Question

如何创建一个规则来读取3个集合中的整数。即，...

1 2 3                   OK, 1 set of 3 ints
1 2 3 4 5 6             OK, 2 sets of 3 ints
1 2 3 4 5               ERROR, 1 set of 3 ints, 1 short for 2nd
1 2 3 4 5 6 7 8 9       OK, 3 sets of 3 ints
1 2 3 4 5 6 7 8 9 10    ERROR, 3 sets of 3 ints, 1 short for 4th

我遇到了融合适应结构的问题（如何使args的数量变量）......

BOOST_FUSION_ADAPT_STRUCT(client::ast::number, n1, n2, n3, n4, n5, n6)

不确定为什么这条规则不起作用。

... = *(int_ >> int_ >> int_);

这是我的尝试...... http://coliru.stacked-crooked.com/a/cb10e8096c95fc55

//#define BOOST_SPIRIT_X3_DEBUG
#include <iostream>
#include <boost/spirit/home/x3.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/fusion/include/io.hpp>

namespace client { 
    namespace ast {

        struct number {
            int n1, n2, n3, n4, n5, n6;
        };

        struct comment {
            std::string text;
            bool dummy;
        };

        struct input {
            std::vector<comment> comments;  
            std::vector<number> numbers;
        };
    } 
}

BOOST_FUSION_ADAPT_STRUCT(client::ast::comment, text, dummy)
BOOST_FUSION_ADAPT_STRUCT(client::ast::number, n1, n2, n3) // , n4, n5, n6) error
BOOST_FUSION_ADAPT_STRUCT(client::ast::input, comments, numbers)

namespace client {      
    namespace parser {

        namespace x3 = boost::spirit::x3;
        using namespace x3;

        typedef std::string::const_iterator It;

        using namespace x3;

        auto const comment = rule<struct _c, ast::comment> {"comment"} = lexeme[*(char_ - eol)] >> attr(false);
        auto const number  = rule<struct _n, ast::number> {"number"}   = int_ >> int_ >> int_;
        // auto const number  = rule<struct _n, ast::number> {"number"}   = *(int_ >> int_ >> int_); error

        auto lines = [](auto p) { return *(p >> eol); };

        auto const input = 
            repeat(1)[comment] >> eol >>
            lines(number);
    }
}

int main() {
    namespace x3 = boost::spirit::x3;

    std::string const iss("any char string here\n1 2 3\n1 2 3 4 5 6");

    auto iter = iss.begin(), eof = iss.end();

    client::ast::input types;

    bool ok = phrase_parse(iter, eof, client::parser::input, x3::blank, types);

    if (iter != eof) {
        std::cout << "Remaining unparsed: '" << std::string(iter, eof) << "'\n";
    }
    std::cout << "Parsed: " << (100.0 * std::distance(iss.begin(), iter) / iss.size()) << "%\n";
    std::cout << "ok = " << ok << std::endl;

    for (auto& item : types.comments) { std::cout << "comment: " << boost::fusion::as_deque(item) << "\n"; }
    for (auto& item : types.numbers)  { std::cout << "number:  " << boost::fusion::as_deque(item) << "\n"; }
}

打印

Parsed: 71.0526%
ok = 1
comment: (any char string here 0)
number:  (1 2 3)

Answer 1

运算符Kleene-star合成一个容器属性（docs show：vector<T>）

您的结构number不是容器属性。如此。

另外，我完全不清楚你想要达到的目标。你的结构应该是6个整数，但你想解析3个组？这些群组的含义是什么？我可能会这样做：

struct number {
    struct group { int n1, n2, n3; };
    std::vector<group> groups;
};

BOOST_FUSION_ADAPT_STRUCT(client::ast::number::group, n1, n2, n3)
BOOST_FUSION_ADAPT_STRUCT(client::ast::number, groups)

然后它与解析器表达式

兼容

 *(int_ >> int_ >> int_)

现场演示

两个注释：

再次需要虚拟attibrute（考虑using number = std::vector<number_group>;）
输入结束时缺少\n（考虑eol | eoi）

<强> Live On Coliru

//#define BOOST_SPIRIT_X3_DEBUG
#include <iostream>
#include <boost/spirit/home/x3.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/fusion/include/io.hpp>

namespace client { 
    namespace ast {

        struct number {
            struct group { int n1, n2, n3; };
            std::vector<group> groups;
            bool dummy;
        };

        struct comment {
            std::string text;
            bool dummy;
        };

        struct input {
            std::vector<comment> comments;  
            std::vector<number> numbers;
        };
    } 
}

BOOST_FUSION_ADAPT_STRUCT(client::ast::comment, text, dummy)
BOOST_FUSION_ADAPT_STRUCT(client::ast::number::group, n1, n2, n3)
BOOST_FUSION_ADAPT_STRUCT(client::ast::number, groups, dummy)
BOOST_FUSION_ADAPT_STRUCT(client::ast::input, comments, numbers)

namespace client {      
    namespace parser {

        namespace x3 = boost::spirit::x3;
        using namespace x3;

        typedef std::string::const_iterator It;

        using namespace x3;

        auto const comment = rule<struct _c, ast::comment> {"comment"} = lexeme[*(char_ - eol)] >> attr(false);
        auto const number  = rule<struct _n, ast::number> {"number"}   = *(int_ >> int_ >> int_) >> attr(false);

        auto lines = [](auto p) { return *(p >> eol); };

        auto const input = 
            repeat(1)[comment] >> eol >>
            lines(number);
    }
}

int main() {
    namespace x3 = boost::spirit::x3;

    std::string const iss("any char string here\n1 2 3\n1 2 3 4 5 6\n");

    auto iter = iss.begin(), eof = iss.end();

    client::ast::input types;

    bool ok = phrase_parse(iter, eof, client::parser::input, x3::blank, types);

    if (iter != eof) {
        std::cout << "Remaining unparsed: '" << std::string(iter, eof) << "'\n";
    }
    std::cout << "Parsed: " << (100.0 * std::distance(iss.begin(), iter) / iss.size()) << "%\n";
    std::cout << "ok = " << ok << std::endl;

    for (auto &item : types.comments) {
        std::cout << "comment: " << boost::fusion::as_deque(item) << "\n";
    }
    for (auto& item : types.numbers) {
        std::cout << "number:  ";
        for (auto& g : item.groups)
            std::cout << boost::fusion::as_deque(g) << " ";
        std::cout << "\n";
    }
}

打印

Parsed: 100%
ok = 1
comment: (any char string here 0)
number:  (1 2 3) 
number:  (1 2 3) (4 5 6)

X3：如何创建解析器以进行读取？

1 个答案:

现场演示