Haskell 中的可变参数绑定
以下代码尝试编写一个如下所示的可变参数函数：
- bind_variadic mx f = mx >>= f
- bind_variadic mx my f = do { x <- mx; y <- 我的； 如果将“其余绑定”表示为变量
k，我可以编写它，但为了编写
类型类，我需要根据另一个函数编写一个函数。准确地说，我想用l0来表达l1，用l1来表达l2，等等。
import Prelude hiding ((>>=), (>>), Monad, return)
-- override the default monad so we don't get confusing
-- instances like "Monad (->)".
class Monad m where
(>>=) :: m a -> (a -> m b) -> m b
(>>) :: m a -> m b -> m b
return :: a -> m a
fail :: String -> m a
h :: Monad m => m a -> (t -> m b) -> (a -> t) -> m b
h mx k f = mx >>= \x -> k (f x)
l0 = h (return 3) id (\x -> return x)
l1 = h (return 3) (h (return 4) id) (\x y -> return x)
l2 = h (return 3) (h (return 4) (h (return 5) id)) (\x y z -> return x)
也许解决方案涉及另一个延续？
编辑
这里有一个需要额外加入的想法...
-- if not using Control.Monad, use this
join :: Monad
The following code is an attempt to write a variadic function that acts like this:
bind_variadic mx f = mx >>= fbind_variadic mx my f = do { x <- mx; y <- my; f x y }
I can write it if one expresses the "rest of binding" as a variable k, but in order to write a typeclass I need to write one function in terms of the other. To be precise, I want to express l1 in terms of l0, l2 in terms of l1, etc.
import Prelude hiding ((>>=), (>>), Monad, return)
-- override the default monad so we don't get confusing
-- instances like "Monad (->)".
class Monad m where
(>>=) :: m a -> (a -> m b) -> m b
(>>) :: m a -> m b -> m b
return :: a -> m a
fail :: String -> m a
h :: Monad m => m a -> (t -> m b) -> (a -> t) -> m b
h mx k f = mx >>= \x -> k (f x)
l0 = h (return 3) id (\x -> return x)
l1 = h (return 3) (h (return 4) id) (\x y -> return x)
l2 = h (return 3) (h (return 4) (h (return 5) id)) (\x y z -> return x)
Perhaps the solution involves another continuation?
edit
here's an idea that requires an additional join...
-- if not using Control.Monad, use this
join :: Monad ???? => ???? (???? α) -> ???? α
join mx = mx >>= id
-- idea: get side effects of evaluating first arguments first
h' mz k f = k f >>= \f' -> mz >>= (return . f')
l1' = h' (return 3) return
unary = join (l1' (\x -> return x))
l2' = h' (return 4) l1'
binary = join (l2' (\x y -> return x))
l3' = h' (return 5) l2'
ternary = join (l3' (\x y z -> return x))
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如果你想表达这一点：
我会使用
Control.Applicative代替。然后：我认为这比任何多元解决方案都更好（更简单，更易于维护）。
If you want to express this:
I would use
Control.Applicativeinstead. Then:I think this is better (simpler, more maintainable) than any polyvariadic solution would be.