提升业力语义动作调用图

Question

我需要将值映射到 std::string（使用以下映射和 BOOST_FUSION_ADAPT_STRUCT）

    std::map< TYPEX, std::string> author2name;
    struct Emp
   {
     std::string name;
     TYPEX author;
   };

我想使用以下代码生成我的输出：

karma::rule< it, std::string()> quote = '"' >> karma::string >> '"';
karma::rule< it, Emp> emp = karma::delimit('\t')[ quite << quite[ author2name[ karma::_1] ]];

Emp x;
karma::generate( std::ostream_iterator<char>(std::cout), emp, x);

但是编译不了。

有什么方法可以让我像这样写 header:

karma::rule< it, std::vector<std::string>()> header = karma::delimit('\t')[ % quote];
karma::rule< it, Emp> emp = header >> karma::eol >> karma::delimit('\t')[ quite << quite[ author2name[ karma::_1] ]];

karma::generate( std::ostream_iterator<char>(std::cout), {"A", "B", "C"},emp, x);

Answer 1

那里有很多小paper-cuts害死你了:)

工作示例：

Live On Coliru

#include <boost/fusion/adapted.hpp>
#include <boost/spirit/include/karma.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <map>

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

enum TYPEX { AUTHOR1, AUTHOR2, AUTHOR3, AUTHOR4 };

std::map<TYPEX, std::string> author2name;
struct Emp {
    std::string name;
    TYPEX author;
};

BOOST_FUSION_ADAPT_STRUCT(Emp, name, author) // boost 1_59
//BOOST_FUSION_ADAPT_STRUCT(Emp, (std::string, name)(std::string, author)) // older boost

int main() {
    using it = boost::spirit::ostream_iterator;

    karma::rule<it, std::string()> quote;
    karma::rule<it, TYPEX()> author;
    karma::rule<it, Emp()> emp;

    {
        using namespace karma;
        quote  %= '"' << string << '"';
        author  = quote [ _1 = phx::ref(author2name)[ _val ] ];
        emp    %= delimit('\t')[ quote << author ];
    }

    Emp x { "one", AUTHOR2 };
    author2name[AUTHOR2] = "TWO!";
    std::cout << karma::format(emp, x);
}

打印：

"one"   "TWO!"

惹事的事情：

建议使用 boost::spirit::ostream_iterator 和 karma::format 更多 user-friendly API
在 emp 上添加缺少的括号：
```
karma::rule<it, Emp()> emp;
```
NOTE: very recent boost (1_59 IIRC) doesn't not require these anymore. Which is why I found out only on Coliru
此处：
```
quote[ author2name[ karma::_1] ]
```
您使用 ... qi::_1 将 [] 索引到 std::map 中。那不能编译。您想要的是调用 operator[] 的 Phoenix 惰性表达式模板。您必须包括 Phoenix header 并强制 author2name 成为 Phoenix 参考演员：
```
quote [ _1 = phx::ref(author2name)[_1] ]
```
Note also, assigning back to _1 is important!
此外，要在存在语义操作的情况下使用 auto-rule，您需要使用 %=（否则 Karma 将抑制所有自动属性传播)

提升业力语义动作调用图

boost karma semantic action call map

boost

boost-spirit-karma