Alternative: you can fool many people some time, and some people many time, but...

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Alternative: you can fool many people some time, and some people many time, but...

Corentin Dupont

Hi guys,
I'm playing with the mysterious "some" and "many" from Control.Applicative.
If I try:

many $ Just 1

It just loops, I understand why:
http://stackoverflow.com/questions/18108608/what-are-alternatives-some-and-many-useful-for
It seems that some and many are usually used in a context where something is consumed, and can be depleted, so the loop ends.

But why doesn't this terminates?

take 3 $ many $ Just 1

It's a recursive call, but the construction of the result should be lazy...


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Re: Alternative: you can fool many people some time, and some people many time, but...

Jake
take 3 $ many $ Just 1

doesn't type check. Did you mean this?

take 3 <$> (many $ Just 1)
I think this may have something to do with the default definition of many in the definition of Alternative:
many :: f a -> f [a]
many v = many_v
  where
    many_v = some_v <|> pure []
    some_v = (fmap (:) v) <*> many_v
many_v and some_v are mutually recursive functions, and it may be that this prevents the thunks from being made available to take in some way. I'm really not sure though, this is just an idea about why this is not quite the same as (take $ repeat 1)

On Thu, Sep 29, 2016 at 3:51 PM Corentin Dupont <[hidden email]> wrote:
Hi guys,
I'm playing with the mysterious "some" and "many" from Control.Applicative.
If I try:

many $ Just 1

It just loops, I understand why:
http://stackoverflow.com/questions/18108608/what-are-alternatives-some-and-many-useful-for
It seems that some and many are usually used in a context where something is consumed, and can be depleted, so the loop ends.

But why doesn't this terminates?

take 3 $ many $ Just 1

It's a recursive call, but the construction of the result should be lazy...

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Re: Alternative: you can fool many people some time, and some people many time, but...

David Menendez-2
On Thu, Sep 29, 2016 at 4:28 PM, Jake <[hidden email]> wrote:
I think this may have something to do with the default definition of many in the definition of Alternative:
many :: f a -> f [a]
many v = many_v
  where
    many_v = some_v <|> pure []
    some_v = (fmap (:) v) <*> many_v
many_v and some_v are mutually recursive functions, and it may be that this prevents the thunks from being made available to take in some way. I'm really not sure though, this is just an idea about why this is not quite the same as (take $ repeat 1)

The problem is that many is creating an infinite sum that’s nested to the left. So you’re trying to compute

(((… <|> Just [1,1,1]) <|> Just [1,1]) <|> Just [1]) <|> Just 1

which will never terminate because Maybe is strict in the first argument to <|>.


As a practical matter, the Alternative instance for Maybe should probably be changed to either call error or return Just (repeat v).

Similarly, we should probably flip the order for many in the instance for [].

--

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Re: Alternative: you can fool many people some time, and some people many time, but...

S D Swierstra-2
In reply to this post by Jake
The type of the last part of the expression is:

many $ Just 1 :: Num a => Maybe [a]

So in order to be able to return the “Just” constructor which inspected by the application of (take 3 <$>) we have somehow to know for sure that all the <*> executions will indeed see a “Just” in both of their arguments. This forces more and more evaluations.
  
 Doaitse


Op 29 sep. 2016, om 22:28  heeft Jake <[hidden email]> het volgende geschreven:

take 3 $ many $ Just 1

doesn't type check. Did you mean this?

take 3 <$> (many $ Just 1)
I think this may have something to do with the default definition of many in the definition of Alternative:
many :: f a -> f [a]
many v = many_v
  where
    many_v = some_v <|> pure []
    some_v = (fmap (:) v) <*> many_v
many_v and some_v are mutually recursive functions, and it may be that this prevents the thunks from being made available to take in some way. I'm really not sure though, this is just an idea about why this is not quite the same as (take $ repeat 1)

On Thu, Sep 29, 2016 at 3:51 PM Corentin Dupont <[hidden email]> wrote:
Hi guys,
I'm playing with the mysterious "some" and "many" from Control.Applicative.
If I try:

many $ Just 1

It just loops, I understand why:
http://stackoverflow.com/questions/18108608/what-are-alternatives-some-and-many-useful-for
It seems that some and many are usually used in a context where something is consumed, and can be depleted, so the loop ends.

But why doesn't this terminates?

take 3 $ many $ Just 1

It's a recursive call, but the construction of the result should be lazy...

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Re: Alternative: you can fool many people some time, and some people many time, but...

Corentin Dupont
Thanks for the replies...
What I'm trying to do is a simple input system:

query :: IO (Maybe String)
query = do
  putStrLn "Enter text or press q:"
  r <- getLine
  return $ if r == "q" then Nothing else Just r


This will ask an input to the user which is returned, unless "q" is pressed.
I want to repeat this query "some" or "many" times:

main = do
  qs <- some $ query
  --
qs <- many $ query
  putStrLn qs

This should ask the query multiple times until "q" is pressed.
The type of qs is Maybe [String].
The expected result is that with "some", returning zero results will not be permitted, while with "many" it is.

Probably I should defined a newtype for IO Maybe:
data IOMaybe a = IOMaybe {getIOMaybe :: IO (Maybe a)}

And define all the instances.

Or use Data.Fucntor.Compose:
type IOMaybe = Compose IO Maybe



On Thu, Sep 29, 2016 at 11:15 PM, Doaitse Swierstra <[hidden email]> wrote:
The type of the last part of the expression is:

many $ Just 1 :: Num a => Maybe [a]

So in order to be able to return the “Just” constructor which inspected by the application of (take 3 <$>) we have somehow to know for sure that all the <*> executions will indeed see a “Just” in both of their arguments. This forces more and more evaluations.
  
 Doaitse


Op 29 sep. 2016, om 22:28  heeft Jake <[hidden email]> het volgende geschreven:

take 3 $ many $ Just 1

doesn't type check. Did you mean this?

take 3 <$> (many $ Just 1)
I think this may have something to do with the default definition of many in the definition of Alternative:
many :: f a -> f [a]
many v = many_v
  where
    many_v = some_v <|> pure []
    some_v = (fmap (:) v) <*> many_v
many_v and some_v are mutually recursive functions, and it may be that this prevents the thunks from being made available to take in some way. I'm really not sure though, this is just an idea about why this is not quite the same as (take $ repeat 1)

On Thu, Sep 29, 2016 at 3:51 PM Corentin Dupont <[hidden email]> wrote:
Hi guys,
I'm playing with the mysterious "some" and "many" from Control.Applicative.
If I try:

many $ Just 1

It just loops, I understand why:
http://stackoverflow.com/questions/18108608/what-are-alternatives-some-and-many-useful-for
It seems that some and many are usually used in a context where something is consumed, and can be depleted, so the loop ends.

But why doesn't this terminates?

take 3 $ many $ Just 1

It's a recursive call, but the construction of the result should be lazy...

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