C ++中的Perl正则表达式替换将Perl打包并解压缩为C ++

问题描述

我有一部分Perl代码：

# In a string,replace all instances of `\A` with `A`,# where `A` is any non-word character.
# Word character include letter,digits and more.
$exampleString =~ s/\\(\w)/$1/g;

# In a string,replace all instances of `\!` with `@33~`,# where `!` is any non-whitespace character,and
# `33` is the ordinal value of the Unicode Code Point.
$exampleString =~ s/\\(\S)/$char1. unpack('C*',$1) . $char2/ge;

然后在$exampleString上进行了一些更改。

然后在末尾：

# This attempts to reverse the aforementioned use of s///.
$exampleString =~ s/$char1(.*?)$char2/pack('C*',$1)/ge;

$char1和$char2是由其ASCII值定义的两个字符。例如，假设char1是@，而char2是~。

此序列从（未显示）中间代码中隐藏\ 转义序列并将其恢复。

我想在C ++中做同样的事情。我想知道是否有人可以提供逐行翻译或代码可以提供的类似功能。

解决方法

这里是一个例子：

#include <iostream>
#include <regex>
#include <string>

std::string compute_ord_value(std::string str,char c1,char c2)
{
    const char c = str.at(0);
    return std::string{ c1 } + std::to_string(int(c)) + std::string{c2};
}

std::string compute_char_value(std::string str)
{
    int ord = std::stoi(str);
    return std::string {R"(\)"} + std::string { (char) ord };
}

void transform_back(std::string str,char c2)
{
    std::string::const_iterator it = str.cbegin();
    std::string::const_iterator end = str.cend();
    std::string result;
    std::regex re { std::string{c1} + R"((\d+))" + std::string{c2} };
    for (
         std::smatch match;
         std::regex_search(it,end,match,re);
         it = match[0].second
    ) {
        result += match.prefix();
        result += compute_char_value(match.str(1));
    }
    result.append(it,end);
    std::cout << result << '\n';
}

int main() {
    std::string str { R"(xx \A \& \B yy)" };
    std::cout << str << '\n';
    std::regex re { R"(\\(\w))" };
    char c1 = '@';
    char c2 = '~';
    std::string result1 = std::regex_replace(str,re,"$1");

    std::string::const_iterator it = result1.cbegin();
    std::string::const_iterator end = result1.cend();
    std::string result2;
    std::regex re2 { R"(\\(\S))" };
    for (
         std::smatch match;
         std::regex_search(it,re2);
         it = match[0].second
    ) {
        result2 += match.prefix();
        result2 += compute_ord_value(match.str(1),c1,c2);
    }
    result2.append(it,end);
    std::cout << result2 << '\n';
    transform_back(result2,c2);
    
    return 0;
}

输出：

xx \A \& \B yy
xx A @38~ B yy
xx A \& B yy