fmap与bind的关系

Question

查阅 Control.Monad 文档后，我对这段话：

The above laws imply:

fmap f xs = xs >>= return . f

他们是怎么暗示的？

Answer 1

Control.Applicative 说

As a consequence of these laws, the Functor instance for f will satisfy
fmap f x = pure f <*> x

Applicative和Monad的关系说

pure = return

(<*>) = ap

ap 说

return f `ap` x1 `ap` ... `ap` xn

is equivalent to

liftMn f x1 x2 ... xn

因此

fmap f x = pure f <*> x
         = return f `ap` x
         = liftM f x
         = do { v <- x; return (f v) }
         = x >>= return . f

Answer 2

Functor instances are unique，如果 F 是一个 Functor 并且你有一个函数 foobar :: (a -> b) -> F a -> F b 使得 foobar id = id（也就是说，它遵循第一函子定律）然后 foobar = fmap。现在，考虑这个函数：

liftM :: Monad f => (a -> b) -> f a -> f b
liftM f xs = xs >>= return . f

那么liftM id xs是什么？

liftM id xs
xs >>= return . id
-- id does nothing, so...
xs >>= return
-- By the second monad law...
xs

liftM id xs = xs；即 liftM id = id。因此，liftM = fmap；或者，换句话说...

fmap f xs  =  xs >>= return . f

，通过Applicative定律，也是得出这个结论的有效途径。

fmap与bind的关系

Relationship between fmap and bind

monads

haskell

functor