A common beginner's mistake is to see return and think that it's a language keyword that exits the current function with a return value. Of course, we know that isn't what it does at all. But I wonder... could we actually make such a function? Purely for argument's sake, at this point.
It seems we're looking for some monad Foo which possesses a function
exit :: x -> Foo x
which will abort the rest of the computation and immediately return x.
Is such a thing constructable? If it could be constructed, would it be useful? Is this even a sane thing to attempt to do?
Yes, it is possible in the Cont monad. The Cont is defined as follows:
newtype Cont r a = Cont {runCont :: (a -> r) -> r}
instance Monad (Cont r) where
return a = Cont ($ a)
m >>= k = Cont $ \c -> runCont m $ \a -> runCont (k a) c
Now we can create exit as follows:
exit = Cont . const
In fact you can do it in almost any monad (for example you could do it in the Either monad but not in the State monad). However the Cont monad is the mother of all monads: http://blog.sigfpe.com/2008/12/mother-of-all-monads.html
Of course it's doable. It's even already in the standard libraries. You just need a slightly saner type signature.
import Control.Monad
exit :: a -> Either a b
exit = Left
main = print $ do
x <- Right 5
exit "foo"
error "this is a good place to crash"
The important thing to note is that the type given when bailing out isn't free - it has to match the type of the Either.
It appears you can just about implement this:
data Foo e x =
Next x |
Done e
instance Monad (Foo e) where
return = Next
(Next x) >>= f = f x
(Done e) >>= f = Done e
exit :: e -> Foo e x
exit = Done
run :: Foo x x -> x
run (Next x) = x
run (Done x) = x
Notice how the monad has an extra type parameter; I don't think it's possible to avoid this. Basically Foo e x produces an x and will "ultimately" produce an e. Note that the run function requires both types to match; you could equally demand that run only works if exit gets called somewhere, or one of a few other possibilities...
So, yes you can construct this, and no, it's not especially useful AFAIK.
If you want to satisfy the monad laws, you certainly can't call it return since we need that return a >>= f = f a.
Also, I don't think it can exist at all with the precise specification given. Suppose x :: a and f :: a -> Foo b, then just following the types, exit a >>= f :: Foo b yet we want it to produce x :: a.
Related
Lets say I got the following very basic example:
f :: Either Int String -> IO ()
as well as a function g :: Int -> IO () and a function g' :: String -> IO () and I basically want to implement f as a "selector" which calls g or g' depending on its input, so that in the future I only have to work with f because I know that my program will only encounter either Int or String.
Does this way of using Either make sense? The convention seems to be to use it mainly for Error and Exception handling.
If it does make sense, what would be a good way or best practice to implement such an example? I've heard/read about case as well as bifunctors.
If it does not: what is the haskell way of handling the possibility of different input types? Or is this something which should be avoided from the beginning?
So that definitely can make sense and one way to implement that is:
f :: Either Int String -> IO ()
f e =
case e of
Left l -> g l
Right r -> g' r
or using either:
import Data.Either (either)
f :: Either Int String -> IO ()
f = either g g'
Note that in the second version I don't assign a variable name to the Either Int String argument. That is called eta conversion / eta reduction. But obviously you could also write f e = either g g' e.
As an armchair Haskell programmer, it seems fine to me.
f :: Either Int String -> IO ()
f (Left n) = g n
f (Right s) = g' s
This is a type declaration of a bind method:
(>>=) :: (Monad m) => m a -> (a -> m b) -> m b
I read this as follows: apply a function that returns a wrapped value, to a wrapped value.
This method was included to Prelude as part of Monad typeclass. That means there are a lot of cases where it's needed.
OK, but I don't understand why it's a typical solution of a typical case at all.
If you already created a function which returns a wrapped value, why that function doesn't already take a wrapped value?
In other words, what are typical cases where there are many functions which take a normal value, but return a wrapped value? (instead of taking a wrapped value and return a wrapped value)
The 'unwrapping' of values is exactly what you want to keep hidden when dealing with monads, since it is this that causes a lot of boilerplate.
For example, if you have a sequence of operations which return Maybe values that you want to combine, you have to manually propagate Nothing if you receive one:
nested :: a -> Maybe b
nested x = case f x of
Nothing -> Nothing
Just r ->
case g r of
Nothing -> Nothing
Just r' ->
case h r' of
Nothing -> Nothing
r'' -> i r''
This is what bind does for you:
Nothing >>= _ = Nothing
Just a >>= f = f a
so you can just write:
nested x = f x >>= g >>= h >>= i
Some monads don't allow you to manually unpack the values at all - the most common example is IO. The only way to get the value from an IO is to map or >>= and both of these require you to propagate IO in the output.
Everyone focuses on IO monad and inability to "unwrap".
But a Monad is not always a container, so you can't unwrap.
Reader r a == r->a such that (Reader r) is a Monad
to my mind is the simplest best example of a Monad that is not a container.
You can easily write a function that can produce m b given a: a->(r->b). But you can't easily "unwrap" the value from m a, because a is not wrapped in it. Monad is a type-level concept.
Also, notice that if you have m a->m b, you don't have a Monad. What Monad gives you, is a way to build a function m a->m b from a->m b (compare: Functor gives you a way to build a function m a->m b from a->b; ApplicativeFunctor gives you a way to build a function m a->m b from m (a->b))
If you already created a function which returns a wrapped value, why that function doesn't already take a wrapped value?
Because that function would have to unwrap its argument in order to do something with it.
But for many choices of m, you can only unwrap a value if you will eventually rewrap your own result. This idea of "unwrap, do something, then rewrap" is embodied in the (>>=) function which unwraps for you, let's you do something, and forces you to rewrap by the type a -> m b.
To understand why you cannot unwrap without eventually rewrapping, we can look at some examples:
If m a = Maybe a, unwrapping for Just x would be easy: just return x. But how can we unwrap Nothing? We cannot. But if we know that we will eventually rewrap, we can skip the "do something" step and return Nothing for the overall operation.
If m a = [a], unwrapping for [x] would be easy: just return x. But for unwrapping [], we need the same trick as for Maybe a. And what about unwrapping [x, y, z]? If we know that we will eventually rewrap, we can execute the "do something" three times, for x, y and z and concat the results into a single list.
If m a = IO a, no unwrapping is easy because we only know the result sometimes in the future, when we actually run the IO action. But if we know that we will eventually rewrap, we can store the "do something" inside the IO action and perform it later, when we execute the IO action.
I hope these examples make it clear that for many interesting choices of m, we can only implement unwrapping if we know that we are going to rewrap. The type of (>>=) allows precisely this assumption, so it is cleverly chosen to make things work.
While (>>=) can sometimes be useful when used directly, its main purpose is to implement the <- bind syntax in do notation. It has the type m a -> (a -> m b) -> m b mainly because, when used in a do notation block, the right hand side of the <- is of type m a, the left hand side "binds" an a to the given identifier and, when combined with remainder of the do block, is of type a -> m b, the resulting monadic action is of type m b, and this is the only type it possibly could have to make this work.
For example:
echo = do
input <- getLine
putStrLn input
The right hand side of the <- is of type IO String
The left hands side of the <- with the remainder of the do block are of type String -> IO (). Compare with the desugared version using >>=:
echo = getLine >>= (\input -> putStrLn input)
The left hand side of the >>= is of type IO String. The right hand side is of type String -> IO (). Now, by applying an eta reduction to the lambda we can instead get:
echo = getLine >>= putStrLn
which shows why >>= is sometimes used directly rather than as the "engine" that powers do notation along with >>.
I'd also like to provide what I think is an important correction to the concept of "unwrapping" a monadic value, which is that it doesn't happen. The Monad class does not provide a generic function of type Monad m => m a -> a. Some particular instances do but this is not a feature of monads in general. Monads, generally speaking, cannot be "unwrapped".
Remember that m >>= k = join (fmap k m) is a law that must be true for any monad. Any particular implementation of >>= must satisfy this law and so must be equivalent to this general implementation.
What this means is that what really happens is that the monadic "computation" a -> m b is "lifted" to become an m a -> m (m b) using fmap and then applied the m a, giving an m (m b); and then join :: m (m a) -> m a is used to squish the two ms together to yield a m b. So the a never gets "out" of the monad. The monad is never "unwrapped". This is an incorrect way to think about monads and I would strongly recommend that you not get in the habit.
I will focus on your point
If you already created a function which returns a wrapped value, why
that function doesn't already take a wrapped value?
and the IO monad. Suppose you had
getLine :: IO String
putStrLn :: IO String -> IO () -- "already takes a wrapped value"
how one could write a program which reads a line and print it twice? An attempt would be
let line = getLine
in putStrLn line >> putStrLn line
but equational reasoning dictates that this is equivalent to
putStrLn getLine >> putStrLn getLine
which reads two lines instead.
What we lack is a way to "unwrap" the getLine once, and use it twice. The same issue would apply to reading a line, printing "hello", and then printing a line:
let line = getLine in putStrLn "hello" >> putStrLn line
-- equivalent to
putStrLn "hello" >> putStrLn getLine
So, we also lack a way to specify "when to unwrap" the getLine. The bind >>= operator provides a way to do this.
A more advanced theoretical note
If you swap the arguments around the (>>=) bind operator becomes (=<<)
(=<<) :: (a -> m b) -> (m a -> m b)
which turns any function f taking an unwrapped value into a function g taking a wrapped
value. Such g is known as the Kleisli extension of f. The bind operator guarantees
such an extension always exists, and provides a convenient way to use it.
Because we like to be able to apply functions like a -> b to our m as. Lifting such a function to m a -> m b is trivial (liftM, liftA, >>= return ., fmap) but the opposite is not necessarily possible.
You want some typical examples? How about putStrLn :: String -> IO ()? It would make no sense for this function to have the type IO String -> IO () because the origin of the string doesn't matter.
Anyway: You might have the wrong idea because of your "wrapped value" metaphor; I use it myself quite often, but it has its limitations. There isn't necessarily a pure way to get an a out of an m a - for example, if you have a getLine :: IO String, there's not a great deal of interesting things you can do with it - you can put it in a list, chain it in a row and other neat things, but you can't get any useful information out of it because you can't look inside an IO action. What you can do is use >>= which gives you a way to use the result of the action.
Similar things apply to monads where the "wrapping" metaphor applies too; For example the point Maybe monad is to avoid manually wrapping and unwrapping values with and from Just all the time.
My two most common examples:
1) I have a series of functions that generate a list of lists, but I finally need a flat list:
f :: a -> [a]
fAppliedThrice :: [a] -> [a]
fAppliedThrice aList = concat (map f (concat (map f (concat (map f a)))))
fAppliedThrice' :: [a] -> [a]
fAppliedThrice' aList = aList >>= f >>= f >>= f
A practical example of using this was when my functions fetched attributes of a foreign key relationship. I could just chain them together to finally obtain a flat list of attributes. Eg: Product hasMany Review hasMany Tag type relationship, and I finally want a list of all the tag names for a product. (I added some template-haskell and got a very good generic attribute fetcher for my purposes).
2) Say you have a series of filter-like functions to apply to some data. And they return Maybe values.
case (val >>= filter >>= filter2 >>= filter3) of
Nothing -> putStrLn "Bad data"
Just x -> putStrLn "Good data"
In most of programming languages that support mutable variables, one can easily implement something like this Java example:
interface Accepter<T> {
void accept(T t);
}
<T> T getFromDoubleAccepter(Accepter<Accepter<T>> acc){
final List<T> l = new ArrayList<T>();
acc.accept(new Accepter<T>(){
#Override
public void accept(T t) {
l.add(t);
}
});
return l.get(0); //Not being called? Exception!
}
Just for those do not understand Java, the above code receives something can can be provided a function that takes one parameter, and it supposed to grape this parameter as the final result.
This is not like callCC: there is no control flow alternation. Only the inner function's parameter is concerned.
I think the equivalent type signature in Haskell should be
getFromDoubleAccepter :: (forall b. (a -> b) -> b) -> a
So, if someone can gives you a function (a -> b) -> b for a type of your choice, he MUST already have an a in hand. So your job is to give them a "callback", and than keep whatever they sends you in mind, once they returned to you, return that value to your caller.
But I have no idea how to implement this. There are several possible solutions I can think of. Although I don't know how each of them would work, I can rate and order them by prospected difficulties:
Cont or ContT monad. This I consider to be easiest.
RWS monad or similar.
Any other monads. Pure monads like Maybe I consider harder.
Use only standard pure functional features like lazy evaluation, pattern-matching, the fixed point contaminator, etc. This I consider the hardest (or even impossible).
I would like to see answers using any of the above techniques (and prefer harder ways).
Note: There should not be any modification of the type signature, and the solution should do the same thing that the Java code does.
UPDATE
Once I seen somebody commented out getFromDoubleAccepter f = f id I realize that I have made something wrong. Basically I use forall just to make the game easier but it looks like this twist makes it too easy. Actually, the above type signature forces the caller to pass back whatever we gave them, so if we choose a as b then that implementation gives the same expected result, but it is just... not expected.
Actually what came up to my mind is a type signature like:
getFromDoubleAccepter :: ((a -> ()) -> ()) -> a
And this time it is harder.
Another comment writer asks for reasoning. Let's look at a similar function
getFunctionFromAccepter :: (((a -> b) -> b) -> b) -> a -> b
This one have an naive solution:
getFunctionFromAccepter f = \a -> f $ \x -> x a
But in the following test code it fails on the third:
exeMain = do
print $ getFunctionFromAccepter (\f -> f (\x -> 10)) "Example 1" -- 10
print $ getFunctionFromAccepter (\f -> 20) "Example 2" -- 20
print $ getFunctionFromAccepter (\f -> 10 + f (\x -> 30)) "Example 3" --40, should be 30
In the failing case, we pass a function that returns 30, and we expect to get that function back. However the final result is in turn 40, so it fails. Are there any way to implement doing Just that thing I wanted?
If this can be done in Haskell there are a lot of interesting sequences. For example, tuples (or other "algebraic" types) can be defined as functions as well, since we can say something like type (a,b) = (a->b->())->() and implement fst and snd in term of this. And this, is the way I used in a couple of other languages that do not have native "tuple" support but features "closure".
The type of accept is void accept(T) so the equivalent Haskell type is t -> IO () (since every function in Java is essentially IO). Thus getFromDoubleAccepted can be directly translated as
import Data.IORef
type Accepter t = t -> IO ()
getFromDoubleAccepter :: Accepter (Accepter a) -> IO a
getFromDoubleAccepter acc = do
l <- newIORef $ error "Not called"
acc $ writeIORef l
readIORef l
If you want an idiomatic, non-IO solution in Haskell, you need to be more specific about what your actual end goal is besides trying to imitate some Java-pattern.
EDIT: regarding the update
getFromDoubleAccepter :: ((a -> ()) -> ()) -> a
I'm sorry, but this signature is in no way equal to the Java version. What you are saying is that for any a, given a function that takes a function that takes an a but doesn't return anything or do any kind of side effects, you want to somehow conjure up a value of type a. The only implementation that satisfies the given signature is essentially:
getFromDoubleAccepter :: ((a -> ()) -> ()) -> a
getFromDoubleAccepter f = getFromDoubleAccepter f
First, I'll transliterate as much as I can. I'm going to lift these computations to a monad because accept returns void (read () in Haskell-land), which is useless unless there is some effect.
type Accepter m t = t -> m ()
getFromDoubleAccepter :: (MonadSomething m) => Accepter m (Accepter m t) -> m t
getFromDoubleAccepter acc = do
l <- {- new mutable list -}
acc $ \t -> add l t
return (head l)
Of course, we can't make a mutable list like that, so we'll have to use some intuitive sparks here. When an action just adds an element to some accumulator, I think of the Writer monad. So maybe that line should be:
acc $ \t -> tell [t]
Since you are simply returning the head of the list at the end, which doesn't have any effects, I think the signature should become:
getFromDoubleAccepter :: Accepter M (Accepter M t) -> t
where M is an appropriate monad. It needs to be able to write [t]s, so that gives us:
type M t = Writer [t]
getFromDoubleAccepter :: Accepter (M t) (Accepter (M t) t) -> t
And now the type of this function informs us how to write the rest of it:
getFromDoubleAccepter acc =
head . execWriter . acc $ \t -> tell [t]
We can check that it does something...
ghci> getFromDoubleAccepter $ \acc -> acc 42
42
So that seems right, I guess. I'm still a bit unclear on what this code is supposed to mean.
The explicit M t in the type signature is a bit aesthetically bothersome to me. If I knew what problem I was solving I would look at that carefully. If you mean that the argument can be a sequence of commands, but otherwise has no computational features available, then you could specialize the type signature to:
getFromDoubleAccepter :: (forall m. (Monad m) => Accepter m (Accepter m t)) -> t
which still works with our example. Of course, this is all a bit silly. Consider
forall m. (Monad m) => Accepter m (Accepter m t))
= forall m. (Monad m) => (t -> m ()) -> m ()
The only thing a function with this type can do is call its argument with various ts in order and then return (). The information in such a function is completely characterized[1] by those ts, so we could just as easily have used
getFromDoubleAccepter :: [t] -> t
getFromDoubleAccepter = head
[1] As long as I'm going on about nothing, I might as well say that that is not quite accurate in the face of infinity. The computation
crazy :: Integer -> Accepter m (Accepter m Integer)
crazy n acc = crazy (n+1) >> acc n
can be used to form the infinite sequence
... >> acc 3 >> acc 2 >> acc 1 >> acc 0
which has no first element. If we tried to interpret this as a list, we would get an infinite loop when trying to find the first element. However this computation has more information than an infinite loop -- if instead of a list, we used the Last monoid to interpret it, we would be able to extract 0 off the end. So really
forall m. (Monad m) => Accepter m (Accepter m t)
is isomorphic to something slightly more general than a list; specifically a free monoid.
Thanks to the above answers, I finally concluded that in Haskell we can do some different things than other languages.
Actually, the motivation of this post is to translate the famous "single axiom classical logic reduction system". I have implemented this in some other languages. It should be no problem to implement the
Axiom: (a|(b|c)) | ((d|(d|d)) | ((e|b) | ((a|e) | (a|e))))
However, since the reduction rule looks like
Rule: a|(b|c), a |-- c
It is necessary to extract the inner parameter as the final result. In other languages, this is done by using side-effects like mutable slots. However, in Haskell we do not have mutable slots and involving IO will be ugly so I keep looking for solutions.
In the first glance (as show in my question), the getFromDoubleAccepter f = f id seems nonsense, but I realise that it actually work in this case! For example:
rule :: (forall r.a -> (b -> c -> r) -> r) -> a -> c
rule abc a = abc a $ flip const
The trick is still the same: since the existential qualification hides r from the caller, and it is up to the callee to pick up a type for it, we can specify c to be r, so we simply apply the given function to get the result. On the other hand, the given function has to use our input to produce the final answer, so it effectively limiting the implementation to what we exactally want!
Putting them together, let's see what we can do with it:
newtype I r a b = I { runI :: a -> b -> r }
rule :: (forall r. I r a (I r b c)) -> a -> c
rule (I abc) a = abc a (I (\b c -> c))
axiom :: I r0 (I r1 a (I r2 b c))
(I r0 (I r3 d (I r3 d d))
(I r4 (I r2 e b) (I r4 (I r1 a e) (I r1 a e))))
axiom = let
a1 (I eb) e = I $ \b c -> eb e b
a2 = I $ \d (I dd) -> dd d d
a3 (I abc) eb = I $ \a e -> abc a (a1 eb e)
a4 abc = I $ \eb aeae -> runI a2 (a3 abc eb) aeae
in I $ \abc (I dddebaeae) -> dddebaeae a2 (a4 abc)
Here I use a naming convention to trace the type signatures: a variable name is combinded by the "effective" type varialbes (means it is not result type - all r* type variable).
I wouldn't repeat the prove represented in the sited essay, but I want to show something. In the above definition of axiom we use some let bindings variables to construct the result. Not surprisingly, those variables themselves can be extracted by using rule and axiom. let's see how:
--Equal to a4
t4 :: I r0 a (I r1 b c) -> I r2 (I r1 d b) (I r2 (I r0 a d) (I r0 a d))
t4 abc = rule axiom abc
--Equal to a3
t3 :: I r0 a (I r1 b c) -> I r1 d b -> I r0 a d
t3 abc eb = rule (t4 abc) eb
--Equal to a2
t2 :: I r a (I r a a)
t2 = rule (t3 axiom (t3 (t4 axiom) axiom)) axiom
--Equal to a1
t1 :: I r a b -> a -> I r b c
t1 ab a = rule (t3 t2 (t3 (t3 t2 t2) ab)) a
One thing left to be proved is that we can use t1 to t4 only to prove all tautologies. I feel it is the case but have not yet proved it.
Compare to other languages, the Haskell salutation seems more effective and expressive.
Since the following do block:
do
x <- foo
y <- bar
return x + y
is desugared to the following form:
foo >>= (\x -> bar >>= (\y -> return x + y))
aren't \x -> ... and y -> ... actually continuations here?
I was wondering if there is a way to capture the continuations in the definition of bind, but I can't get the types right. I.e:
data Pause a = Pause a | Stop
instance Monad Pause where
return x = Stop
m >>= k = Pause k -- this doesn't work of course
Now I tried muddling around with the types:
data Pause a = Pause a (a -> Pause ???) | Stop
------- k ------
But this doesn't work either. Is there no way to capture these implicit continuations?
Btw, I know about the Cont monad, I'm just experimenting and trying out stuff.
OK I'm not really sure, but let me put a few thoughts. I'm not quite sure what it
ought to mean to capture the continuation. You could, for example, capture the whole
do block in a structure:
{-# LANGUAGE ExistentialQuantification #-}
import Control.Monad
data MonadExp b = Return b | forall a. Bind (MonadExp a) (a -> MonadExp b)
instance Monad MonadExp where
return x = Return x
f >>= g = Bind f g
For example:
block :: MonadExp Int
block = do
x <- return 1
y <- return 2
return $ x + y
instance Show (MonadExp a) where
show (Return _) = "Return _"
show (Bind _ _) = "Bind _ _"
print block
>> Bind _ _
And then evaluate the whole thing:
finish :: MonadExp a -> a
finish (Return x) = x
finish (Bind f g) = finish $ g (finish f)
print $ finish block
>> 3
Or step through it and see the parts
step :: MonadExp a -> MonadExp a
step (Return _) = error "At the end"
step (Bind f g) = g $ finish f
print $ step block
>> Bind _ _
print $ step $ step block
>> Return _
Well, now that I think about it more, that's probably not what you're asking. But
maybe it'll help you think.
Well, I don't know if your lambdas are continuations in the strictest sense of the term, but they also look similar to this concept in my eye.
But note that if they are continuations, then the desugared monadic code is already written in continuation passing style (CPS). The usual notion of a control operator that "captures" a continuation is based on direct-style programs; the "captured" continuation is only implicit in the direct-style program, but the CPS transformation makes it explicit.
Since the desugared monadic code is already in CPS or something like it, well, maybe a way to frame your question is whether monadic code can express some of the control flow tricks that CPS code can. Typically, those tricks boil down to the idea that while under the CPS regime it is conventional for a function to finish by invoking its continuation, a function can choose to replace its continuation with another of its choosing. This replacement continuation can be constructed with a reference to the original continuation, so that it can in turn "restore" that original one if it chooses. So for example, coroutines are implemented as a mutual "replace/restore" cycle.
And looked at in this light, I think your answer is mostly no; CPS requires that in foo >>= bar, foo must be able to choose whether bar will be invoked at all, and foo must be abble to supply a substitute for bar, but (>>=) by itself does not offer a mechanism for foo to do this, and more importantly, (>>=) is in control of the execution flow, not foo. Some specific monads implement parts or all of it (for example, the Maybe monad allows foo to forego execution of bar by producing a Nothing result), but others don't.
Closest I could get is to forego (>>=) and use this instead:
-- | Execute action #foo# with its "continuation" #bar#.
callCC :: Monad m => ((a -> m b) -> m b) -> (a -> m b) -> m b
foo `callCC` bar = foo bar
Here foo can choose whether bar will be used at all. But notice that this callCC is really just ($)!
Continuing quest to make sense of ContT and friends. Please consider the (absurd but illustrative) code below:
v :: IO (Either String [String])
v = return $ Left "Error message"
doit :: IO (Either String ())
doit = (flip runContT return) $ callCC $ \k -> do
x <- liftIO $ v
x2 <- either (k . Left) return x
when True $ k (Left "Error message 2")
-- k (Left "Error message 3")
return $ Right () -- success
This code does not compile. However, if the replace the when with the commented k call below it, it compiles. What's going on?
Alternatively, if I comment out the x2 line, it also compiles. ???
Obviously, this is a distilled version of the original code and so all of the elements serve a purpose. Appreciate explanatory help on what's going on and how to fix it. Thanks.
The problem here has to do with the types of when and either, not anything particular to ContT:
when :: forall (m :: * -> *). (Monad m) => Bool -> m () -> m ()
either :: forall a c b. (a -> c) -> (b -> c) -> Either a b -> c
The second argument needs to be of type m () for some monad m. The when line of your code could thus be amended like so:
when True $ k (Left "Error message 2") >> return ()
to make the code compile. This is probably not what you want to do, but it gives us a hint as to what might be wrong: k's type has been inferred to be something unpalatable to when.
Now for the either signature: notice that the two arguments to either must be functions which produce results of the same type. The type of return here is determined by the type of x, which is in turn fixed by the explicit signature on v. Thus the (k . Left) bit must have the same type; this in turn fixes the type of k at (GHC-determined)
k :: Either String () -> ContT (Either String ()) IO [String]
This is incompatible with when's expectations.
When you comment out the x2 line, however, its effect on the type checker's view of the code is removed, so k is no longer forced into an inconvenient type and is free to assume the type
k :: Either [Char] () -> ContT (Either [Char] ()) IO ()
which is fine in when's book. Thus, the code compiles.
As a final note, I used GHCi's breakpoints facility to obtain the exact types of k under the two scenarios -- I'm nowhere near expert enough to write them out by hand and be in any way assured of their correctness. :-) Use :break ModuleName line-number column-number to try it out.