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使用至少一个空格来增强spirit skip解析器

boost-spirit boost

pschulz · 技术社区 · 6 年前

在我要实现的语法中,有一些元素被空格隔开。使用skip解析器,元素之间的空格将自动跳过,但这也允许空间,这不是我想要的。当然,我可以显式地编写包含这些空格的语法,但在我看来(凭借spirit提供的复杂性和灵活性),有更好的方法可以做到这一点。有?

#include <cstdlib>
#include <iostream>
#include <string>

#include <boost/spirit/include/qi.hpp>    

namespace qi = boost::spirit::qi;

int main(int argc, char** argv)
{
    if(argc != 2)
    {
        std::exit(1);
    }
    std::string str = argv[1];
    auto iter = str.begin();
    bool r = qi::phrase_parse(iter, str.end(), qi::char_ >> qi::char_, qi::blank);

    if (r && iter == str.end())
    {
        std::cout << "parse succeeded\n";
    }
    else
    {
        std::cout << "parse failed. Remaining unparsed: " << std::string(iter, str.end()) << '\n';
    }
}

这允许 ab 以及 a b . 我只希望后者被允许。

one   = char_("X") >> repeat(2)[omit[+blank] >> +alnum] >> qi::omit[+qi::blank] >> +alnum;
two   = char_("Y") >> repeat(3)[omit[+blank] >> +alnum];
three = char_("Z") >> repeat(4)[omit[+blank] >> +alnum] >> qi::omit[+qi::blank] >> +alnum;

main = one | two | three;

这使得语法很混乱,我想避免。

1 回复 | 直到 6 年前

sehe 6 年前

首先,我通常在RFCs中看到的语法规范是(总是?)RFCs。在99%的情况下没有问题,例如:

 myrule = skip(space) [ uint_ >> uint_ ];

Boost.Spirit qi value sequence vector ).

这样一来,根据定义,skipper应用了零次或多次,因此没有一种方法可以通过现有的有状态指令(如 skip() http://stackoverflow.com/questions/17072987/boost-spirit-skipper-issues/17073965#17073965 或者 docs -低于 lexeme [no_]skip 和 skip_flag::dont_postskip ).

看看你具体的语法,我会这么做:

bool r = qi::phrase_parse(iter, end, token >> token, qi::blank);

在这里,您可以在lexeme中添加一个否定的lookahead断言,以断言“到达了令牌的末尾”——在解析器中,这将被强制为 !qi::graph

    auto token = qi::copy(qi::lexeme [ qi::char_ >> !qi::graph ]);

查看演示:

Live On Coliru

#include <iostream>
#include <iomanip>
#include <boost/spirit/include/qi.hpp>

namespace qi = boost::spirit::qi;

int main() {
    for (std::string const str : { "ab", " ab ", " a b ", "a b" }) {
        auto iter = str.begin(), end = str.end();

        auto token = qi::copy(qi::lexeme [ qi::char_ >> !qi::graph ]);

        bool r = qi::phrase_parse(iter, end, token >> token, qi::blank);

        std::cout << " --- " << std::quoted(str) << " --- ";
        if (r) {
            std::cout << "parse succeeded.";
        } else {
            std::cout << "parse failed.";
        }

        if (iter != end) {
            std::cout << " Remaining unparsed: " << std::string(iter, str.end());
        }

        std::cout << std::endl;
    }
}

印刷品

 --- "ab" --- parse failed. Remaining unparsed: ab
 --- " ab " --- parse failed. Remaining unparsed:  ab 
 --- " a b " --- parse succeeded.
 --- "a b" --- parse succeeded.

奖金审查说明

你的船长应该是语法的责任。令人难过的是,所有的Qi样本都让人们相信你需要让打电话的人来决定
结束迭代器检查相等错误检查。很有可能 parse things correctly without consuming all input . 这就是为什么在解析失败的情况下不应该只报告“剩余输入”的原因。
如果尾随未分析的输入 spell it out :

Live On Coliru

#include <iostream>
#include <iomanip>
#include <boost/spirit/include/qi.hpp>

namespace qi = boost::spirit::qi;

int main() {
    for (std::string const str : { "ab", " ab ", " a b ", "a b happy trees are trailing" }) {
        auto iter = str.begin(), end = str.end();

        auto token = qi::copy(qi::lexeme [ qi::char_ >> !qi::graph ]);

        bool r = qi::parse(iter, end, qi::skip(qi::space) [ token >> token >> qi::eoi ]);

        std::cout << " --- " << std::quoted(str) << " --- ";
        if (r) {
            std::cout << "parse succeeded.";
        } else {
            std::cout << "parse failed.";
        }

        if (iter != end) {
            std::cout << " Remaining unparsed: " << std::quoted(std::string(iter, str.end()));
        }

        std::cout << std::endl;
    }
}

印刷品

 --- "ab" --- parse failed. Remaining unparsed: "ab"
 --- " ab " --- parse failed. Remaining unparsed: " ab "
 --- " a b " --- parse succeeded.
 --- "a b happy trees are trailing" --- parse failed. Remaining unparsed: "a b happy trees are trailing"