Question

我的语法有各种以通用名称开头的条目。确定类型之后，我想使用期望运算符来创建解析错误。

rule1=name >> (type1 > something);
rule2=name >> (type2 > something);

我已经知道我不能混合两个运算符＆gt;和＆gt;＆gt; - 这就是括号的原因。我的猜测是括号会导致创建一个元组。如何在语义操作中访问元组的元素？以下肯定是错误的，但应该澄清我想要完成的事情。

rule1=(name >> (type1 > something))[qi::_val = boost::phoenix::bind(
    create,
    qi::_1,
    std::get<0>(qi::_2), 
    std::get<1>(qi::_2))];

感谢

Answer 1

直接解决问题：

using px::at_c;
rule1 = (name >> (type1 > something)) [_val = px::bind(create, _1, at_c<0>(_2), at_c<1>(_2))];

但是，我会使用qi::eps这个小技巧来避免复杂性：

rule2 = (name >> type1 >> (eps > something)) [_val = px::bind(create, _1, _2, _3)];

最后，请查看boost::phoenix::function<>：

px::function<decltype(&create)> create_(create); // or just decltype(create) if it's a function object
rule3 = (name >> type1 >> (eps > something)) [_val = create_(_1, _2, _3)];

这样你甚至可以拥有可读代码！

样本

只是为了证明这三个人都有相同的行为¹

<强> Live On Coliru

#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/fusion/include/at_c.hpp>

namespace qi = boost::spirit::qi;
namespace px = boost::phoenix;

static int create(char n, char t, char s) {
    assert(n=='n' && t=='t' && s=='s');
    return 42;
}

int main() {
    using It = std::string::const_iterator;

    // fake rules just for demo
    qi::rule<It, char()>
        name = qi::char_("n"),
        type1 = qi::char_("t"),
        something = qi::char_("s");

    //using boost::fusion::at_c;
    qi::rule<It, int(), qi::space_type> rule1, rule2, rule3;

    {
        using namespace qi;
        using px::at_c;

        rule1 = (name >> (type1 > something))        [_val = px::bind(create, _1, at_c<0>(_2), at_c<1>(_2))];
        rule2 = (name >> type1 >> (eps > something)) [_val = px::bind(create, _1, _2, _3)];

        px::function<decltype(&create)> create_(create); // or just decltype(create) if it's a function object
        rule3 = (name >> type1 >> (eps > something)) [_val = create_(_1, _2, _3)];
    }

    for(auto& parser : { rule1, rule2, rule3 }) {
        for(std::string const input : { "n t s", "n t !" }) {
            std::cout << "Input: '" << input << "'\n";
            auto f = input.begin(), l = input.end();
            int data;

            try {
                bool ok = qi::phrase_parse(f, l, parser, qi::space, data);
                if (ok) {
                    std::cout << "Parsing result: " << data << '\n';
                } else {
                    std::cout << "Parsing failed\n";
                }
            } catch(qi::expectation_failure<It> const& e) {
                std::cout << "Expectation failure: " << e.what() << " at '" << std::string(e.first, e.last) << "'\n";
            }

            if (f!=l) {
                std::cout << "Remaining unparsed: '" << std::string(f,l) << "'\n";
            }

            std::cout << "-------------------------------------------\n";
        }
    }
}

打印3x相同的输出：

Input: 'n t s'
Parsing result: 42
-------------------------------------------
Input: 'n t !'
Expectation failure: boost::spirit::qi::expectation_failure at '!'
Remaining unparsed: 'n t !'
-------------------------------------------
Input: 'n t s'
Parsing result: 42
-------------------------------------------
Input: 'n t !'
Expectation failure: boost::spirit::qi::expectation_failure at '!'
Remaining unparsed: 'n t !'
-------------------------------------------
Input: 'n t s'
Parsing result: 42
-------------------------------------------
Input: 'n t !'
Expectation failure: boost::spirit::qi::expectation_failure at '!'
Remaining unparsed: 'n t !'
-------------------------------------------

¹PS让这个作为如何在您的问题中创建SSCCE代码示例的示例

如何在语义动作中访问元组的元素

1 个答案:

样本