Question

我想用Boost.Spirit x3解析一个递归语法，但它失败了，模板实例化深度问题。

语法如下：

value: int | float | char | tuple
int: "int: " int_
float: "float: " real_ 
char: "char: " char_
tuple: "tuple: [" value* "]"

这是一个包含的例子：

#include <boost/fusion/adapted.hpp>
#include <boost/spirit/home/x3.hpp>
#include <string>
#include <vector>
#include <variant>

struct value: std::variant<int,float,std::vector<value>>
{ 
    using std::variant<int,float,std::vector<value>>::variant;

    value& operator=(float) { return *this; } 
    value& operator=(int) { return *this; } 
    value& operator=(std::vector<value>) { return *this; } 
};

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

using x3::skip;
using x3::int_;
using x3::real_parser;
using x3::char_;

x3::rule<class value_, value> const value_ = "value";
x3::rule<class o_tuple_, std::vector<value>> o_tuple_ = "tuple";

using float_p = real_parser<float, x3::strict_real_policies<float>>;


const auto o_tuple__def = "tuple: " >> skip(boost::spirit::x3::space)["[" >> value_ % "," >> "]"];
BOOST_SPIRIT_DEFINE(o_tuple_)

const auto value__def
    = ("float: " >> float_p())
    | ("int: " >> int_)
    | o_tuple_
    ;

BOOST_SPIRIT_DEFINE(value_)

int main()
{
  std::string str;
  value val;

  using boost::spirit::x3::parse;
  auto first = str.cbegin(), last = str.cend();
  bool r = parse(first, last, value_, val);
}

如果注释了行| o_tuple_（例如没有递归），则此方法有效。

Answer 1

这是X3中递归的常见问题。它尚未解决。

我想我理解这个问题是因为x3::skip改变了上下文对象¹。确实，删除它会使事情编译，并成功解析一些琐碎的测试用例：

"float: 3.14",
"int: 3.14",
"tuple: [float: 3.14,int: 3]",

但是，如果没有船长，显然以下内容不会解析：

// the following _should_ have compiled with the original skip() configuration:
"tuple: [ float: 3.14,\tint: 3 ]",

现在，我敢说你可以通过在顶层应用队长来解决问题（这意味着上下文对于实例化“循环”中涉及的所有规则是相同的）。如果这样做，您将立即开始接受输入中更灵活的空格：

// the following would not have parsed with the original skip() configuration:
"float:3.14",
"int:3.14",
"tuple:[float: 3.14,int: 3]",
"tuple:[float:3.14,int:3]",
"tuple: [ float:3.14,\tint:3 ]",

即使已经成功编译，这些都不会用原始方法解析。

需要什么

以下是我对代码进行的一些调整。

删除了无效的赋值运算符value::operator=（我不知道你为什么会这样做）

添加代码以打印任何value的调试转储：

friend std::ostream& operator<<(std::ostream& os, base_type const& v) {
    struct {
        std::ostream& operator()(float const& f) const { return _os << "float:" << f; }
        std::ostream& operator()(int const& i)   const { return _os << "int:" << i; }
        std::ostream& operator()(std::vector<value> const& v) const { 
            _os << "tuple: [";
            for (auto& el : v) _os << el << ",";
            return _os << ']';
        }
        std::ostream& _os;
    } vis { os };

    return std::visit(vis, v);
}

删除队长并从: interpunction中分离出关键字：

namespace x3 = boost::spirit::x3;

x3::rule<struct value_class, value> const value_ = "value";
x3::rule<struct o_tuple_class, std::vector<value> > o_tuple_ = "tuple";

x3::real_parser<float, x3::strict_real_policies<float> > float_;

const auto o_tuple__def = "tuple" >> x3::lit(':') >> ("[" >> value_ % "," >> "]");

const auto value__def
    = "float" >> (':' >> float_)
    | "int" >> (':' >> x3::int_)
    | o_tuple_
    ;

BOOST_SPIRIT_DEFINE(value_, o_tuple_)

现在，关键步骤：在toplevel添加队长：

const auto entry_point = x3::skip(x3::space) [ value_ ];

创建好的测试驱动程序main()：

int main()
{
    for (std::string const str : {
            "",
            "float: 3.14",
            "int: 3.14",
            "tuple: [float: 3.14,int: 3]",
            // the following _should_ have compiled with the original skip() configuration:
            "tuple: [ float: 3.14,\tint: 3 ]",
            // the following would not have parsed with the original skip() configuration:
            "float:3.14",
            "int:3.14",
            "tuple:[float: 3.14,int: 3]",
            "tuple:[float:3.14,int:3]",
            "tuple: [ float:3.14,\tint:3 ]",
            // one final show case for good measure
            R"(
            tuple: [
               int  : 4,
               float: 7e9,
               tuple: [float: -inf],


               int: 42
            ])"
    }) {
        std::cout << "============ '" << str << "'\n";

        //using boost::spirit::x3::parse;
        auto first = str.begin(), last = str.end();
        value val;

        if (parse(first, last, parser::entry_point, val))
            std::cout << "Parsed '" << val << "'\n";
        else
            std::cout << "Parse failed\n";

        if (first != last)
            std::cout << "Remaining input: '" << std::string(first, last) << "'\n";
    }
}

现场演示

查看 Live On Coliru

//#define BOOST_SPIRIT_X3_DEBUG
#include <iostream>
#include <boost/fusion/adapted.hpp>
#include <boost/spirit/home/x3.hpp>
#include <string>
#include <vector>
#include <variant>

struct value: std::variant<int,float,std::vector<value>>
{ 
    using base_type = std::variant<int,float,std::vector<value>>;
    using base_type::variant;

    friend std::ostream& operator<<(std::ostream& os, base_type const& v) {
        struct {
            std::ostream& operator()(float const& f) const { return _os << "float:" << f; }
            std::ostream& operator()(int const& i)   const { return _os << "int:" << i; }
            std::ostream& operator()(std::vector<value> const& v) const { 
                _os << "tuple: [";
                for (auto& el : v) _os << el << ",";
                return _os << ']';
            }
            std::ostream& _os;
        } vis { os };

        return std::visit(vis, v);
    }
};

namespace parser {
    namespace x3 = boost::spirit::x3;

    x3::rule<struct value_class, value> const value_ = "value";
    x3::rule<struct o_tuple_class, std::vector<value> > o_tuple_ = "tuple";

    x3::real_parser<float, x3::strict_real_policies<float> > float_;

    const auto o_tuple__def = "tuple" >> x3::lit(':') >> ("[" >> value_ % "," >> "]");

    const auto value__def
        = "float" >> (':' >> float_)
        | "int" >> (':' >> x3::int_)
        | o_tuple_
        ;

    BOOST_SPIRIT_DEFINE(value_, o_tuple_)

    const auto entry_point = x3::skip(x3::space) [ value_ ];
}

int main()
{
    for (std::string const str : {
            "",
            "float: 3.14",
            "int: 3.14",
            "tuple: [float: 3.14,int: 3]",
            // the following _should_ have compiled with the original skip() configuration:
            "tuple: [ float: 3.14,\tint: 3 ]",
            // the following would not have parsed with the original skip() configuration:
            "float:3.14",
            "int:3.14",
            "tuple:[float: 3.14,int: 3]",
            "tuple:[float:3.14,int:3]",
            "tuple: [ float:3.14,\tint:3 ]",
            // one final show case for good measure
            R"(
            tuple: [
               int  : 4,
               float: 7e9,
               tuple: [float: -inf],


               int: 42
            ])"
    }) {
        std::cout << "============ '" << str << "'\n";

        //using boost::spirit::x3::parse;
        auto first = str.begin(), last = str.end();
        value val;

        if (parse(first, last, parser::entry_point, val))
            std::cout << "Parsed '" << val << "'\n";
        else
            std::cout << "Parse failed\n";

        if (first != last)
            std::cout << "Remaining input: '" << std::string(first, last) << "'\n";
    }
}

打印

============ ''
Parse failed
============ 'float: 3.14'
Parsed 'float:3.14'
============ 'int: 3.14'
Parsed 'int:3'
Remaining input: '.14'
============ 'tuple: [float: 3.14,int: 3]'
Parsed 'tuple: [float:3.14,int:3,]'
============ 'tuple: [ float: 3.14, int: 3 ]'
Parsed 'tuple: [float:3.14,int:3,]'
============ 'float:3.14'
Parsed 'float:3.14'
============ 'int:3.14'
Parsed 'int:3'
Remaining input: '.14'
============ 'tuple:[float: 3.14,int: 3]'
Parsed 'tuple: [float:3.14,int:3,]'
============ 'tuple:[float:3.14,int:3]'
Parsed 'tuple: [float:3.14,int:3,]'
============ 'tuple: [ float:3.14,  int:3 ]'
Parsed 'tuple: [float:3.14,int:3,]'
============ '
            tuple: [
               int  : 4,
               float: 7e9,
               tuple: [float: -inf],


               int: 42
            ]'
Parsed 'tuple: [int:4,float:7e+09,tuple: [float:-inf,],int:42,]'

¹其他指令也是如此，例如x3::with<>。问题是上下文在每个实例化级别上获得扩展，而不是“修改”以获得原始上下文类型，并结束实例化循环。

Spirit.X3中的递归规则

1 个答案:

需要什么

现场演示