatch _ = error "Error: invalid args"
getPointsFile :: String -> IO ([(Double, Double)])
getPointsFile path = do
handle <- openFile path ReadMode
contents <- hGetContents handle
let points_str = lines contents
let points = foldl (\l d -> l ++ [tuplify2 $ splitOn ";" d]) [] points_str
hClose handle
return points
I get this: do-notation in pattern Possibly caused by a missing 'do'?` after `if "file" `elem` argList.
I'm also worried about the binding issue, assuming that I have another flag that says which method will be used to process the points. Obviously it waits for points, but I don't know how to make points visible not only in if then else, constructs. In imperative languages I would write something like:
init points
if ... { points = a}
else points = b
some actions with points
How I can do something similar in Haskell?
Here's a fairly minimal example that I've done half a dozen times when I'm writing something quick and dirty, don't have a complicated argument structure, and so can't be bothered to do a proper job of setting up one of the usual command-line parsing libraries. It doesn't explain what went wrong with your approach -- there's an existing good answer there -- it's just an attempt to show what this kind of thing looks like when done idiomatically.
import System.Environment
import System.Exit
import System.IO
main :: IO ()
main = do
args <- getArgs
pts <- case args of
["--help"] -> usage stdout ExitSuccess
["--file", f] -> getPointsFile f
[] -> getPointsNoFile
_ -> usage stderr (ExitFailure 1)
print (frobnicate pts)
usage :: Handle -> ExitCode -> IO a
usage h c = do
nm <- getProgName
hPutStrLn h $ "Usage: " ++ nm ++ " [--file FILE]"
hPutStrLn h $ "Frobnicate the points in FILE, or from stdin if no file is supplied."
exitWith c
getPointsFile :: FilePath -> IO [(Double, Double)]
getPointsFile = {- ... -}
getPointsNoFile :: IO [(Double, Double)]
getPointsNoFile = {- ... -}
frobnicate :: [(Double, Double)] -> Double
frobnicate = {- ... -}
if in Haskell doesn't inherently have anything to do with control flow, it just switches between expressions. Which, in Haskell, happen to include do blocks of statements (if we want to call them that), but you still always need to make that explicit, i.e. you need to say both then do and else do if there are multiple statements in each branch.
Also, all the statements in a do block need to be indented to the same level. So in your case
if "file" `elem` argList
...
if "help" `elem` argList
Or alternatively, if the help check should only happen in the else branch, it needs to be indented to the statements in that do block.
Independent of all that, I would recommend to avoid parsing anything in an IO context. It is usually much less hassle and easier testable to first parse the strings into a pure data structure, which can then easily be processed by the part of the code that does IO. There are libraries like cmdargs and optparse-applicative that help with the parsing part.
I am beginner in programming via Haskell language. I am interested at writing function in Haskell, which has side effects like this (for educational purposes, it can have another side effect):
-- function foo behaves like random numbers generator
foo :: Int
{- foo = some magic -}
-- It can print True or False
print $ show (foo /= foo)
As I understood, I have to implement some type instance of Monad with such behaviour (like this: IO Why are side-effects modeled as monads in Haskell?). But I really don't understand how to write simple example of function for this purpose.
If foo really behaves like some RNG, it must have some monadic type, or take as an additional input the state of the RNG.
Here's an example, using the system RNG (which lives in the IO monad):
import System.Random
foo :: IO Int
foo = randomRIO (1,5) -- random number between 1 and 5
main :: IO ()
main = do
x <- foo
y <- foo
print $ x /= y
In main, we simply call foo twice, binding its results to x,y, which have type Int. Then, we can compare them as usual.
Note that we can not use foo /= foo since foo :: IO Int, an action returning an Int, and we can't compare actions, only basic data like integers. This is why we first execute the actions, and then compare their results.
The main could be written in one line exploiting some more advanced combinators
main = print =<< ((/=) <$> foo <*> foo)
-- or
main = print =<< liftA2 (/=) foo foo
but I would not focus on these at the beginning.
A more general example of a side-effecful computation:
foo :: IORef String -> IO Int
foo s = do
x <- readIORef s -- reading a "variable"
print x -- output
w <- getLine -- input
writeIORef s ("qwerty" ++ x) -- writing a variable
length <$> readIORef s
main :: IO ()
main = do
s <- newIORef "dummy"
print =<< foo s
Often times I found myself in need of skipping the rest of the iteration (like continue in C) in Haskell:
forM_ [1..100] $ \ i ->
a <- doSomeIO
when (not $ isValid1 a) <skip_rest_of_the_iteration>
b <- doSomeOtherIO a
when (not $ isValid2 b) <skip_rest_of_the_iteration>
...
However, I failed to find an easy way to do so. The only way I am aware of is probably the Trans.Maybe, but is it necessary to use a monad transform to achieve something so trivial?
Remember that loops like this in Haskell are not magic...they're just normal first-class things that you can write yourself.
For what it's worth, I don't think it's too useful to think of MaybeT as a Monad transformer. To me, MaybeT is just a newtype wrapper to give an alternative implementation of (>>=)...just like how you use Product, Sum, First, And, etc. to give alternative implementations of mappend and mempty.
Right now, (>>=) for you is IO a -> (a -> IO b) -> IO b. But it'd be more useful to have (>>=) here be IO (Maybe a) -> (a -> IO (Maybe b) -> IO (Maybe b). As soon as you get to the first action that returns a Nothing, it's really impossible to "bind" any further. That's exactly what MaybeT gives you. You also get a "custom instance" of guard, guard :: Bool -> IO (Maybe a), instead of guard :: IO a.
forM_ [1..100] $ \i -> runMaybeT $ do
a <- lift doSomeIO
guard (isValid1 a)
b <- lift $ doSomeOtherIO a
guard (isValid2 b)
...
and that's it :)
MaybeT is not magic either, and you can achieve basically the same effect by using nested whens. It's not necessary, it just makes things a lot simpler and cleaner :)
Here's how you would do it using bare-bones recursion:
loop [] = return () -- done with the loop
loop (x:xs) =
do a <- doSomeIO
if ...a...
then return () -- exit the loop
else do -- continuing with the loop
b <- doSomeMoreIO
if ...b...
then return () -- exit the loop
else do -- continuing with the loop
...
loop xs -- perform the next iteration
and then invoke it with:
loop [1..100]
You can tidy this up a bit with the when function from Control.Monad:
loop [] = return ()
loop (x:xs) =
do a <- doSomeIO
when (not ...a...) $ do
b <- doSomeMoreIO
when (not ...b...) $ do
...
loop xs
There is also unless in Control.Monad which you might prefer to use.
Using #Ørjan Johansen 's helpful advice, here is an simple example:
import Control.Monad
loop [] = return ()
loop (x:xs) = do
putStrLn $ "x = " ++ show x
a <- getLine
when (a /= "stop") $ do
b <- getLine
when (b /= "stop") $ do
print $ "iteration: " ++ show x ++ ": a = " ++ a ++ " b = " ++ b
loop xs
main = loop [1..3]
If you want to loop over a list or other container to perform actions and/or produce a summary value, and you're finding the usual convenience tools like for_ and foldM aren't good enough for the job, you might want to consider foldr, which is plenty strong enough for the job. When you're not really looping over a container, you can use plain old recursion or pull in something like https://hackage.haskell.org/package/loops or (for a very different flavor) https://hackage.haskell.org/package/machines or perhaps https://hackage.haskell.org/package/pipes.
I'm fairly new to Haskell and have been trying to find a way to pass multiple IO-tainted values to a function to deal with a C library. Most people seem to use the <- operator inside a do block, like this:
g x y = x ++ y
interactiveConcat1 = do {x <- getLine;
y <- getLine;
putStrLn (g x y);
return ()}
This makes me feel like I'm doing C, except emacs can't auto-indent. I tried to write this in a more Lispy style:
interactiveConcat2 = getLine >>= (\x ->
getLine >>= (\y ->
putStrLn (g x y) >>
return () ))
That looks like a mess, and has a string of closed parentheses you have to count at the end (except again, emacs can reliably assist with this task in Lisp, but not in Haskell). Yet another way is to say
import Control.Applicative
interactiveConcat3 = return g <*> getLine <*> getLine >>= putStrLn
which looks pretty neat but isn't part of the base language.
Is there any less laborious notation for peeling values out of the IO taint boxes? Perhaps there is a cleaner way using a lift* or fmap? I hope it isn't too subjective to ask what is considered "idiomatic"?
Also, any tips for making emacs cooperate better than (Haskell Ind) mode would be greatly appreciated. Thanks!
John
Edit: I stumbled across https://wiki.haskell.org/Do_notation_considered_harmful and realized that the nested parentheses in the lambda chain I wrote is not necessary. However it seems the community (and ghc implementors) have embraced the Applicative-inspired style using , <*>, etc, which seems to make the code easier to read in spite of the headaches with figuring out operator precedence.
Note: This post is written in literate Haskell. You can save it as Main.lhs and try it in your GHCi.
A short remark first: you can get rid of the semicolons and the braces in do. Also, putStrLn has type IO (), so you don't need return ():
interactiveConcat1 = do
x <- getLine
y <- getLine
putStrLn $ g x y
We're going to work with IO, so importing Control.Applicative or Control.Monad will come in handy:
> module Main where
> import Control.Applicative
> -- Repeat your definition for completeness
> g :: [a] -> [a] -> [a]
> g = (++)
You're looking for something like this:
> interactiveConcat :: IO ()
> interactiveConcat = magic g getLine getLine >>= putStrLn
What type does magic need? It returns a IO String, takes a function that returns an String and takes usual Strings, and takes two IO Strings:
magic :: (String -> String -> String) -> IO String -> IO String -> IO String
We can probably generalize this type to
> magic :: (a -> b -> c) -> IO a -> IO b -> IO c
A quick hoogle search reveals that there are already two functions with almost that type: liftA2 from Control.Applicative and liftM2 from Control.Monad. They're defined for every Applicative and – in case of liftM2 – Monad. Since IO is an instance of both, you can choose either one:
> magic = liftA2
If you use GHC 7.10 or higher, you can also use <$> and <*> without import and write interactiveConcat as
interactiveConcat = g <$> getLine <*> getLine >>= putStrLn
For completeness, lets add a main so that we can easily check this functionality via runhaskell Main.lhs:
> main :: IO ()
> main = interactiveConcat
A simple check shows that it works as intended:
$ echo "Hello\nWorld" | runhaskell Main.lhs
HelloWorld
References
Applicative in the Typeclassopedia
The section "Some useful monadic functions" of LYAH's chapter "For a Few Monads More".
You can use liftA2 (or liftM2 from Control.Monad):
import Control.Applicative (liftA2)
liftA2 g getLine getLine >>= putStrLn
I know that the following "do" notation's "bind" function is equivalent to getLine >>= \line -> putStrLn
do line <- getLine
putStrLn line
But how is the following notation equivalent to bind function?
do line1 <- getLine
putStrLn "enter second line"
line2 <- getLine
return (line1,line2)
I take it you are trying to see how to bind the result of "putStrLn". The answer is in the type of putStrLn:
putStrLn :: String -> IO ()
Remember that "()" is the unit type, which has a single value (also written "()"). So you can bind this in exactly the same way. But since you don't use it you bind it to a "don't care" value:
getLine >>= \line1 ->
putStrLn "enter second line" >>= \_ ->
getline >>= \line2 ->
return (line1, line2)
As it happens, there is an operator already defined for ignoring the return value, ">>". So you could just rewrite this as
getLine >>= \line1 ->
putStrLn "enter second line" >>
getline >>= \line2 ->
return (line1, line2)
I'm not sure if you are also trying to understand how bind operators are daisy-chained. To see this, let me put the implicit brackets and extra indentation in the example above:
getLine >>= (\line1 ->
putStrLn "enter second line" >> (
getline >>= (\line2 ->
return (line1, line2))))
Each bind operator links the value to the left with a function to the right. That function consists of all the rest of the lines in the "do" clause. So the variable being bound through the lambda ("line1" in the first line) is in scope for the whole of the rest of the clause.
For this specific example you can actually avoid both do and >>= by using combinators from Control.Applicative:
module Main where
import Control.Applicative ((<$>), (<*>), (<*))
getInput :: IO (String, String)
getInput = (,) <$> getLine <* putStrLn "enter second line" <*> getLine
main = print =<< getInput
Which works as expected:
travis#sidmouth% ./Main
hello
enter second line
world
("hello","world")
It looks a little weird at first, but in my opinion the applicative style feels very natural once you're used to it.
I would strongly suggest you to read the chapter Desugaring of Do-blocks in the book Real-World haskell. It tells you, that you all are wrong. For a programmer, it's the natural way to use a lambda, but the do-block is implemented using functions which - if a pattern maching failuire occurs - will call the fail implementation of the according monad.
For instance, your case is like:
let f x =
putStrLn "enter second line" >>
let g y = return (x,y)
g _ = fail "Pattern mismatched"
in getLine >>= g
f _ = fail "Pattern mismatched"
in getLine >>= f
In a case like this, this may be completely irrelevant. But consider some expression that involves pattern-matching. Also, you can use this effect for some special stuff, eg, you can do something like this:
oddFunction :: Integral a => [a] -> [a]
oddFunctiond list = do
(True,y) <- zip (map odd list) list
return y
What will this function do? You can read this statement as a rule for working with the elements of the list. The first statement binds an element of the list to the var y, but only if y is odd. If y is even, a pattern matching failure occurs and fail will be called. In the monad instance for Lists, fail is simply []. Thus, the function strips all even elements from the list.
(I know, oddFunction = filter odd would do this better, but this is just an example)
getLine >>= \line1 ->
putStrLn "enter second line" >>
getLine >>= \line2 ->
return (line1, line2)
Generally foo <- bar becomes bar >>= \foo -> and baz becomes baz >> (unless it's the last line of the do-block, in which case it just stays baz).