c++ - 分配给引用参数会使对象无效

Question

我遇到了一些误导性问题。我有一个将字符串作为引用的函数，使用正则表达式将其拆分，将一部分保存回给定字符串并返回另一部分。但是，将一个匹配项分配给引用参数似乎会使匹配对象无效(看起来像是一种误用的移动分配)，而将其分配给局部变量则可以正常工作。这是代码:

std::string ParseVar(std::string & szExp)
{
    static std::regex oRegex("^([[:alpha:]][[:alnum:]_]*)\\s*<-\\s*(.+)$");
    std::smatch oMatch;

    if (std::regex_match(szExp, oMatch, oRegex))
    {
        if (oMatch.size() != 3)
            throw std::runtime_error("internal error");

        std::cout << "Match before: " << oMatch.str(2) << std::endl;
        szExp = oMatch.str(2);
        std::cout << "Match after: " << oMatch.str(2) << std::endl;
        return oMatch.str(1);
    }

    return "";
}

打印(对于 szExp = "foo <- 5+5+5+5"):

Match before: 5+5+5+5
Match after:  +5+5+5

返回值似乎也被破坏了，但是 szExp 包含正确的字符串。

将其更改为:

std::string ParseVar(std::string & szExp)
{
    static std::regex oRegex("^([[:alpha:]][[:alnum:]_]*)\\s*<-\\s*(.+)$");
    std::smatch oMatch;
    std::string save1, save2;

    if (std::regex_match(szExp, oMatch, oRegex))
    {
        if (oMatch.size() != 3)
            throw std::runtime_error("internal error");

        save1 = oMatch.str(1);
        save2 = oMatch.str(2);

        std::cout << "Match before: " << oMatch.str(2) << std::endl;
        szExp = save2;
        std::cout << "Match after: " << oMatch.str(2) << std::endl;
        return save1;
    }

    return "";
}

打印相同的东西，但至少返回值和 szExp 都可以。

这是怎么回事？

Answer 1

std::smatch 对象是 std::match_results 模板的实例化。它在 cppreference 上的条目包含以下段落:

This is a specialized allocator-aware container. It can only be default created, obtained from std::regex_iterator, or modified by std::regex_search or std::regex_match. Because std::match_results holds std::sub_matches, each of which is a pair of iterators into the original character sequence that was matched, it's undefined behavior to examine std::match_results if the original character sequence was destroyed or iterators to it were invalidated for other reasons.

因为修改原始字符串(这是您通过分配给 szExp 所做的事情)会使迭代器在其字符序列中无效，所以您违反了上述规定并导致未定义的行为。