为什么boost regex会耗尽堆栈空间？

Question

#include <boost/regex.hpp>
#include <string>
#include <iostream>

using namespace boost;
static const regex regexp(
        "std::vector<"
            "(std::map<"
                   "(std::pair<((\\w+)(::)?)+, (\\w+)>,?)+"
             ">,?)+"
        ">");

std::string errorMsg =
"std::vector<"
        "std::map<"
                "std::pair<Test::Test, int>,"
                "std::pair<Test::Test, int>,"
                "std::pair<Test::Test, int>"
        ">,"
        "std::map<"
                "std::pair<Test::Test, int>,"
                "std::pair<Test::Test, int>,"
                "std::pair<Test::Test, int>"
        ">"
">";
int main()
{
    smatch result;
    if(regex_match(errorMsg, result, regexp))
    {  
        for (unsigned i = 0; i < result.size(); ++i)
        {  
            std::cout << result[i] << std::endl;
        }
    }

//    std::cout << errorMsg << std::endl;

    return 0;
}

这会产生:

terminate called after throwing an instance of 'boost::exception_detail::clone_impl<boost::exception_detail::error_info_injector<std::runtime_error>
>'   what():  Ran out of stack space trying to match the regular expression.

用.编译

g++ regex.cc -lboost_regex

编辑

我的平台:

g++ (Ubuntu/Linaro 4.4.4-14ubuntu5) 4.4.5
libboost-regex1.42
Intel(R) Core(TM)2 Quad CPU Q9400 @ 2.66GHz
So the latest Ubuntu 64 bit

Answer 1

((\\w+)(::)?)+它是所谓的"病态"正则表达式之一 - 它将采用指数时间,因为你有两个表达式,它们彼此依赖于彼此.也就是说,它由于灾难性的回溯而失败.

考虑我们是否遵循链接的示例,并将"更复杂的东西"减少为"x".让我们这样做\\w:

• ((x+)(::)?)+

我们也假设我们的输入永远不会有::.这实际上使正则表达式更复杂,所以如果我们抛弃复杂性,那么我们真的应该让事情更简单,如果没有别的:

• (x+)+

现在你有一个教科书嵌套量词问题,如上面链接中详述的那样.

有几种方法可以解决这个问题,但最简单的方法可能就是禁止使用原子组修饰符 " (?>" 对内部匹配进行回溯:

• ((?>\\w+)(::)?)+