下面的代码(改编自 spirit qi mini_xml 示例)无法编译。存在与具有递归 boost::variant
属性的规则 brac
相关的错误。
但是,所有注释掉的 brac
版本都可以编译。
我很好奇是什么让简单的字符串解析器在这种情况下如此特别:
#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/spirit/include/phoenix_fusion.hpp>
#include <boost/spirit/include/phoenix_stl.hpp>
#include <boost/spirit/include/phoenix_object.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/variant/recursive_variant.hpp>
#include <string>
#include <vector>
namespace client
{
namespace fusion = boost::fusion;
namespace phoenix = boost::phoenix;
namespace qi = boost::spirit::qi;
namespace ascii = boost::spirit::ascii;
struct ast_node;
typedef boost::variant<
boost::recursive_wrapper<ast_node>,
std::string
> ast_branch;
struct ast_node
{
std::string text;
std::vector<ast_branch> children;
};
}
BOOST_FUSION_ADAPT_STRUCT(
client::ast_node,
(std::string, text)
(std::vector<client::ast_branch>, children)
)
namespace client
{
template <typename Iterator>
struct ast_node_grammar
: qi::grammar<Iterator, ast_branch(), ascii::space_type>
{
ast_node_grammar()
: ast_node_grammar::base_type(brac)
{
using qi::_1;
using qi::_val;
using ascii::char_;
using ascii::string;
name %= *char_;
brac %= string("no way") ;
// brac = string("works")[_val = _1] ;
// brac %= string("this") | string("works");
// brac %= name ; // works
// brac %= *char_ ; // works
}
qi::rule<Iterator, std::string()> name;
qi::rule<Iterator, ast_branch(), ascii::space_type> brac;
};
}
int main(int argc, char **argv)
{
typedef client::ast_node_grammar<std::string::const_iterator> ast_node_grammar;
ast_node_grammar gram;
client::ast_branch ast;
std::string text("dummy");
using boost::spirit::ascii::space;
std::string::const_iterator iter = text.begin();
std::string::const_iterator end = text.end();
bool r = phrase_parse(iter, end, gram, space, ast);
return r ? 0 : 1;
}
部分错误信息:
/usr/include/boost/spirit/home/qi/detail/assign_to.hpp:38:17: error: No match for ‘boost::variant<
boost::recursive_wrapper<client::ast_node>, basic_string<char>
>::variant(
const __normal_iterator<const char *, basic_string<char> > &, const __normal_iterator<
const char *, basic_string<char> > &)’
提前致谢。
最佳答案
我认为问题出在属性兼容性上。与the documentation相反, ascii::string 解析器似乎暴露了一个迭代器范围而不是一个字符串。
name = string("no way");
没问题,因为可以毫无困难地将 ascii::string 公开的属性强制转换为规则的属性类型。
但是,brac
规则的属性类型是 ast_branch
,它只是一个变体,它可能包含一种类型。因此,ast_branch
类型有多个构造函数,Spirit 不清楚哪一个适合这个特定的转换。
有几种方法(除了您已经展示的方法):
使用
attr_cast
brac = qi::attr_cast( string("no way") );
使用
as_string
brac = qi::as_string[ string("no way") ];
使用自定义点
namespace boost { namespace spirit { namespace traits { template <typename It> struct assign_to_attribute_from_iterators<client::ast_branch, It> { static void call(It const& f, It const& l, client::ast_branch& val) { val = std::string(f, l); } }; }}}
这些都具有相同的作用:让Spirit意识到使用什么属性转换。
这是一个完整的工作示例,显示了所有三个:
// #define BOOST_SPIRIT_ACTIONS_ALLOW_ATTR_COMPAT
#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/adapted.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/variant/recursive_variant.hpp>
#include <string>
#include <vector>
namespace client
{
namespace fusion = boost::fusion;
namespace phoenix = boost::phoenix;
namespace qi = boost::spirit::qi;
namespace ascii = boost::spirit::ascii;
struct ast_node;
typedef boost::variant<
boost::recursive_wrapper<ast_node>,
std::string
> ast_branch;
struct ast_node
{
std::string text;
std::vector<ast_branch> children;
};
}
namespace boost { namespace spirit { namespace traits {
template <typename It>
struct assign_to_attribute_from_iterators<client::ast_branch, It>
{
static void call(It const& f, It const& l, client::ast_branch& val)
{
val = std::string(f, l);
}
};
}}}
BOOST_FUSION_ADAPT_STRUCT(
client::ast_node,
(std::string, text)
(std::vector<client::ast_branch>, children)
)
namespace client
{
template <typename Iterator>
struct ast_node_grammar : qi::grammar<Iterator, ast_branch(), ascii::space_type>
{
ast_node_grammar()
: ast_node_grammar::base_type(brac)
{
using qi::_1;
using qi::_val;
using ascii::char_;
using ascii::string;
name %= *char_;
brac = string("works");
brac = string("works")[_val = _1] ;
brac %= string("this") | string("works");
brac %= name ; // works
brac %= *char_ ; // works
brac = qi::as_string[ string("no way") ];
brac = qi::attr_cast( string("no way") );
}
qi::rule<Iterator, std::string()> name;
qi::rule<Iterator, ast_branch(), ascii::space_type> brac;
};
}
int main(int argc, char **argv)
{
typedef client::ast_node_grammar<std::string::const_iterator> ast_node_grammar;
ast_node_grammar gram;
client::ast_branch ast;
std::string text("dummy");
using boost::spirit::ascii::space;
std::string::const_iterator iter = text.begin();
std::string::const_iterator end = text.end();
bool r = phrase_parse(iter, end, gram, space, ast);
return r ? 0 : 1;
}
关于c++ - 带有 boost 变体递归包装器的字符串解析器,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11421430/