Question

说我有这样的结构：

struct MyStruct
{
    int a;
    int b; 
    int c;
}

BOOST_FUSION_ADAPT_STRUCT
(
    MyStruct,
    (int, a)
    (int, b)
    (int, c)
)

然后如果我有一个简单的发电机：

struct MyStructGenerator
    : boost::spirit::karma::grammar<boost::spirit::ostream_iterator, MyStruct()>
{
    MyStructGenerator() : MyStructGenerator::base_type(start_)
    {

        namespace bsk = boost::spirit::karma;

        start_ = '<' 
            << bsk::int_
            << ','
            << bsk::int_
            << ','
            << bsk::int_
            << '>';
    }

    ~MyStructGenerator() = default;
    boost::spirit::karma::rule<boost::spirit::ostream_iterator, MyStruct()> start_;
};

我运行以下内容：

int main()
{
    MyStruct ms = { 3, 2, 1 };
    std::cout << boost::spirit::karma::format(MyStructGenerator(), ms) << std::endl;
}

我当然希望看到<3, 2, 1>。我无法弄清楚如何改变规则内的顺序？如果我想查看<1, 2, 3>或即使我想查看<2, 1, 3>该怎么办？

另外，无论如何我可以在没有BOOST_FUSION_ADAPT_STRUCT的情况下执行此操作吗？

Answer 1

Q1 如果我想看到＆lt; 1,2,3＆gt;或者即使我想看{4}}

只需更改适应的订单：

简化 Live On Wandbox ：

<2, 1, 3>

打印#include <boost/fusion/adapted/struct.hpp> struct MyStruct { int a, b, c; }; BOOST_FUSION_ADAPT_STRUCT(MyStruct, a, b, c) #include <boost/spirit/include/karma.hpp> namespace bsk = boost::spirit::karma; template <typename It = boost::spirit::ostream_iterator> struct MyGen : bsk::grammar<It, MyStruct()> { MyGen() : MyGen::base_type(start_) { using namespace bsk; start_ = '<' << int_ << ',' << int_ << ',' << int_ << '>'; } private: bsk::rule<It, MyStruct()> start_; }; int main() { MyGen<> gen; std::cout << format(gen, MyStruct { 3, 2, 1 }) << "\n"; }，但

<3,2,1>

打印BOOST_FUSION_ADAPT_STRUCT(MyStruct, c, b, a)。

Q2没有适应？

好。我可以向您展示一些您可能感兴趣的事情：

： Live On Wandbox ：

<1,2,3>

打印

#include <boost/fusion/adapted/struct.hpp>

struct MyStruct { int a, b, c; };
BOOST_FUSION_ADAPT_STRUCT(MyStruct, c, b, a)

#include <iostream>
#include <boost/fusion/include/io.hpp>
#include <boost/fusion/include/as_vector.hpp>
using boost::fusion::as_vector;

int main() {
    MyStruct ms { 3, 2, 1 };
    std::cout << as_vector(ms) << "\n";

    std::cout 
        << boost::fusion::tuple_open("<")
        << boost::fusion::tuple_delimiter(",")
        << boost::fusion::tuple_close(">");

    std::cout << as_vector(ms) << "\n";
}

命名改编：您可以使用(1 2 3) <1,2,3>宏同时调整的不同订单。这是一个使用Fusion IO和Karma生成器演示的演示。

注意我稍微修改了结构，因此更容易跟踪哪个字段为*_NAMED，'a'或'b'。

查看 Live On Wandbox ：

'c'

打印

#include <boost/fusion/adapted/struct.hpp> struct MyStruct { char a, b, c; }; BOOST_FUSION_ADAPT_STRUCT_NAMED(MyStruct, AsABC, a, b, c) BOOST_FUSION_ADAPT_STRUCT_NAMED(MyStruct, AsBCA, b, c, a) BOOST_FUSION_ADAPT_STRUCT_NAMED(MyStruct, AsCBA, c, b, a) #include <boost/spirit/include/karma.hpp> #include <boost/spirit/include/qi.hpp> namespace bsk = boost::spirit::karma; template <typename Attr, typename It = boost::spirit::ostream_iterator> struct MyGen : bsk::grammar<It, Attr()> { MyGen() : MyGen::base_type(start_) { using namespace bsk; start_ = '<' << auto_ << ',' << auto_ << ',' << auto_ << '>'; } private: bsk::rule<It, Attr()> start_; }; #include <iostream> #include <boost/fusion/include/io.hpp> #include <boost/fusion/include/as_vector.hpp> using boost::fusion::as_vector; template <typename Attr> void do_tests(Attr const& ms) { std::cout << as_vector(ms) << "\n"; std::cout << format(MyGen<Attr>{}, ms) << "\n"; } int main() { std::cout << boost::fusion::tuple_open("<") << boost::fusion::tuple_delimiter(",") << boost::fusion::tuple_close(">"); MyStruct ms { 'a', 'b', 'c' }; using namespace boost::fusion::adapted; do_tests(AsABC{ms}); do_tests(AsCBA{ms}); do_tests(AsBCA{ms}); }

是的，你可以不做适应（不要）：

Live On Wandbox （由于编译时限制而评论部分）

<a,b,c> <a,b,c> <c,b,a> <c,b,a> <b,c,a> <b,c,a>

打印

struct MyStruct { char a, b, c; }; #include <boost/spirit/include/karma.hpp> #include <boost/spirit/include/phoenix.hpp> namespace bsk = boost::spirit::karma; namespace phx = boost::phoenix; template <typename It = boost::spirit::ostream_iterator> struct MyGen : bsk::grammar<It, MyStruct()> { MyGen() : MyGen::base_type(start_) { using boost::proto::deep_copy; using namespace bsk; auto A = deep_copy(char_[ _1 = phx::bind(&MyStruct::a, _val) ]); auto B = deep_copy(char_[ _1 = phx::bind(&MyStruct::b, _val) ]); auto C = deep_copy(char_[ _1 = phx::bind(&MyStruct::c, _val) ]); start_ = '<' << A << ',' <' << eol << '<' << A << ',' << C << ',' <' << eol << '<' <' << eol << '<' << C << ',' << A << ',' <' << eol << '<' <' << eol << '<' << C << ',' <' << eol ; } private: bsk::rule<It, MyStruct()> start_; }; int main() { std::cout << format(MyGen<>{}, MyStruct { 'a', 'b', 'c' }); }

Answer 2

您可以简单地使用phoenix和bind来访问成员值。因此，对于您交换成员的示例，它将是这样的：

#include <iostream>
#include <boost/spirit/include/karma.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/phoenix/bind/bind_member_variable.hpp>
using namespace std;
struct MyStruct {
    int a;
    int b; 
    int c;
};

struct MyStructGenerator : boost::spirit::karma::grammar<boost::spirit::ostream_iterator, MyStruct()>
{
    MyStructGenerator() : MyStructGenerator::base_type(start_)
    {
        namespace bsk = boost::spirit::karma;
        start_ = '<' 
            << bsk::int_[bsk::_1 = boost::phoenix::bind(&MyStruct::c, bsk::_val)]
            << ','
            << bsk::int_[bsk::_1 = boost::phoenix::bind(&MyStruct::b, bsk::_val)]
            << ','
            << bsk::int_[bsk::_1 = boost::phoenix::bind(&MyStruct::a, bsk::_val)]
            << '>';
    }

    ~MyStructGenerator() = default;
    boost::spirit::karma::rule<boost::spirit::ostream_iterator, MyStruct()> start_;
};

int main() {
    MyStruct ms = { 3, 2, 1 };
    std::cout << boost::spirit::karma::format(MyStructGenerator(), ms) << std::endl;
}

提升Karma，重新排序结构元素

2 个答案:

Q1 如果我想看到＆lt; 1,2,3＆gt;或者即使我想看{4}}

Q2没有适应？