为什么解析器组合器& quot; seq&"会出现错误.用&" bind&"定义和“返回"? [英] Why is parser combinator "seq" defined with "bind" and "return"?

问题描述

我正在阅读有关解析器组合器的文章，但没有了解以下内容:

I am reading this article about parser combinators and didn't understand the following:

他们说，使用 seq (见下文)导致解析器将嵌套元组作为结果，操作起来很麻烦.

They say that using seq (see below) leads to parsers with nested tuples as results, which are messy to manipulate.

 seq :: Parser a -> Parser b -> Parser (a,b)
 p ‘seq‘ q = \inp -> [((v,w),inp’’) | (v,inp’) <- p inp, (w,inp’’) <- q inp’]

为了避免嵌套元组的这种问题，他们引入了 monadic bind 和 return 进行解析，然后定义 seq 如下:

In order to avoid this problem of nested tuples they introduce monadic bind and return for parsers and then define seq as follows:

 p ‘seq‘ q = p ‘bind‘ \x -> q ‘bind‘ \y -> result (x,y)

不幸的是，我看不到嵌套元组的问题是什么，为什么 seq 的第二个实现比第一个更好.你能帮我理解吗?

Unfortunately I don not see what the problem of nested tuples is and why the 2-nd implementation of seq is better then the 1-st one. Could you help me to understand it?

推荐答案

第一个示例扩展到类型((a，b)，(c，d，e)):

The first example extendend to the type ((a,b),(c,d,e)):

seq232 ((p,q),(r,s,t) = \inp ->
     [ (((v,w),(x,y,z)),inp’’''')
     | (v, inp’) <- p inp
     , (w, inp’’) <- q inp’
     , (x, inp''') <- r inp''
     , (y, inp'''') <- s inp'''
     , (z, inp''''') <- t imp''''
     ]

第二个示例扩展为类型((a，b)，(c，d，e)):

The second example extended to the type ((a,b),(c,d,e)):

seq232 ((p,q),(r,s,t)) =
    p ‘bind‘ \v ->
    q ‘bind‘ \w ->
    r `bind` \x ->
    s `bind` \y ->
    t `bind` \z ->
    result ((v,w),(x,y,z))

虽然不是很多 更好，但我认为您可以看到第二个更加干净.

While it isn't a a lot better, I think you can see that the second is a bit cleaner.

这篇关于为什么解析器组合器& quot; seq&"会出现错误.用&" bind&"定义和“返回"?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！