用“气”解析帝王价值观
Parsing imperial values using boost spirit (qi)
我是一个精神初学者
我想使用spirit将一个帝国字符串值解析为一个结构体。
输入应该接受以下语法:
<>之前5 ' 3"1/25 ' 1/23"1/2之前 struct imp_constant
看起来是这样的,请注意下面的流操作符,我将打印结果作为这个操作符:
struct imp_constant
{
explicit imp_constant(unsigned int feet=0
,unsigned int inch=0
,unsigned int fracn=0
,unsigned int fracd=1)
:feet_(feet),inches_(inch),fracn_(fracn),fracd_(fracd){}
unsigned int feet_,inches_,fracn_,fracd_;
};
std::ostream& operator<<(std::ostream& os, imp_constant const& cst)
{
if (cst.feet_)
os << cst.feet_ << ''';
if (cst.inches_)
os << cst.inches_ << '"';
if (cst.fracn_)
os << cst.fracn_ << '/' << cst.fracd_;
return os;
}
我的伪语法很简单,看起来像这样:
myrule = (
(
(qi::uint_ >> L''')
||
(qi::uint_ >> L'"')
)
>> -(
qi::uint_
>> L'/' >>
qi::uint_
)
);
这是我第一次尝试填充我的结构体:
我已经添加了BOOST_FUSION_ADAPT_STRUCT
宏到我的struct imp_constant
然后尝试以下语法:
qi::rule<std::string::const_iterator, imp_constant()>
impconst = qi::lexeme[ //not sure this is required since no skipper precised
(
(qi::uint_[phx::at_c<0>(qi::_val)=qi::_1] >> L''')
||
(qi::uint_[phx::at_c<1>(qi::_val)=qi::_1] >> L'"')
)
>> -(
qi::uint_[phx::at_c<2>(qi::_val)=qi::_1]
>> L'/' >>
qi::uint_[phx::at_c<3>(qi::_val)=qi::_1]
)
];
的结果是:
input:5'3"1/2 ==> output:5'3"1/2 (ok)
input:5'1/2 ==> output:5'1"1/2 (__nok__)
我想我不明白_1
占位符在这种情况下是如何表现的。
由于我是精神世界的初学者,欢迎任何建议
Thank you very much
这是完整的代码,这应该有帮助
#define BOOST_SPIRIT_DONT_USE_MPL_ASSERT_MSG 1
//#define BOOST_SPIRIT_DEBUG << uncomment to enable debug
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/variant/recursive_wrapper.hpp>
#include <boost/fusion/adapted.hpp>
namespace qi = boost::spirit::qi;
namespace phx = boost::phoenix;
struct imp_constant
{
explicit imp_constant(unsigned int feet=0
,unsigned int inch=0
,unsigned int fracn=0
,unsigned int fracd=1)
:feet_(feet),inches_(inch),fracn_(fracn),fracd_(fracd){}
unsigned int feet_,inches_,fracn_,fracd_;
};
std::ostream& operator<<(std::ostream& os, imp_constant const& cst)
{
if (cst.feet_)
os << cst.feet_ << ''';
if (cst.inches_)
os << cst.inches_ << '"';
if (cst.fracn_)
os << cst.fracn_ << '/' << cst.fracd_;
return os;
}
BOOST_FUSION_ADAPT_STRUCT(imp_constant,
(unsigned int, feet_)
(unsigned int, inches_)
(unsigned int, fracn_)
(unsigned int, fracd_))
int _tmain(int argc, _TCHAR* argv[])
{
std::string input;
std::cout << "n----------------------n> ";
while (std::getline(std::cin, input))
{
if (input.empty() || input[0] == 'q' || input[0] == 'Q')
break;
std::string::const_iterator f(input.begin()),l(input.end());
try
{
imp_constant result;
std::cout << "parsing: " << input << "n";
bool ok;
qi::rule<std::string::const_iterator, imp_constant()>
impconst = qi::lexeme[ //not sure this is required since
//no skipper precised
(
(qi::uint_[phx::at_c<0>(qi::_val)=qi::_1]
>> L''')
||
(qi::uint_[phx::at_c<1>(qi::_val)=qi::_1]
>> L'"')
)
>> -(
qi::uint_[phx::at_c<2>(qi::_val)=qi::_1]
>> L'/' >>
qi::uint_[phx::at_c<3>(qi::_val)=qi::_1]
)
];
ok=qi::phrase_parse(f,l,impconst ,qi::space,result);
if (!ok)
std::cerr << "invalid inputn";
else
{
std::cout << "n---------------------------n";
std::cout << "result="<< result;
}
}
catch (const qi::expectation_failure<const char *>& e)
{
std::cerr << "expect failure at '"
<< std::string(e.first, e.last) << "'n";
}
catch (...)
{
std::cerr << "parse errorn";
}
if (f!=l) std::cerr << "unparsed: '" << std::string(f,l) << "'n";
std::cout << "n-----------------------n> ";
}
std::getchar();
return 0;
}
作为她的回答(你应该接受)的替代方案:
如果你把你的规则改为:
,你的解决方案就会起作用。qi::rule<std::string::const_iterator, imp_constant()>
impconst =
(
(qi::uint_ >> L''')[phx::at_c<0>(qi::_val)=qi::_1]
||
(qi::uint_ >> L'"')[phx::at_c<1>(qi::_val)=qi::_1]
)
>>
-(qi::uint_ >> L'/' >> qi::uint_)
[phx::at_c<2>(qi::_val)=qi::_1,phx::at_c<3>(qi::_val)=qi::_2]
;
我该怎么做:
稍微改变imp_constant
:
struct fraction
{
unsigned int n_,d_;
};
struct imp_constant
{
unsigned int feet_,inches_;
fraction frac_;
};
BOOST_FUSION_ADAPT_STRUCT(fraction,
(unsigned int, n_)
(unsigned int, d_)
)
BOOST_FUSION_ADAPT_STRUCT(imp_constant,
(unsigned int, feet_)
(unsigned int, inches_)
(fraction , frac_)
那么规则就是:
qi::rule<std::string::const_iterator,unsigned int()> feet = (qi::uint_ >> L''') | qi::attr(0);
qi::rule<std::string::const_iterator,unsigned int()> inches = (qi::uint_ >> L'"') | qi::attr(0);
qi::rule<std::string::const_iterator,fraction()> fract = (qi::uint_ >> L'/' >> qi::uint_) | (qi::attr(0)>> qi::attr(1));
qi::rule<std::string::const_iterator, imp_constant()> impconst=feet>>inches>>fract;
LWS示例
你在看回溯。
首先将5'1/2
中的'1'解析为潜在英寸分支中的qi::uint_
,从而分配属性。
只有然后是遇到的'/',这导致从'inch分支'回溯。但是属性已经设置好了。在这种情况下,显式默认值或qi::hold
可以修复它。
这是我自己可能会用的表达方式:
qi::rule<std::string::const_iterator, unsigned(char)>
component = qi::hold [ qi::uint_ >> qi::lit(qi::_r1) ];
qi::rule<std::string::const_iterator, imp_constant()>
impconst =
component(L''') ||
component(L'"') ||
component(L'/') ||
qi::uint_
;
BOOST_SPIRIT_DEBUG_NODE(component);
BOOST_SPIRIT_DEBUG_NODE(impconst);
我认为这可能是一个很好的起点?
进一步指出:- test
f!=l
检测未解析的剩余输入 - 完整代码示例中的语义动作是冗余的
- 确实,
lexeme
在那里是多余的
----------------------
> parsing: 5'1/2
<impconst>
<try>5'1/2</try>
<component>
<try>5'1/2</try>
<success>1/2</success>
<attributes>[5, ']</attributes>
</component>
<component>
<try>1/2</try>
<fail/>
</component>
<component>
<try>1/2</try>
<success>2</success>
<attributes>[1, /]</attributes>
</component>
<success></success>
<attributes>[[5, 0, 1, 2]]</attributes>
</impconst>
---------------------------
result=5'1/2
-----------------------
>