为什么联接是独立的，而不是Monad类型类的最小实现的一部分? [英] Why is join standalone, instead of part of the minimal implementation of the Monad typeclass?

问题描述

我宁愿通过join函数而不是>>=来定义Monad的实例...从这种等效性开始:

I would prefer defining an instance of Monad via the join function instead of >>= ... Starting from that equivalence:

x >>= f = join (fmap f x)

您只能通过联接定义Monad实例吗?

Could you define a Monad instance only with join?

我希望加入{-# MINIMAL (>>=)| join #-}

为什么join在顶层而不是Monad类型类中?

Why is join at the top level and not even in the Monad typeclass?

推荐答案

遗憾的是，由于与广义新类型派生和角色系统相关的技术限制，join不在GHC标准库中的Monad类型类中.长话短说，考虑到某些新类型newtype T m a = MkT (m a)，GHC不够聪明，无法弄清楚如何证明m (m a)和m (T m a)之间的表示相等性，这对于证明join的第一个参数(这是证明这是必不可少的)的类型为m (m a) -> m a).

Sadly, join is not a part of the Monad typeclass in GHC’s standard library because of technical restrictions related to generalized newtype deriving and the roles system. Long story short, given some newtype newtype T m a = MkT (m a), GHC is not smart enough to figure out how to prove representational equality between m (m a) and m (T m a), which is necessary for proving representational equality for the first argument of join (which has type m (m a) -> m a).

幸运的是，最近对GHC Haskell的扩展QuantifiedConstraints可能使角色系统足够智能以支持此功能.有关问题及其潜在解决方案的更详细处理，请参阅Ryan Scott的博客文章

Fortunately, a recent extension to GHC Haskell, QuantifiedConstraints, might make it possible to make the roles system smart enough to support this. For a more detailed treatment of both the problem and its potential solution, see Ryan Scott’s blog post, How QuantifiedConstraints can let us put join back in Monad.

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