升压::精神::齐.如何将内联的解析器表达式转换为独立的语法,以及如何解包它们生成的元组？

Question

我正在使用QI和Phoenix,我想编写一个返回4个bool的小语法,它将用作语义动作中函数调用的参数.

我有几个需要这些东西的函数,到目前为止我已经使用过这种方法:

( qi::_bool >>  qi::_bool >>  qi::_bool >>  qi::_bool)
[px::bind(&Bool4Function, spirit::_val, spirit::_1, spirit::_2, spirit::_3, spirit::_4)]

虽然它可以自己使用它,但是在整个地方使用它只是简单的丑陋和混乱,即使使用'命名空间部分.

这就是为什么我想把这个表达式提取成一个独立的语法.

所以我尝试了这个(信用证转到ildjarn测试床):

///// grammar implementation /////
#include <boost/fusion/include/vector10.hpp>
#include <boost/spirit/include/qi_bool.hpp>
#include <boost/spirit/include/qi_char_.hpp>
#include <boost/spirit/include/qi_grammar.hpp>
#include <boost/spirit/include/qi_operator.hpp>
#include <boost/spirit/include/qi_rule.hpp>
#include <boost/spirit/include/qi_string.hpp>

struct FourBools : boost::spirit::qi::grammar<
    char const*,
    boost::fusion::vector4<bool, bool, bool, bool>()
>
{
    typedef boost::fusion::vector4<bool, bool, bool, bool> attribute_type;

    FourBools() : base_type(start_)
    {
        using boost::spirit::bool_;

        start_
            =   "4bools:"
            >> bool_ >> ','
            >> bool_ >> ','
            >> bool_ >> ','
            >> bool_ >> ';'
            ;
    }

private:
    boost::spirit::qi::rule<
        base_type::iterator_type,
        base_type::sig_type
    > start_;
};
FourBools const fourBools;


///// demonstration of use /////
#include <string>
#include <ios>
#include <iostream>
#include <boost/fusion/include/at_c.hpp>
#include <boost/spirit/include/phoenix_bind.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/qi_action.hpp>
#include <boost/spirit/include/qi_parse.hpp>



void noDice(bool a, bool b, bool c, bool d) 
{

}

void worksFine(boost::fusion::vector4<bool, bool, bool, bool> a)
{

}
int main()
{
    namespace phx = boost::phoenix;
    namespace spirit = boost::spirit;

    std::string const input("4bools:true,true,true,false;");


    char const* first = input.c_str();
    char const* const last = first + input.size();
    bool const success = spirit::qi::parse(
        first, last,
        fourBools[phx::bind(&noDice, spirit::_1)]
    );


    if (!success)
        std::cout << "parse() failed\n";
    else if (first != last)
        std::cout << "didn't consume all input\n";
    std::cout.flush();
}

除非fourBools[phx::bind(&noDice, spirit::_1)]替换为,否则不会编译fourBools[phx::bind(&worksFine, spirit::_1)].

这意味着,我的问题是解包参数以匹配要调用的函数的签名,因为参数的数量在签名级别上有所不同(四个bool的一个元组,而它们自己的四个bool).

是否可以直接使用phoenix占位符解压缩,而不是编写包装器将元组转换为我需要单独的现有函数的单个参数？如果是,那将是什么语法？毕竟,内联版本( qi::_bool >> qi::_bool >> qi::_bool >> qi::_bool)在spirit::_1 - spirit::_4,占位符"解压缩"时工作正常.

这让我看起来好像这个版本也返回了一个元组,并且在某种程度上可以通过上述方法解压缩,而不像返回一个语法.

我该如何处理？

Answer 1

如果你没有发表完整,连贯的复制品,那么几乎不可能诊断你的问题; 它可能是一个语法错误,它可能是一个缺失#include,谁知道..？

这是一个工作示范; 希望你可以用它来作为参考,找出你的代码有什么问题:

///// grammar implementation /////
#include <boost/fusion/include/vector10.hpp>
#include <boost/spirit/include/qi_bool.hpp>
#include <boost/spirit/include/qi_char_.hpp>
#include <boost/spirit/include/qi_grammar.hpp>
#include <boost/spirit/include/qi_operator.hpp>
#include <boost/spirit/include/qi_rule.hpp>
#include <boost/spirit/include/qi_string.hpp>

struct FourBools : boost::spirit::qi::grammar<
    char const*,
    boost::fusion::vector4<bool, bool, bool, bool>()
>
{
    typedef boost::fusion::vector4<bool, bool, bool, bool> attribute_type;

    FourBools() : base_type(start_)
    {
        using boost::spirit::bool_;

        start_
            =   "4bools:"
                >> bool_ >> ','
                >> bool_ >> ','
                >> bool_ >> ','
                >> bool_ >> ';'
            ;
    }

private:
    boost::spirit::qi::rule<
        base_type::iterator_type,
        base_type::sig_type
    > start_;
};
FourBools const fourBools;


///// demonstration of use /////
#include <string>
#include <ios>
#include <iostream>
#include <boost/fusion/include/at_c.hpp>
#include <boost/spirit/include/phoenix_bind.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/qi_action.hpp>
#include <boost/spirit/include/qi_parse.hpp>

typedef FourBools::attribute_type attr_t;

struct verify_same
{
    explicit verify_same(attr_t const& expected) : expected_(expected) { }

    void verify(attr_t const& actual) const
    {
        using boost::fusion::at_c;

        std::cout << std::boolalpha
            << "same as expected: " << (actual == expected_)
            << "\nactual values: "
            << at_c<0>(actual) << ' '
            << at_c<1>(actual) << ' '
            << at_c<2>(actual) << ' '
            << at_c<3>(actual) << '\n';
    }

private:
    attr_t expected_;
};

int main()
{
    namespace phx = boost::phoenix;
    namespace spirit = boost::spirit;

    std::string const input("4bools:true,true,true,false;");
    verify_same const vs(attr_t(true, true, true, false));

    char const* first = input.c_str();
    char const* const last = first + input.size();
    bool const success = spirit::qi::parse(
        first, last,
        fourBools[phx::bind(&verify_same::verify, phx::cref(vs), spirit::_1)]
    );
    if (!success)
        std::cout << "parse() failed\n";
    else if (first != last)
        std::cout << "didn't consume all input\n";
    std::cout.flush();
}

另外,我认为使用纯粹同类型的元组是奇怪的; 就个人而言,我将语法的合成属性更改为boost::array<bool, 4>.

编辑(回应OP的编辑):有好消息,坏消息和更多好消息.

这是个好消息:Boost.Fusion具有使用最少代码完成您想做的功能:boost::fusion::fused<>.这将采用可调用类型(包括自由函数指针和成员函数指针),它接受多个参数并将该可调用类型包装在采用Fusion序列的仿函数中; 当调用此仿函数时,它会获取Fusion序列并将其解包,将元组的各个元素作为单独的参数转发到包装的可调用类型.

所以,鉴于我已经发布的语法和以下内容:

#include <string>
#include <ios>
#include <iostream>
#include <boost/fusion/include/at_c.hpp>
#include <boost/fusion/include/make_fused.hpp>
#include <boost/spirit/include/phoenix_bind.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_fusion.hpp>
#include <boost/spirit/include/qi_action.hpp>
#include <boost/spirit/include/qi_parse.hpp>

typedef FourBools::attribute_type attr_t;

void free_func_taking_tuple(attr_t const& tup)
{
    using boost::fusion::at_c;

    std::cout << std::boolalpha
        << "inside free_func_taking_tuple() :: "
        << at_c<0>(tup) << ' '
        << at_c<1>(tup) << ' '
        << at_c<2>(tup) << ' '
        << at_c<3>(tup) << '\n';
}

void free_func_taking_bools(
    bool const a, bool const b,
    bool const c, bool const d
)
{
    std::cout << std::boolalpha
        << "inside free_func_taking_bools() :: "
        << a << ' '
        << b << ' '
        << c << ' '
        << d << '\n';
}

boost::spirit::qi::parse() 可以像这样调用:

namespace phx = boost::phoenix;
namespace spirit = boost::spirit;
using boost::fusion::make_fused;

// calls free_func_taking_tuple, nothing new here
spirit::qi::parse(
    first, last,
    fourBools[phx::bind(free_func_taking_tuple, spirit::_1)]
);

// calls free_func_taking_bools, using boost::fusion::fused<> to unpack the tuple
// into separate arguments
spirit::qi::parse(
    first, last,
    fourBools[phx::bind(make_fused(&free_func_taking_bools), spirit::_1)]
);

这是坏消息:Boost.Fusion的可调用类型包装器依赖于TR1/C++ 11 result_of协议,而Boost.Phoenix v2实现了Boost.Lambda result_of协议 - 这些不兼容.因此,您必须自己解压缩元组元素:

namespace phx = boost::phoenix;
namespace spirit = boost::spirit;

spirit::qi::parse(
    first, last,
    fourBools[phx::bind(
        free_func_taking_bools,
        phx::at_c<0>(spirit::_1),
        phx::at_c<1>(spirit::_1),
        phx::at_c<2>(spirit::_1),
        phx::at_c<3>(spirit::_1)
    )]
);

呸!但是,还有更多好消息:Boost.Phoenix v3将在Boost 1.47中发布,它实现了TR1/C++ 11 result_of协议.因此,从Boost 1.47开始,您将能够使用boost::fusion::fused<>并节省一些繁琐的样板.