问题
I have some haskell code Im trying to work my way thourgh but I dont have understand what is going in it.
type Bag a = a -> Int
emptyB :: Bag a
emptyB = \e -> 0
countB :: Eq a => Bag a -> a -> Int
countB b e = b e
I understand that the Bag type is a function that takes in a generic object and returns a Int and countB is basically a wrapper for Bag that gets the number of generic objects in that Bag. But I dont really understand anything past that. How do I modify whats in the bag? Or the bag itself? From what I figure adding to the bag would be something like
addB :: Eq a => Bag a -> a -> Bag a
addB bag num = bag (num+bag)
But this returns a int, when the add function requires a bag be returned. Can anyone explain to me how this works?
回答1:
Terms and Discussion
type Bag a = a -> Int
Here Bag is not an object. It is just a type - an alias for a -> Int. If you have a value of type a it will compute and return a value of type Int. That's it. There is no Bag, no structure to which you can add things. It would be better to not even call this a Bag.
emptyB :: Bag a
emptyB = \e -> 0
A function from any type to the constant number zero.
countB :: Eq a => Bag a -> a -> Int
countB b e = b e
In short, this is just function application. Apply the function named b to the input e.
Rewriting for fun and learning
I appreciate that you can use functions to imitate structures - it's a common programming language class assignment. You can take a Bag a and another Bag a then union them, such as returning a new countB by adding the counts of the two individual bags - cool.
... but this seems too much. Before moving on with your assignment (did I guess that right?) you should probably become slightly more comfortable with the basics.
It might be easier if you rewrite the functions without the type alias:
emptyB :: a -> Int
emptyB = \e -> 0
-- or: emptyB e = 0
-- or: emptyB _ = 0
-- or: emptyB = const 0
Bag or no bag, it's just a function.
countB :: Eq a => (a -> Int) -> a -> Int
countB b e = b e
A function that takes an a and produces an Int can... be given a value (the variable e is of type a) and produce an Int.
来源：https://stackoverflow.com/questions/54995025/dont-understand-whats-on-in-this-haskell-code