c – 我无法获取令牌的字符串值

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我尝试用Boost Spirit实现一种使用一种编程语言的Lexer.

我得到一个令牌的值,我得到一个bad_get异常:

terminate called after throwing an instance of ‘boost::bad_get’
what(): boost::bad_get: Failed value get using boost::get Aborted

在这样做时我得到这个例外:

std::string contents = "void";

base_iterator_type first = contents.begin();
base_iterator_type last = contents.end();

SimpleLexer<lexer_type> lexer;

iter = lexer.begin(first,last);
end = lexer.end();

std::cout << "Value = " << boost::get<std::string>(iter->value()) << std::endl;

我的词法分类是这样定义的:

typedef std::string::iterator base_iterator_type;
typedef boost::spirit::lex::lexertl::token<base_iterator_type,boost::mpl::vector<unsigned int,std::string>> Tok;
typedef lex::lexertl::actor_lexer<Tok> lexer_type;

template<typename L>
class SimpleLexer : public lex::lexer<L> {
    private:

    public:
        SimpleLexer() {
            keyword_for = "for";
            keyword_while = "while";
            keyword_if = "if";
            keyword_else = "else";
            keyword_false = "false";
            keyword_true = "true";
            keyword_from = "from";
            keyword_to = "to";
            keyword_foreach = "foreach";

            word = "[a-zA-Z]+";
            integer = "[0-9]+";
            litteral = "...";

            left_parenth = '('; 
            right_parenth = ')'; 
            left_brace = '{'; 
            right_brace = '}'; 

            stop = ';';
            comma = ',';

            swap = "<>";
            assign = '=';
            addition = '+';
            subtraction = '-';
            multiplication = '*';
            division = '/';
            modulo = '%';

            equals = "==";
            not_equals = "!=";
            greater = '>';
            less = '<';
            greater_equals = ">=";
            less_equals = "<=";

            whitespaces = "[ \\t\\n]+";
            comments = "\\/\\*[^*]*\\*+([^/*][^*]*\\*+)*\\/";

            //Add keywords
            this->self += keyword_for | keyword_while | keyword_true | keyword_false | keyword_if | keyword_else | keyword_from | keyword_to | keyword_foreach;
            this->self += integer | litteral | word;

            this->self += equals | not_equals | greater_equals | less_equals | greater | less ;
            this->self += left_parenth | right_parenth | left_brace | right_brace;
            this->self += comma | stop;
            this->self += assign | swap | addition | subtraction | multiplication | division | modulo;

            //Ignore whitespaces and comments
            this->self += whitespaces [lex::_pass = lex::pass_flags::pass_ignore];
            this->self += comments [lex::_pass = lex::pass_flags::pass_ignore]; 
        }

        lex::token_def<std::string> word,litteral,integer;

        lex::token_def<lex::omit> left_parenth,right_parenth,left_brace,right_brace;

        lex::token_def<lex::omit> stop,comma;

        lex::token_def<lex::omit> assign,swap,addition,subtraction,multiplication,division,modulo;
        lex::token_def<lex::omit> equals,not_equals,greater,less,greater_equals,less_equals;

        //Keywords
        lex::token_def<lex::omit> keyword_if,keyword_else,keyword_for,keyword_while,keyword_from,keyword_to,keyword_foreach;
        lex::token_def<lex::omit> keyword_true,keyword_false;

        //Ignored tokens
        lex::token_def<lex::omit> whitespaces;
        lex::token_def<lex::omit> comments;
};

有没有办法获得令牌的价值?

解决方法

您可以随时使用“默认”令牌数据(这是源迭代器类型的iterator_range).
std::string tokenvalue(iter->value().begin(),iter->value().end());

在boost库中研究了测试用例之后,我发现了一些事情:

>这是设计
>有一个更简单的方法
>在Lex语义动作(例如使用_1)和在Qi中使用词法分析器令牌时,更容易的方式自动化;该赋值将自动转换为Qi属性类型
>这个(确实)得到了文档中提到的“懒惰,一次性,评估”的语义

令牌是令牌数据是变体,它作为原始输入迭代器范围开始.只有在“a”强制分配之后,转换的属性才会被缓存在变体中.你可以看到过渡:

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

assert(0 == iter->value().which());
std::cout << "Value = " << boost::get<boost::iterator_range<base_iterator_type> >(iter->value()) << std::endl;

std::string s;
boost::spirit::traits::assign_to(*iter,s);
assert(1 == iter->value().which());
std::cout << "Value = " << s << std::endl;

如您所见,属性分配在这里被强制直接使用assign_to trait实现.

全程演示:

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

#include <iostream>
#include <string>

namespace lex = boost::spirit::lex;

typedef std::string::iterator base_iterator_type;
typedef boost::spirit::lex::lexertl::token<base_iterator_type,boost::mpl::vector<int,std::string>> Tok;
typedef lex::lexertl::actor_lexer<Tok> lexer_type;

template<typename L>
class SimpleLexer : public lex::lexer<L> {
    private:

    public:
        SimpleLexer() {
            word = "[a-zA-Z]+";
            integer = "[0-9]+";
            literal = "...";

            this->self += integer | literal | word;
        }

        lex::token_def<std::string> word,literal;
        lex::token_def<int> integer;
};

int main(int argc,const char* argv[]) {
    SimpleLexer<lexer_type> lexer;

    std::string contents = "void";

    base_iterator_type first = contents.begin();
    base_iterator_type last = contents.end();

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

    assert(0 == iter->value().which());
    std::cout << "Value = " << boost::get<boost::iterator_range<base_iterator_type> >(iter->value()) << std::endl;

    std::string s;
    boost::spirit::traits::assign_to(*iter,s);
    assert(2 == iter->value().which());
    std::cout << "Value = " << s << std::endl;

    return 0;
}

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