Trivial parsec示例产生一个类型错误 [英] Trivial parsec example produces a type error
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问题描述
我试图让这个简单的parsec代码编译
I'm trying to get this trivial parsec code to compile
import Text.Parsec
simple = letter
但我不断收到此错误
No instance for (Stream s0 m0 Char)
arising from a use of `letter'
Possible fix: add an instance declaration for (Stream s0 m0 Char)
In the expression: letter
In an equation for `simple': simple = letter
推荐答案
我认为你已经违反了单态限制。这个限制意味着：如果一个变量没有明确的参数声明，它的类型必须是单形的。这强制类型检查器选择 Stream 的特定实例，但它无法做出决定。
I think you have ran against the monomorphism restriction. This restriction means: If a variable is declared with no explicit arguments, its type has to be monomorphic. This forces the typechecker to pick a particular instance of Stream, but it can't decide.
有两个打击它的方法：
$ b
There are two ways to fight it:
-
为
简单显式签名：
simple :: Stream s m Char => ParsecT s u m Char
simple = letter
禁用单态限制：
Disable the monorphism restriction:
{-# LANGUAGE NoMonomorphismRestriction #-}
import Text.Parsec
simple = letter
请参阅什么是单态限制？有关单态限制的更多信息。
See What is the monomorphism restriction? for more information on the monomorphism restriction.
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