C ++中的字符串标记生成器,允许多个分隔符

时间:2010-04-16 15:19:03

标签: c# c++ string tokenize

有没有办法在C ++中用多个分隔符标记字符串?在C#中我会做到:

string[] tokens = "adsl, dkks; dk".Split(new [] { ",", " ", ";" }, StringSplitOptions.RemoveEmpty);

3 个答案:

答案 0 :(得分:3)

使用boost :: tokenizer。它支持多个分隔符。

事实上,你甚至不需要boost :: tokenizer。如果您想要的只是一个分割,请使用boost :: split。文档有一个例子: http://www.boost.org/doc/libs/1_42_0/doc/html/string_algo/usage.html#id1718906

答案 1 :(得分:2)

这样的事情会发生:

void tokenize_string(const std::string &original_string, const std::string &delimiters, std::vector<std::string> *tokens)
{
        if (NULL == tokens) return;

        size_t pos_start = original_string.find_first_not_of(delimiters);
        size_t pos_end   = original_string.find_first_of(delimiters, pos_start);

        while (std::string::npos != pos_start)
        {
                tokens->push_back(original_string.substr(pos_start, pos_end - pos_start));
                pos_start = original_string.find_first_not_of(delimiters, pos_end);
                pos_end   = original_string.find_first_of(delimiters, pos_start);
        }
}

答案 2 :(得分:0)

这是我的版本(未经过严格测试):

std::vector<std::string> split(std::string const& s,
    std::vector<std::string> const& delims)
{
    std::vector<std::string> parts;

    std::vector<std::pair<std::string::size_type, std::string::size_type>> poss;
    poss.reserve(delims.size());

    std::string::size_type beg = 0;

    for(;;)
    {
        poss.clear();

        std::string::size_type idx = 0;
        for(auto const& delim: delims)
        {
            if(auto end = s.find(delim, beg) + 1)
                poss.emplace_back(end - 1, idx);
            ++idx;
        }

        if(poss.empty())
            break;

        std::sort(std::begin(poss), std::end(poss));

        auto old_beg = beg;

        for(auto pos: poss)
        {
            parts.emplace_back(std::begin(s) + beg,
                std::begin(s) + old_beg + pos.first);
            beg = pos.first + delims[pos.second].size();
        }
    }

    if(beg < s.size())
        parts.emplace_back(std::begin(s) + beg, std::end(s));

    return parts;
}