为什么Tuple没有Monad实例? [英] Why does Tuple not have a Monad instance?

问题描述

我注意到的一件事是Tuple没有Monad实例.

One thing I noticed was that Tuple does not have a Monad instance.

元组确实具有一个Applicative实例:

Tuple does however have an Applicative instance:

instance Monoid a => Applicative ((,) a)

已经非常严格地限制了我们使Monad实例成为什么样子.

Which already extremely heavily restricts what we can make the Monad instance be.

让我们看看我们将获得的连接类型签名:

Lets look at the type signature we would get for join:

instance Monoid a => Monad ((,) a)

join :: Monad m => m (m a) -> m a

join :: Monoid a => (a, (a, b)) -> (a, b)

我们还可以查看Monad法则:

We can also look at the Monad laws:

join $ f <$> pure x == f x
join $ f <$> (mempty, x) == f x
join (mempty, f x) == f x
join (mempty, (a, b)) == (a, b)

join $ pure <$> x = x
join $ pure <$> (a, b) = (a, b)
join (a, (mempty, b)) = (a, b)

至此，我们知道以任何一种方式组合mempty和x都会导致x.我们唯一有关x的类型信息是它是Monoid.因此，基本上仅有的两种实现是:

At this point we know that combining mempty and x in either way results in x. And the only type information we have about x is that it is a Monoid. So basically the only two implementations are:

join (a, (b, x)) = (a <> b, x)

和:

join (a, (b, x)) = (b <> a, x)

其中第二个使ap和<*>不同.

所以现在我们知道((,) a)唯一有效的Monad实例是:

So now we know that the only valid Monad instance for ((,) a) is:

instance Monoid a => Monad ((,) a) where
    (a, c) >>= f = let (b, c') = f c in (a <> b, c')

那为什么现在不是这种情况呢?

So why is that not currently the case?

推荐答案

这个问题的答案似乎是我只需要使用ghc-8.0.1，它确实为Tuple提供了一个Monad实例.感谢@Michael指出了这一点.

Well it seems like the answer to this question is that I just have to use ghc-8.0.1, which does give Tuple a Monad instance. Credit to @Michael for pointing this out.

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