I have a program like bellow.

incVal :: Num a => a -> a
incVal x = x   1

incVal' :: (Monad m, Num a) => a -> m a
incVal' x = return (x   1)

incVal'' :: (Monad m, Num a) => a -> a -> m a
incVal'' x y = return (x   y)

incVal''' :: (Monad m, Num a) => a -> a -> a -> m a
incVal''' x y z = return (x   y   z)

main = do
    print(Just 9 >>= incVal')

I can Invoke incVal' with >>= (as shown in the above code). But I am not understanding how to invoke incVal'' and incVal''' with >>=.

CodePudding user response：

You can use Applicative in combination with join.

First, you can apply the function to the args using <$> and <*>:

print (incVal'' <$> Just 5 <*> Just 37)

But this is not quite enough, because the result of this will be Maybe (Maybe Int), due to how <*> works, so the line above would print "Just (Just 42)".

So as a last step, to "unnest" the value, you can use the join function, which is a standard Monad operation:

print (join (incVal'' <$> Just 5 <*> Just 37))

And, of course, the same approach works for incVal''':

print (join (incVal''' <$> Just 1 <*> Just 4 <*> Just 37))

Incidentally, even incVal' call can be expressed in the same terms, because a combination of <$> (aka fmap) and join is equivalent to >>=:

print (join (incVal' <$> Just 42))