Haskell中MonadPlus的默认类型评估是什么? [英] What is the default type evaluation of MonadPlus in Haskell?

问题描述

我有以下代码:

import Control.Monad

coin :: MonadPlus m => m Int
coin = return 0 `mplus` return 1

如果我在解释器上求值coin :: Maybe Int，则将其插入Just 0.这是正常的，因为可能将Maybe作为MonadPlus的实例来实现.

If I evaluate coin :: Maybe Int on the interpreter, it prits Just 0. That's normal because of the implementation of Maybe as instance of MonadPlus.

如果我在解释器上求值coin :: [Int]，它会显示[0, 1]，因为列表中mplus的实现是append.

If I evaluate coin :: [Int]on the interpreter, it prints [0, 1], because the implementation of mplus on list is an append.

但是，如果我评估coin，而没有任何类型修饰符，它将显示0.为什么?解释器将'c6>转换为coin的哪种类型?

But if I evaluate coin, without any type decorators, it prints 0. Why? What type does the interpreter 'converts' coin to evaluate it?

此代码摘自: http://homes.sice.indiana. edu/ccshan/rational/S0956796811000189a.pdf

推荐答案

是的，这不是ghci记录得非常清楚的角落.在ghci中输入表达式时，它会使用表达式的类型来决定要执行的操作:

Yeah, this is a not super-well documented corner of ghci. When you enter an expression into ghci, it uses the expression's type to decide what to do:

1. IO ():运行操作，什么都不做.
2. Show a => IO a:运行操作并print结果.
3. 其他任何IO a:执行操作，再无其他操作.
4. 其他:用print包装整个表达式.

1. IO (): Run the action, do nothing further.
2. Show a => IO a: Run the action and print the result.
3. any other IO a: Run the action, do nothing further.
4. anything else: wrap the whole expression with print.

它如何确定事物具有哪种类型?简单:它试图统一您的表达式的类型，而上述每个签名又依次解决所有产生的约束. (对于cognoscenti:这是扩展默认规则的 ！这解释了为什么它似乎是默认m的原因，即使标准默认规则和扩展默认规则都没有说明默认值是什么使用.)

How does it decide which of these types a thing has? Easy: it tries to unify the type of your expression which each of the above signatures in turn and solve all resulting constraints. (For the cognoscenti: this is in addition to the extended defaulting rules! This explains why it appears to be defaulting the m, even though neither the standard defaulting rules nor the extended defaulting rules say what default to use.)

因此，由于表达式不与IO ()统一，但与Show a => IO a统一，因此ghci在统一期间找到m ~ IO(和a ~ Int)，发现存在MonadPlus IO(和)实例来解决约束，运行操作并打印结果.

So, since your expression does not unify with IO () but does unify with Show a => IO a, ghci finds m ~ IO (and a ~ Int) during unification, discovers that there is a MonadPlus IO (and a Show Int) instance to resolve the constraints, runs your action, and prints the result.

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