在 C 或 C++ 中使用 GNU 正则表达式函数

Question

谁能给我完整的示例程序，如何在 gcc C 或 C++ (http://docs.freebsd.org/info/regex/regex.info.GNU_Regex_Functions.html) 中使用 GNU 正则表达式函数， re_pattern_buffer, re_compile_fastmap?

例如，翻译这个 Python 小程序：

import re
unlucky = re.compile('1\d*?3')

nums = ("13", "31", "777", "10003")

for n in nums:
    if unlucky.search(n) is None:
        print "lucky"
    else:
        print "unlucky"

谢谢！

Answer 1

好的，在深入研究代码之前，我应该提到您可能想要使用更高级别的库。您确实说过 C++，因此您可以使用 Boost.Regex 等。即使你想留在C，也有更好的选择。我发现POSIX functions 更干净一些，更不用说更便携了。

// Tell GNU to define the non-standard APIs
#define _GNU_SOURCE 
// This is actually the same header used for the POSIX API.
// Except then you obviously don't need _GNU_SOURCE
#include <regex.h> 
#include <stdio.h>
#include <string.h>

int main()
{
    struct re_pattern_buffer pat_buff; // Put a re_pattern_buffer on the stack
    // The next 4 fields must be set.  

    // If non-zero, applies a translation function to characters before 
    // attempting match (http://www.delorie.com/gnu/docs/regex/regex_51.html)
    pat_buff.translate = 0; 
    // If non-zero, optimization technique.  Don't know details.
    // See http://www.delorie.com/gnu/docs/regex/regex_45.html
    pat_buff.fastmap = 0;
    // Next two must be set to 0 to request library allocate memory
    pat_buff.buffer = 0;
    pat_buff.allocated = 0;
    char pat_str[] = "1[^3]*3";
    // This is a global (!) used to set the regex type (note POSIX APIs don't use global for this)
    re_syntax_options = RE_SYNTAX_EGREP; 
    // Compile the pattern into our buffer
    re_compile_pattern(pat_str, sizeof(pat_str) - 1, &pat_buff); 
    char* nums[] = {"13", "31", "777", "10003"}; // Array of char-strings
    for(int i = 0; i < sizeof(nums) / sizeof(char*); i++)
    {
        int match_ret;
        // Returns number of characters matches (may be 0, but if so there's still a match)
        if((match_ret = re_match(&pat_buff, nums[i], strlen(nums[i]), 0, NULL)) >= 0) 
        {
            printf("unlucky\n");
        }
        else if(match_ret == -1) // No match
        {
            printf("lucky\n");
        }
        // Anything else (though docs say -2) is internal library error
        else 
    {
            perror("re_match");
        }
    }
    regfree(&pat_buff);
}

编辑：我添加了对必填字段和 regfree 的更多解释。我之前有幸运/不幸倒退，这解释了部分差异。另一部分是我认为这里可用的任何正则表达式语法都不支持惰性运算符（*？）。在这种情况下，有一个简单的修复方法，使用"1[^3]*3"。