逃避一个C ++字符串 [英] Escaping a C++ string

问题描述

将C ++ std :: string转换为另一个std :: string，其中所有不可打印字符都转义的最简单的方法是什么？

What's the easiest way to convert a C++ std::string to another std::string, which has all the unprintable characters escaped?

例如，对于两个字符[0x61,0x01]的字符串，结果字符串可能是a\x01或a％01 p>

For example, for the string of two characters [0x61,0x01], the result string might be "a\x01" or "a%01".

推荐答案

看看Boost的字符串算法库。您可以使用其 is_print 分类器（与其运算符！overload）一起挑选不可打印的字符，其 find_format（）函数可以用任何您想要的格式替换。

Take a look at the Boost's String Algorithm Library. You can use its is_print classifier (together with its operator! overload) to pick out nonprintable characters, and its find_format() functions can replace those with whatever formatting you wish.

#include <iostream>
#include <boost/format.hpp>
#include <boost/algorithm/string.hpp>

struct character_escaper
{
    template<typename FindResultT>
    std::string operator()(const FindResultT& Match) const
    {
        std::string s;
        for (typename FindResultT::const_iterator i = Match.begin();
             i != Match.end();
             i++) {
            s += str(boost::format("\\x%02x") % static_cast<int>(*i));
        }
        return s;
    }
};

int main (int argc, char **argv)
{
    std::string s("a\x01");
    boost::find_format_all(s, boost::token_finder(!boost::is_print()), character_escaper());
    std::cout << s << std::endl;
    return 0;
}

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