使用局部变量增强精神 [英] Boost spirit using local variables

查看:108
本文介绍了使用局部变量增强精神的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我想基于先前解析的值i定义一个规则. e.输入字符串具有以下结构:D <double number>I <integer number>.无论第一个读取字符是D还是I,我都保留在本地布尔变量中.完整的代码是:

I would like to define a rule based on a previously parsed value, i. e. the input string has the following structure: D <double number> or I <integer number>. I keep in a local boolean variable whether the first read character is D or I. The complete code is:

#define BOOST_SPIRIT_USE_PHOENIX_V3
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <string>

namespace qi = boost::spirit::qi;
namespace spirit = boost::spirit;
namespace ascii = boost::spirit::ascii;
using boost::phoenix::ref;

template <typename Iterator>
struct x_grammar : public qi::grammar<Iterator, std::string(), ascii::space_type>
{
public:
    x_grammar() : x_grammar::base_type(start_rule, "x_grammar")
    {
        using namespace qi;
        bool is_int = false;
        start_rule = lit("I")[ref(is_int) = true] | lit("D")[ref(is_int) = false] > digit_rule;
        if(ref(is_int)()) {
            digit_rule = int_[std::cout << "int " << _1 << ".\n"];
        } else {
            digit_rule = double_[std::cout << "double " << _1 << ".\n"];
        }
    }
private:
    qi::rule<Iterator, std::string(), ascii::space_type> start_rule;
    qi::rule<Iterator, std::string(), ascii::space_type> digit_rule;
};

int main()
{
    typedef std::string::const_iterator iter;
    std::string storage("I 5");
    iter it_begin(storage.begin());
    iter it_end(storage.end());
    std::string read_data;
    using boost::spirit::ascii::space;
    x_grammar<iter> g;
    try {
        bool r = qi::phrase_parse(it_begin, it_end, g, space, read_data);
        if(r) {
            std::cout << "Pass!\n";
        } else {
            std::cout << "Fail!\n";
        }
    } catch (const qi::expectation_failure<iter>& x) {
        std::cout << "Fail!\n";
    }
    return 0;
}                                                                                                                                                    

输出为:double Pass!!它既不识别if语句,也不打印已解析的数字!

The output is: double Pass! !! It neither recognizes the if statement, nor prints the parsed number!

注意:我知道还有其他简单的方法可以解析上述示例.我必须解析的实际字符串看起来很复杂,此示例仅说明了我想要实现的目标.总体目标是使用局部变量并基于这些变量定义其他规则.

Note: I know that there are other straightforward ways to parse the example above. The actual string I have to parse looks quite complicated, and this example just illustrates what I want to achieve. The general goal is to use local variables and define other rules based on those variables.

我使用了4.6.1和Boost 1.55版本.

I have used 4.6.1 and Boost 1.55 versions.

推荐答案

    if(ref(is_int)()) {

在这里您可以评估施工过程中的状况.这不是它的工作方式.规则将始终采用相同的分支.

here you evaluate condition during construction. This is not how it works. The rule will always take the same branch.

相反,请看一下Nabialek技巧: http://boost-spirit.com/home/articles/qi-example/nabialek-trick/

Instead, look at the Nabialek trick: http://boost-spirit.com/home/articles/qi-example/nabialek-trick/

以下是完整的Nabialek技巧已应用于您的示例 在Coliru直播 :

Here's the full Nabialek Trick applied to your sample Live On Coliru:

  1. 您需要使std::cout << "int" 懒惰演员(至少包装phx::ref(std::cout)phx::val("int")作为Phoenix演员)

  1. you needed to make std::cout << "int" lazy actors (by wrapping at least phx::ref(std::cout) or phx::val("int") as a Phoenix Actor)

您仍然没有使用属性传播(std::string()),因为在存在语义动作的情况下禁用了该属性传播(请参阅上一个答案).不过,您可以传播Nabialek子规则中的值:

you still have no use for the attribute propagation (std::string()) since it is disabled in the presence of Semantic Actions (see the previous answer). You can propagate values from Nabialek subrules, though:

#define BOOST_SPIRIT_USE_PHOENIX_V3
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <string>

namespace qi = boost::spirit::qi;
namespace spirit = boost::spirit;
namespace ascii = boost::spirit::ascii;
namespace phx = boost::phoenix;
using boost::phoenix::ref;


template <typename Iterator>
struct x_grammar : public qi::grammar<Iterator, ascii::space_type, qi::locals<qi::rule<Iterator, ascii::space_type>*> >
{
public:
    x_grammar() : x_grammar::base_type(start_rule, "x_grammar")
    {
        using namespace qi;

        int_rule = int_   [std::cout << phx::val("int ") << _1 << ".\n"];
        dbl_rule = double_[std::cout << phx::val("double ") << _1 << ".\n"];
        subrules.add
            ("I", &int_rule)
            ("D", &dbl_rule);

        start_rule = subrules[_a = _1] >> lazy(*_a);
    }
private:
    typedef qi::rule<Iterator, ascii::space_type> subrule;

    qi::symbols<char, subrule*> subrules;
    qi::rule<Iterator, ascii::space_type, qi::locals<subrule*> > start_rule;
    qi::rule<Iterator, ascii::space_type> int_rule, dbl_rule;
};

int main()
{
    typedef std::string::const_iterator iter;
    std::string storage("I 5");
    iter it_begin(storage.begin());
    iter it_end(storage.end());
    using boost::spirit::ascii::space;
    x_grammar<iter> g;
    try {
        bool r = qi::phrase_parse(it_begin, it_end, g, space);
        if (r) {
            std::cout << "Pass!\n";
        }
        else {
            std::cout << "Fail!\n";
        }
    }
    catch (const qi::expectation_failure<iter>&) {
        std::cout << "Fail!\n";
    }
    return 0;
}

这篇关于使用局部变量增强精神的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆