为什么这个正则表达式不会编译？ [英] Why this regular expression will not compile?

问题描述

我想从这里使用正则表达式：

I would like to use regular expression from here:

https://tools.ietf.org/html/rfc3986#appendix-B

我想编译它这：

#include <regex.h>
...
regex_t regexp;
if((regcomp(&regexp, "^(([^:/?#]+):)?(//([^/?#]*))?([^?#]*)(\?([^#]*))?(#(.*))?", REG_EXTENDED)) != 0){
    return SOME_ERROR:
}

但我遇到了regcomp的返回值：

But I am stuck with return value of regcomp:

REG_BADRPT

根据 man 它意味着：

无效使用重复运算符，例如使用 * 作为第一个字符。

Invalid use of repetition operators such as using * as the first character.

此的相似含义男子：

？， * 或 + 前面没有有效的正则表达式

?, * or + is not preceded by valid regular expression

我使用自己的正则表达式写了解析器，测试这个，因为它正式在rfc。我不打算使用它进行验证。

I wrote parser using my own regular expression, but I would like to test this one too, since its officially in rfc. I do no intend to use it for validation though.

推荐答案

正如Oli Charlesworth所建议的，您需要避开反斜杠 \\ 为问号 \？。有关详情，请参阅C ++ 转义序列。

As Oli Charlesworth suggested, you need to escape backslash \\ for the question marks \?. See C++ escape sequences for more information.

测试程序

#include <regex.h>
#include <iostream>

void test_regcomp(char *rx){
 regex_t regexp;
 if((regcomp(&regexp, rx, REG_EXTENDED)) != 0){
    std::cout << "ERROR :" << rx <<"\n";
 }
 else{
   std::cout <<  "   OK :"<< rx <<"\n";
 }
}

int main()
{

  char *rx1 = "^(([^:/?#]+):)?(//([^/?#]*))?([^?#]*)(\?([^#]*))?(#(.*))?" ;
  char *rx2 = "^(([^:/\?#]+):)\?(//([^/\?#]*))\?([^\?#]*)(\\\?([^#]*))\?(#(.*))\?" ;

   test_regcomp(rx1);
   test_regcomp(rx2);

   return 0;
}

输出

ERROR :^(([^:/?#]+):)?(//([^/?#]*))?([^?#]*)(?([^#]*))?(#(.*))?
   OK :^(([^:/?#]+):)?(//([^/?#]*))?([^?#]*)(\?([^#]*))?(#(.*))?

---

在您的regex是REG_BADRPT错误的源。它会转换为？。如果你替换它 \\？，regcomp将能够编译你的正则表达式。

---

The \?in your regex is the source of the REG_BADRPT error. It gets converted to ?. If you replace it by \\?, regcomp will be able to compile your regex.

"^(([^:/?#]+):)?(//([^/?#]*))?([^?#]*)(\\?([^#]*))?(#(.*))?"

   OK :^(([^:/?#]+):)?(//([^/?#]*))?([^?#]*)(\?([^#]*))?(#(.*))?

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