拒绝PARSE规则以匹配至少2#[无]的首次出现 [英] Rebol PARSE rule to match to first occurrence of at least 2 #[none]s

问题描述

请参见解决方案

我只能展示Red和Rebol2的解决方案.由于规则中的单词会自动减少，因此您必须屏蔽它们.

红色

>> parse [x x x x _ a _ _ b] [to [ '_ '_] y: ]
== false
>> 
>> y
== [_ _ b]

Rebol2

>> parse [x x x x _ a _ _ b] [some [r: [ '_ '_  ] (y: r) | skip]   ]
== true
>> y
== [_ _ b]

由 HostileFork 编辑问题后，Red的解决方案就这样

>> parse [x x x x #[none] a #[none] #[none] b] [to [none! none!] y:] 
== false
>> y
== [none none b]
>> 

根据 giuliolunati

注释中的问题的

例子

>> parse  [x x x x 0 a 1 2 b]  [to [integer! integer!] y:]
== false
>> y
== [1 2 b]

See similar question for string case.

In R3-Alpha, I tried to adapt the @sqlab response to block case:

parse [x x x x #[none] a #[none] #[none] b] [to [none! none!] ??]

I expect ??: [#[none] #[none] b], but get

** Script error: PARSE - invalid rule or usage of rule: none!

It's the result right and my expectation wrong? Or it's a bug?

解决方案

I can just show a solution for Red and Rebol2. As the words in the rule are reduced automatic, you have to shield them.

Red

>> parse [x x x x _ a _ _ b] [to [ '_ '_] y: ]
== false
>> 
>> y
== [_ _ b]

Rebol2

>> parse [x x x x _ a _ _ b] [some [r: [ '_ '_  ] (y: r) | skip]   ]
== true
>> y
== [_ _ b]

After the editing of the question by HostileFork the solution for Red looks like that

>> parse [x x x x #[none] a #[none] #[none] b] [to [none! none!] y:] 
== false
>> y
== [none none b]
>> 

example according the question in the comment of giuliolunati

>> parse  [x x x x 0 a 1 2 b]  [to [integer! integer!] y:]
== false
>> y
== [1 2 b]

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