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I am having difficulty understanding how a function can be a monad. Function (->) r is a monad according to a declaration in Control.Monad.Instances : instance Monad ((->) r) where return x = \_ -> x h >>= f = \w -> f (h w) w Even what Miran Lipovača says about it makes me confused: The implementation for >>= seems a bit cryptic, but it's really not all that. When we use >>= to feed a monadic value to a function, the result is always a monadic value. So in this case, when we feed a function to another function, the result is a function as well. That's why the result starts off as a lambda. All of the implementations of >>= so far always somehow isolated the result from the monadic value and then applied the function f to that result. The same thing happens here. To get the result from a function, we have to apply it to something, which is why we do (h w) here to get the result from the function and then we apply f to that. f returns a monadic value, which is a function in our case, so we apply it to w as well. The type signature of (>>=) is this: (>>=) :: m a -> (a -> m b) -> m b So I take that h is typed as m a and f as (a -> m b) . If a function is m a , does it return an a type value? or does it return something else taking an a type? If the non-monad value of h is fed to f , then we get: f (h w) Looks fine. Since f is a function and has taken its sole argument, it is already a value, no? Since it's a monadic function the value is also a monadic value. Why then does it need another value w</cod

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