Haskell:将列表理解为组合式 [英] Haskell: List Comprehension to Combinatory

问题描述

受到此文章的启发.我正在将翻译功能从列表理解转换为组合样式.我发现了一些有趣的东西.

Inspired by this article. I was playing with translating functions from list comprehension to combinatory style. I found something interesting.

-- Example 1: List Comprehension 
*Main> [x|(x:_)<-["hi","hello",""]]
"hh"

-- Example 2: Combinatory
*Main> map head ["hi","hello",""]
"hh*** Exception: Prelude.head: empty list

-- Example 3: List Comprehension (translated from Example 2)
*Main> [head xs|xs<-["hi","hello",""]]
"hh*** Exception: Prelude.head: empty list 

示例1没有引发异常似乎很奇怪，因为(x:_)模式与head的定义之一匹配.使用列表推导时是否暗含filter (not . null)?

It seems strange that example 1 does not throw an exception, because (x:_) pattern matches one of the definitions of head. Is there an implied filter (not . null) when using list comprehensions?

推荐答案

请参阅有关列表理解的部分在Haskell报告中.所以基本上

See the section on list comprehensions in the Haskell report. So basically

[x|(x:_)<-["hi","hello",""]]

被翻译为

let ok (x:_) = [ x ]
    ok _     = [ ]
in concatMap ok ["hi","hello",""]

P.S.由于列表解析可以转换为do表达式，因此do表达式也会发生类似的事情，如

P.S. Since list comprehensions can be translated into do expressions, a similar thing happens with do expressions, as detailed in the section on do expressions. So the following will also produce the same result:

do (x:_)<-["hi","hello",""]
   return x

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