haskell - 理解 Writer Monad 的示例

Question

我正在通过《Learn You A Haskell》一书了解 Writer Monad。

这是一段代码:

import Control.Monad.Writer

logNumber :: Int -> Writer [String] Int
logNumber num = writer (num, ["Got number: " ++ show num])

multWithLog :: Writer [String] Int
multWithLog = do
  a <- logNumber 3
  b <- logNumber 5
  return (a * b)

运行multWithLog时，结果如下:

*Main> runWriter multWithLog
(15,["Got number: 3","Got number: 5"])

在这一行:

a <- logNumber 3
b <- logNumber 5

很容易看出a = 3和b = 5，因为它们都在return函数上相乘。

我不明白的是为什么这些值为 3 和 5。 a 和 b 不应该是 Writer Monad 中包含的值吗？在这种情况下元组？

例如，对于Maybe Monad，a和b将是3和5:

do
  a <- Just 3
  b <- Just 5
  return (a * b)

在这种情况下，这对我来说是有意义的，因为 a 和 b 接收 Just 内的内容。但在最初的示例中，a 和 b 仅接收部分值。

Answer 1

It is easy to see that a = 3 and b = 5, since both of them are being multiplied on the return function. What I don't understand is why those values are 3 and 5. Shouldn't a and b be the values that contain inside the Writer Monad? In this case tuples?

没有。我认为回答这个问题的最简单方法就是实现 Writer输入并研究其 Monad类实例:

newtype Writer w a = Writer { runWriter :: (a, w) }

instance Functor (Writer w) where
  fmap f (Writer (a, w)) = Writer (f a, w)

instance Monoid w => Applicative (Writer w) where
  pure a = Writer (a, mempty)
  Writer (f, w) <*> Writer (a, w') = Writer (f a, w <> w')

instance Monoid w => Monad (Writer w) where
  return = pure
  Writer (a, w) >>= f = let (b, w') = runWriter (f a)
                        in Writer (b, w <> w')

tell :: w -> Writer w ()
tell w = Writer ((), w)

正如您在 >>= 的实例方法中看到的那样，函数f应用于a值，而不是整个元组。语法a <- logNumber 3使用 >>= 脱糖，因此绑定(bind)到 a 的值将是 Writer 的元组的第一个元素环绕。