On Thu, Jan 15, 2015 at 9:24 AM, Dimitri DeFigueiredo < defigueiredo@ucdavis.edu> wrote:

I would not say that the problem is with the guard check. The problem is with 'null'. It's type is

Prelude> :t null null :: [a] -> Bool

So, it expects a list of something, rather than an IO of something, whence the complaint.

While that's the source of the error, the problem is the combination of the guard check and where to bind a value in the IO monad. Guard checks must have a value of Bool. getDBRecord returns something of type IO [Int]. Where just binds a name, so you either need a way to extract the [Int] from the return value before binding it in the where, or a function of type IO [Int] -> Bool for the guard. Note that this isn't an IO issue but a monad issue. There isn't a monad method that returns a value not in the monad, so you can't write either of the two options above using monad methods. The best solution is the one already proposed - write a function from [Int] -> IO String, and use bind (>>=) on that function to handle things. You could also use the do sugaring of <- to get a less functional version. The last option is to use the IO-specific function unsafePerformIO to write something like nullIO = null . unsafePerformIO. But it's called UNSAFE and tucked away in a module of similar operations for a reason. Using bind is much preferred.

On 15/01/15 09:51, Miro Karpis wrote:

Hi,

please is there a way to have guards with 'where' that communicates with IO? Or is there some other more elegant way? I can do this with classic if/else,...but I just find it nicer with guards.

I have something like this (just an example):

f :: Int -> IO String f x | null dbOutput = return "no db record" | otherwise = return "we got some db records" where dbOutput = getDBRecord x

getDBRecord :: Int -> IO [Int] getDBRecord recordId = do putStrLn $ "checking dbRecord" ++ show recordId --getting data from DB return [1,2]

problem is that db dbOutput is IO and the guard check does not like it:

Couldn't match expected type ‘[a0]’ with actual type ‘IO [Int]’ In the first argument of ‘null’, namely ‘dbOutput’ In the expression: null dbOutput

Cheers, Miro

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f has the wrong type. It needs to return a value in the IO monad. For instance: f: String -> IO String f s = return (s ++ "!") g :: IO String g = readLn >>= f You could also let f return a value of a type other than String, say: f :: String -> IO () f = print The type of >>= tells you this: Monad m => m a -> (a -> m b) -> m b. The second argument has type a -> m b. String -> IO () or String -> IO String both match that. As a warning, readLn expects a Haskell value, and f requires that it be a string. So you have to type in a string the way you would in a program, as "Hello" or ['H', 'e', 'l', 'l', 'o'] or you'll get a parse failure. On Thu, Jan 15, 2015 at 2:44 PM, Miro Karpis wrote:

Many thanks everyone,.. but I'm not so confident with monads. If I understand I could translate with them for example IO String to String? If this is true I'm having troubles to achieve this.

Here is my simple test (only to test the logic), which ends with error:

Couldn't match type ‘[]’ with ‘IO’ Expected type: IO Char Actual type: String In the expression: f a In the second argument of ‘(>>=)’, namely ‘(\ a -> f a)’

g = readLn >>= (\a -> f a)

f :: String -> String f s_ = s_ ++ "!"

Cheers, Miro

On Thu, Jan 15, 2015 at 5:42 PM, Mike Meyer wrote:

...

On Thu, Jan 15, 2015 at 9:24 AM, Dimitri DeFigueiredo < defigueiredo@ucdavis.edu> wrote:

...

I would not say that the problem is with the guard check. The problem is with 'null'. It's type is

Prelude> :t null null :: [a] -> Bool

So, it expects a list of something, rather than an IO of something, whence the complaint.

While that's the source of the error, the problem is the combination of the guard check and where to bind a value in the IO monad.

Guard checks must have a value of Bool. getDBRecord returns something of type IO [Int]. Where just binds a name, so you either need a way to extract the [Int] from the return value before binding it in the where, or a function of type IO [Int] -> Bool for the guard.

Note that this isn't an IO issue but a monad issue. There isn't a monad method that returns a value not in the monad, so you can't write either of the two options above using monad methods. The best solution is the one already proposed - write a function from [Int] -> IO String, and use bind (>>=) on that function to handle things. You could also use the do sugaring of <- to get a less functional version.

The last option is to use the IO-specific function unsafePerformIO to write something like nullIO = null . unsafePerformIO. But it's called UNSAFE and tucked away in a module of similar operations for a reason. Using bind is much preferred.

...

On 15/01/15 09:51, Miro Karpis wrote:

Hi,

please is there a way to have guards with 'where' that communicates with IO? Or is there some other more elegant way? I can do this with classic if/else,...but I just find it nicer with guards.

I have something like this (just an example):

f :: Int -> IO String f x | null dbOutput = return "no db record" | otherwise = return "we got some db records" where dbOutput = getDBRecord x

getDBRecord :: Int -> IO [Int] getDBRecord recordId = do putStrLn $ "checking dbRecord" ++ show recordId --getting data from DB return [1,2]

problem is that db dbOutput is IO and the guard check does not like it:

Couldn't match expected type ‘[a0]’ with actual type ‘IO [Int]’ In the first argument of ‘null’, namely ‘dbOutput’ In the expression: null dbOutput

Cheers, Miro

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Going back to an earlier question: a monad is a bit like a roach motel. You can check in but you can't leave. (This isn't true of every Monad, but the point is there's no guarantees.) In particular, you can't go from IO String to String _at all_. But you can, through Functor, pass it to a function that takes a plain String. And through Monad, you can turn IO (IO String) back to IO String. Hope this helps. On Thursday, January 15, 2015, Marcin Mrotek wrote:

Hello,

A list ([]) is also a monad, and a String is defined as a list of characters ([Char]). So in your example, it's as if you were trying to use (>>=) operator on two different monads ([] and IO), which is impossible. To make a pure value a monadic value, you need to use return:

g = readLn >>= (\a -> return (f a))

which is equivalent to composing f with return:

g = readLn >>= return.f

Regards, Marcin _______________________________________________ Beginners mailing list Beginners@haskell.org javascript:; http://www.haskell.org/mailman/listinfo/beginners

-- Sent from an iPhone, please excuse brevity and typos.

Not really, for the same reason: a guard needs a Bool, and you can't get a Bool from IO Bool. On Thursday, January 15, 2015, Miro Karpis wrote:

many thanks,....but then I unfortunately don't understand how can I fix my initial problem:

to use IO check in guards - is that possible?

Regards, Miro

On Thu, Jan 15, 2015 at 10:12 PM, Julian Birch javascript:_e(%7B%7D,'cvml','julian.birch@gmail.com');> wrote:

...

Going back to an earlier question: a monad is a bit like a roach motel. You can check in but you can't leave. (This isn't true of every Monad, but the point is there's no guarantees.) In particular, you can't go from IO String to String _at all_. But you can, through Functor, pass it to a function that takes a plain String. And through Monad, you can turn IO (IO String) back to IO String.

Hope this helps.

On Thursday, January 15, 2015, Marcin Mrotek javascript:_e(%7B%7D,'cvml','marcin.jan.mrotek@gmail.com');> wrote:

...

Hello,

A list ([]) is also a monad, and a String is defined as a list of characters ([Char]). So in your example, it's as if you were trying to use (>>=) operator on two different monads ([] and IO), which is impossible. To make a pure value a monadic value, you need to use return:

g = readLn >>= (\a -> return (f a))

which is equivalent to composing f with return:

g = readLn >>= return.f

Regards, Marcin _______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

-- Sent from an iPhone, please excuse brevity and typos.

-- Sent from an iPhone, please excuse brevity and typos.

many thanks! I will take a look at it. cheers and thanks to all :-) On Thu, Jan 15, 2015 at 10:28 PM, Christopher Allen wrote:

I haven't seen anybody explain the real reasons you can't yank `a` out of something of type `IO a`, so I thought I'd attempt to clear a few things up.

It's true that you cannot *in general* extract a value of type `a` from something of type `Monad m => m a` but those reasons are *not* why you can't in the case of IO. You can't with Monad in general because the typeclass doesn't provide a method with the type `m a -> a`, in fact, it only provides the opposite: `a -> m a`.

The real reason you can't pattern match on or otherwise extract values from IO is that it's an abstract datatype - on purpose. The constructor is not exported, the internals are hidden.

You are using a non-strict programming language. The IO datatype is how the implementation knows not to replace the thunk with its value when evaluated.

Want to see what I mean?

Play with some code that uses https://hackage.haskell.org/package/time-1.3/docs/Data-Time-Clock.html#v:get... to get the current time. Note how each time you force evaluation of the `IO UTCTime` you're getting a new UTCTime each time. Then wrap it in `unsafePerformIO` - it'll only get executed once. This is *definitely* not the semantics you want, you're too new to know when you'd want the unsafe* functions for now.

If you aren't comfortable with monads, no amount of thrashing is going to let you avoid using Functor/Applicative/Monad if you want to interact with values produced in IO.

I wrote a guide for learning Haskell, it covers Functor/Applicative/Monad which are *everywhere* - not just for use with IO. This is the guide: https://github.com/bitemyapp/learnhaskell

There are people used to teaching and assisting with the courses recommended in my guide (cis194 & NICTA course) on Freenode IRC at #haskell-beginners. There are tons of people equipped to help you in #haskell as well.

--- Chris Allen

On Thu, Jan 15, 2015 at 3:12 PM, Julian Birch wrote:

...

Going back to an earlier question: a monad is a bit like a roach motel. You can check in but you can't leave. (This isn't true of every Monad, but the point is there's no guarantees.) In particular, you can't go from IO String to String _at all_. But you can, through Functor, pass it to a function that takes a plain String. And through Monad, you can turn IO (IO String) back to IO String.

Hope this helps.

On Thursday, January 15, 2015, Marcin Mrotek wrote:

...

Hello,

A list ([]) is also a monad, and a String is defined as a list of characters ([Char]). So in your example, it's as if you were trying to use (>>=) operator on two different monads ([] and IO), which is impossible. To make a pure value a monadic value, you need to use return:

g = readLn >>= (\a -> return (f a))

which is equivalent to composing f with return:

g = readLn >>= return.f

Regards, Marcin _______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

-- Sent from an iPhone, please excuse brevity and typos.

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