提升正则表达式迭代器返回空字符串

Question

我是 c++ 正则表达式的初学者，我想知道为什么这段代码：

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

int main() {

   std::string s = "? 8==2 : true ! false";
   boost::regex re("\?\s+(.*)\s*:\s*(.*)\s*\!\s*(.*)");

   boost::sregex_token_iterator p(s.begin(), s.end(), re, -1);  // sequence and that reg exp
   boost::sregex_token_iterator end;    // Create an end-of-reg-exp
                                        // marker
   while (p != end)
      std::cout << *p++ << '\n';
}

打印空字符串。我把正则表达式放在 regexTester 中，它正确地匹配了字符串，但是在这里当我尝试遍历匹配时它 returns 什么都没有。

Answer 1

我认为分词器实际上是用一些定界符分割文本，而定界符不包括在内。与 std::regex_token_iterator 比较：

std::regex_token_iterator is a read-only LegacyForwardIterator that accesses the individual sub-matches of every match of a regular expression within the underlying character sequence. It can also be used to access the parts of the sequence that were not matched by the given regular expression (e.g. as a tokenizer).

确实，您确实按照 the docs:

调用了此模式

if submatch is -1, then enumerates all the text sequences that did not match the expression re (that is to performs field splitting).

（强调我的）。

所以，修正一下：

for (boost::sregex_token_iterator p(s.begin(), s.end(), re), e; p != e;
     ++p)
{
    boost::sub_match<It> const& current = *p;
    if (current.matched) {
        std::cout << std::quoted(current.str()) << '\n';
    } else {
        std::cout << "non matching" << '\n';
    }
}

其他观察结果

所有贪婪的 Kleene-stars 都是麻烦的根源。您永远不会找到第二个匹配项，因为最后一个匹配项的 .* 将 根据定义 吞噬所有剩余的输入。

相反，让它们成为非贪婪的 (.*?) 和/或更精确的（比如隔离一些字符集，或强制使用非 space 字符？）。

boost::regex re(R"(\?\s+(.*?)\s*:\s*(.*?)\s*\!\s*(.*?))");

// Or, if you don't want raw string literals:
boost::regex re("\?\s+(.*?)\s*:\s*(.*?)\s*\!\s*(.*?)");

Live Demo

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

int main() {
    using It = std::string::const_iterator;
    std::string const s = 
        "? 8==2 : true ! false;"
        "? 9==3 : 'book' ! 'library';";
    boost::regex re(R"(\?\s+(.*?)\s*:\s*(.*?)\s*\!\s*(.*?))");

    {
        std::cout << "=== regex_search:\n";
        boost::smatch results;
        for (It b = s.begin(); boost::regex_search(b, s.end(), results, re); b = results[0].end()) {
            std::cout << results.str() << "\n";
            std::cout << "remain: " << std::quoted(std::string(results[0].second, s.end())) << "\n";
        }
    }

    std::cout << "=== token iteration:\n";
    for (boost::sregex_token_iterator p(s.begin(), s.end(), re), e; p != e;
         ++p)
    {
        boost::sub_match<It> const& current = *p;
        if (current.matched) {
            std::cout << std::quoted(current.str()) << '\n';
        } else {
            std::cout << "non matching" << '\n';
        }
    }
}

版画

=== regex_search:
? 8==2 : true ! 
remain: "false;? 9==3 : 'book' ! 'library';"
? 9==3 : 'book' ! 
remain: "'library';"
=== token iteration:
"? 8==2 : true ! "
"? 9==3 : 'book' ! "

奖励：解析器表达式

您可以生成一个解析器，而不是滥用正则表达式来进行解析，例如使用 Boost Spirit：

Live On Coliru

#include <boost/spirit/home/x3.hpp>
#include <boost/fusion/adapted.hpp>
#include <iomanip>
#include <iostream>
namespace x3 = boost::spirit::x3;

int main() {
    std::string const s = 
        "? 8==2 : true ! false;"
        "? 9==3 : 'book' ! 'library';";

    using expression = std::string;
    using ternary = std::tuple<expression, expression, expression>;
    std::vector<ternary> parsed;

    auto expr_ = x3::lexeme [+(x3::graph - ';')];
    auto ternary_ = "?" >> expr_ >> ":" >> expr_ >> "!" >> expr_;

    std::cout << "=== parser approach:\n";
    if (x3::phrase_parse(begin(s), end(s), *x3::seek[ ternary_ ], x3::space, parsed)) {

        for (auto [cond, e1, e2] : parsed) {
            std::cout
                << " condition " << std::quoted(cond) << "\n"
                << " true expression " << std::quoted(e1) << "\n"
                << " else expression " << std::quoted(e2) << "\n"
                << "\n";
        }
    } else {
        std::cout << "non matching" << '\n';
    }
}

版画

=== parser approach:
 condition "8==2"
 true expression "true"
 else expression "false"

 condition "9==3"
 true expression "'book'"
 else expression "'library'"

这更具可扩展性，将轻松支持递归语法，并且能够合成语法树的类型化表示，而不是只留下零散的字符串。

提升正则表达式迭代器返回空字符串

boost regex iterator returning empty string

c++

boost

boost-regex

其他观察结果

奖励：解析器表达式