如何在Flex中编写以下正则表达式？ [英] How to write the following regex in Flex?

问题描述

我试图在flex中定义一个将捕获多行字符串的规则。

多行字符串是以三个撇号开头的字符串：'' 以三个撇号结尾，并且可以跨越多行。

例如：

 '''这是
一个例子
a multiline 
 string'''
  

所以我的尝试是这样的：

 ％{
＃include< iostream> ; 
 using std :: cout; 
 using std :: endl; 
 
％} 
 
 MULTI_LN_STR'''（。| \\\
）*'''
 
 %% 
 
 {MULTI_LN_STR} {cout<GotIt！;} 
 
 %% 
 
 int main（int argc，char * argv []）{
 
 yyin = fopen（test.txt，r）; 
 
 if（！yyin）{
 cout<<yyin is NULL<< endl; 
 return 1; 
} 
 
 yylex（）; 
 return 0; 
} 
  

这适用于输入：

 '''这是
a multi 
 line 
 string！'''
 
这是
一些随机
文本
  

输出为：

  GotIt！ 
 
这是
一些随机的
文本
  

但对此输入无效（或更准确，产生错误输出）：

 '''这是
a multi 
 line 
 string！'''
 
这是
一些随机的
文本
 
'''and this 
 is another 
 multiline 
 string'''
  

其产生：

  GotIt！这个原因是因为我的规则是：
 
扫描三个撇号，后跟任何可能的字符，后跟三个撇号，
 
，而是应该说：
 
扫描三个撇号，后跟任何可能的字符，除了三个撇号
 
 
 
 
解决方案
对于这样的简单否定，构造正则表达式相对容易：
  ''（[^'] |'[^'] |[^']）*'''
  
 
I'm trying to define a rule in flex that will capture a "multiline string".

A multiline string is a string that starts with three apostrophes: ''', ends with three apostrophes, and can span over multiple lines.

For example:  
'''This is
an example of
a multiline
string'''
So my attempt at this was this:  
%{
#include<iostream>
using std::cout;
using std::endl;

%}

MULTI_LN_STR    '''(.|\n)*'''

%%

{MULTI_LN_STR}  {cout<<"GotIt!";}   

%%

int main(int argc, char* argv[]) {

    yyin=fopen("test.txt", "r");

    if (!yyin) {
        cout<<"yyin is NULL"<<endl;
        return 1;
    }

    yylex();
    return 0;
}
Which works for the input:
'''This is
a multi
line
string!'''

This is
some random
text
The output is:
GotIt!

This is
some random
text
but does not work (or, to be more accurate, produces wrong output) for this input:  
'''This is
a multi
line
string!'''

This is
some random
text

'''and this
is another
multiline
string''' 
Which produces:
GotIt!
This reason is because my rule says:

"scan for three apostrophes, followed by any possible character, followed by three apostrophes",

but rather, it should say:

"scan for three apostrophes, followed by any possible character except three apostrophes, followed by three apostrophes".

How can I do that?
解决方案
For a simple negation like this, it is relatively easy to construct a regular expression:
"'''"([^']|'[^']|''[^'])*"'''"


                        
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