与升压:: ::精神和法功放的烦恼;空白 [英] Troubles with boost::spirit::lex & whitespace

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问题描述

我尝试学习使用的boost ::精神。要做到这一点,我想创造一些简单的词法分析,将它们组合起来,然后开始使用精神分析。我试图修改的例子,但预期不运行(结果r是不是真的)。​​

I try learning to use boost::spirit. To do that, I wanted to create some simple lexer, combine them and then start parsing using spirit. I tried modifying the example, but it doesn't run as expected (the result r isn't true).

下面是词法分析器:

#include <boost/spirit/include/lex_lexertl.hpp>

namespace lex = boost::spirit::lex;

template <typename Lexer>
struct lexer_identifier : lex::lexer<Lexer>
{
    lexer_identifier()
        : identifier("[a-zA-Z_][a-zA-Z0-9_]*")
        , white_space("[ \\t\\n]+")
    {
        using boost::spirit::lex::_start;
        using boost::spirit::lex::_end;

        this->self = identifier;
        this->self("WS") = white_space;
    }
    lex::token_def<> identifier;
    lex::token_def<> white_space;
    std::string identifier_name;
};

这是我试图运行示例:

And this is the example I'm trying to run:

#include "stdafx.h"

#include <boost/spirit/include/lex_lexertl.hpp>
#include "my_Lexer.h"

namespace lex = boost::spirit::lex;

int _tmain(int argc, _TCHAR* argv[])
{
    typedef lex::lexertl::token<char const*,lex::omit, boost::mpl::false_> token_type;
    typedef lex::lexertl::lexer<token_type> lexer_type;

    typedef lexer_identifier<lexer_type>::iterator_type iterator_type;

    lexer_identifier<lexer_type> my_lexer;

    std::string test("adedvied das934adf dfklj_03245");

    char const* first = test.c_str();
    char const* last = &first[test.size()];

    lexer_type::iterator_type iter = my_lexer.begin(first, last);
    lexer_type::iterator_type end = my_lexer.end();

    while (iter != end && token_is_valid(*iter))
    {
        ++iter;
    }

    bool r = (iter == end);

    return 0;
}

,因为这里仅有一个字符串内部令牌r是只要真。为什么会出现这样的情况?

r is true as long as there is only one token inside the string. Why is this the case?

问候
托比亚斯

Regards Tobias

推荐答案

您已经创建了一个词法分析器第二状态,但永远不会调用它。

You have created a second lexer state, but never invoked it.

在大多数情况下,能有预期的效果最简单的方法是使用单态上可跳过的令牌的 pass_ignore 标志乐星:

For most cases, the easiest way to have the desired effect would be to use single-state lexing with a pass_ignore flag on the skippable tokens:

    this->self += identifier
                | white_space [ lex::_pass = lex::pass_flags::pass_ignore ];

请注意,这需要一个 actor_lexer 来允许语义动作:

Note that this requires an actor_lexer to allow for the semantic action:

typedef lex::lexertl::actor_lexer<token_type> lexer_type;

全样本:

#include <boost/spirit/include/lex_lexertl.hpp>
#include <boost/spirit/include/lex_lexertl.hpp>
namespace lex = boost::spirit::lex;

template <typename Lexer>
struct lexer_identifier : lex::lexer<Lexer>
{
    lexer_identifier()
        : identifier("[a-zA-Z_][a-zA-Z0-9_]*")
        , white_space("[ \\t\\n]+")
    {
        using boost::spirit::lex::_start;
        using boost::spirit::lex::_end;

        this->self += identifier
                    | white_space [ lex::_pass = lex::pass_flags::pass_ignore ];
    }
    lex::token_def<> identifier;
    lex::token_def<> white_space;
    std::string identifier_name;
};

int main(int argc, const char *argv[])
{
    typedef lex::lexertl::token<char const*,lex::omit, boost::mpl::false_> token_type;
    typedef lex::lexertl::actor_lexer<token_type> lexer_type;

    typedef lexer_identifier<lexer_type>::iterator_type iterator_type;

    lexer_identifier<lexer_type> my_lexer;

    std::string test("adedvied das934adf dfklj_03245");

    char const* first = test.c_str();
    char const* last = &first[test.size()];

    lexer_type::iterator_type iter = my_lexer.begin(first, last);
    lexer_type::iterator_type end = my_lexer.end();

    while (iter != end && token_is_valid(*iter))
    {
        ++iter;
    }

    bool r = (iter == end);
    std::cout << std::boolalpha << r << "\n";
}

打印

true

WS作为船长的状态


也可能您在使用第二解析器状态船长的样本来( ::法tokenize_and_phrase_parse )。让我花一分钟或10来创建一个工作样本。

"WS" as a Skipper state


It is also possible you came across a sample that uses the second parser state for the skipper (lex::tokenize_and_phrase_parse). Let me take a minute or 10 to create a working sample for that.

更新我花了一点时间超过10分钟(waaaah):)下面是一个对比试验,展示了如何在词法分析器状态交互,以及如何使用精神船长解析调用第二分析器状态:

Update Took me a bit more than 10 minutes (waaaah) :) Here's a comparative test, showing how the lexer states interact, and how to use Spirit Skipper parsing to invoke the second parser state:

#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/lex_lexertl.hpp>
namespace lex = boost::spirit::lex;
namespace qi  = boost::spirit::qi;

template <typename Lexer>
struct lexer_identifier : lex::lexer<Lexer>
{
    lexer_identifier()
        : identifier("[a-zA-Z_][a-zA-Z0-9_]*")
        , white_space("[ \\t\\n]+")
    {
        this->self       = identifier;
        this->self("WS") = white_space;
    }
    lex::token_def<> identifier;
    lex::token_def<lex::omit> white_space;
};

int main()
{
    typedef lex::lexertl::token<char const*, lex::omit, boost::mpl::true_> token_type;
    typedef lex::lexertl::lexer<token_type> lexer_type;

    typedef lexer_identifier<lexer_type>::iterator_type iterator_type;

    lexer_identifier<lexer_type> my_lexer;

    std::string test("adedvied das934adf dfklj_03245");

    {
        char const* first = test.c_str();
        char const* last = &first[test.size()];

        // cannot lex in just default WS state:
        bool ok = lex::tokenize(first, last, my_lexer, "WS");
        std::cout << "Starting state WS:\t" << std::boolalpha << ok << "\n";
    }

    {
        char const* first = test.c_str();
        char const* last = &first[test.size()];

        // cannot lex in just default state either:
        bool ok = lex::tokenize(first, last, my_lexer, "INITIAL");
        std::cout << "Starting state INITIAL:\t" << std::boolalpha << ok << "\n";
    }

    {
        char const* first = test.c_str();
        char const* last = &first[test.size()];

        bool ok = lex::tokenize_and_phrase_parse(first, last, my_lexer, *my_lexer.self, qi::in_state("WS")[my_lexer.self]);
        ok = ok && (first == last); // verify full input consumed
        std::cout << std::boolalpha << ok << "\n";
    }
}

的输出是

Starting state WS:  false
Starting state INITIAL: false
true

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