Question

我有一个非常简单的路径构造，我试图用boost spirit.lex解析。

我们有以下语法：

token := [a-z]+
path := (token : path) | (token)

所以我们这里只讨论冒号分隔的小写ASCII字符串。

我有三个例子“xyz”，“abc：xyz”，“abc：xyz：”。

前两项应视为有效。第三个，有一个尾随结肠，不应被视为有效。不幸的是，解析器我已经认识到这三个都是有效的。语法不应该允许空令牌，但显然精神就是这样做的。为了让第三个被拒绝，我错过了什么？

此外，如果您阅读下面的代码，则在注释中还有另一个版本的解析器要求所有路径以分号结尾。当我激活这些行时，我可以得到适当的行为（即拒绝“abc：xyz：;”），但这不是我想要的。

有人有什么想法吗？

感谢。

#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/lex_lexertl.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>

#include <iostream>
#include <string>

using namespace boost::spirit;
using boost::phoenix::val;

template<typename Lexer>
struct PathTokens : boost::spirit::lex::lexer<Lexer>
{
      PathTokens()
      {
         identifier = "[a-z]+";
         separator = ":";

         this->self.add
            (identifier)
            (separator)
            (';')
            ;
      }
      boost::spirit::lex::token_def<std::string> identifier, separator;
};


template <typename Iterator>
struct PathGrammar 
   : boost::spirit::qi::grammar<Iterator> 
{
      template <typename TokenDef>
      PathGrammar(TokenDef const& tok)
         : PathGrammar::base_type(path)
      {
         using boost::spirit::_val;
         path
            = 
            (token >> tok.separator >> path)[std::cerr << _1 << "\n"]
            |
            //(token >> ';')[std::cerr << _1 << "\n"]
            (token)[std::cerr << _1 << "\n"]
             ; 

          token 
             = (tok.identifier) [_val=_1]
          ;

      }
      boost::spirit::qi::rule<Iterator> path;
      boost::spirit::qi::rule<Iterator, std::string()> token;
};


int main()
{
   typedef std::string::iterator BaseIteratorType;
   typedef boost::spirit::lex::lexertl::token<BaseIteratorType, boost::mpl::vector<std::string> > TokenType;
   typedef boost::spirit::lex::lexertl::lexer<TokenType> LexerType;
   typedef PathTokens<LexerType>::iterator_type TokensIterator;
   typedef std::vector<std::string> Tests;

   Tests paths;
   paths.push_back("abc");
   paths.push_back("abc:xyz");
   paths.push_back("abc:xyz:");
   /*
     paths.clear();
     paths.push_back("abc;");
     paths.push_back("abc:xyz;");
     paths.push_back("abc:xyz:;");
   */
   for ( Tests::iterator iter = paths.begin(); iter != paths.end(); ++iter )
   {
      std::string str = *iter;
      std::cerr << "*****" << str << "*****\n";

      PathTokens<LexerType> tokens;
      PathGrammar<TokensIterator> grammar(tokens);

      BaseIteratorType first = str.begin();
      BaseIteratorType last = str.end();

      bool r = boost::spirit::lex::tokenize_and_parse(first, last, tokens, grammar);

      std::cerr << r << " " << (first==last) << "\n";
   }
}

Answer 1

除了llonesmiz已经说过的内容之外，这里有一个使用qi::eoi的技巧，我有时会使用它：

path = (
           (token >> tok.separator >> path) [std::cerr << _1 << "\n"]
         | token                           [std::cerr << _1 << "\n"]
    ) >> eoi;

这使得语法需要 eoi（输入结束）在成功匹配结束时。这导致了期望的结果：

http://liveworkspace.org/code/23a7adb11889bbb2825097d7c553f71d

*****abc*****
abc
1 1
*****abc:xyz*****
xyz
abc
1 1
*****abc:xyz:*****
xyz
abc
0 1

Answer 2

问题在于您致电first后last和tokenize_and_parse的含义。 first==last检查您的字符串是否已完全标记化，您无法推断语法的任何内容。如果您像这样隔离解析，则获得预期结果：

  PathTokens<LexerType> tokens;
  PathGrammar<TokensIterator> grammar(tokens);

  BaseIteratorType first = str.begin();
  BaseIteratorType last = str.end();

  LexerType::iterator_type lexfirst = tokens.begin(first,last);
  LexerType::iterator_type lexlast = tokens.end();


  bool r = parse(lexfirst, lexlast, grammar);

  std::cerr << r << " " << (lexfirst==lexlast) << "\n";

Answer 3

这就是我最终的结果。它使用了@sehe和@llonesmiz的建议。请注意转换为std :: wstring以及语法定义中的操作的使用，这些操作在原始帖子中不存在。

#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/lex_lexertl.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/bind.hpp>

#include <iostream>
#include <string>

//
// This example uses boost spirit to parse a simple
// colon-delimited grammar.
//
// The grammar we want to recognize is:
//    identifier := [a-z]+
//    separator = :
//    path= (identifier separator path) | identifier
//
// From the boost spirit perspective this example shows
// a few things I found hard to come by when building my
// first parser.
//    1. How to flag an incomplete token at the end of input
//       as an error. (use of boost::spirit::eoi)
//    2. How to bind an action on an instance of an object
//       that is taken as input to the parser.
//    3. Use of std::wstring.
//    4. Use of the lexer iterator.
//

// This using directive will cause issues with boost::bind
// when referencing placeholders such as _1.
// using namespace boost::spirit;

//! A class that tokenizes our input.
template<typename Lexer>
struct Tokens : boost::spirit::lex::lexer<Lexer>
{
      Tokens()
      {
         identifier = L"[a-z]+";
         separator = L":";

         this->self.add
            (identifier)
            (separator)
            ;
      }
      boost::spirit::lex::token_def<std::wstring, wchar_t> identifier, separator;
};

//! This class provides a callback that echoes strings to stderr.
struct Echo
{
      void echo(boost::fusion::vector<std::wstring> const& t) const
      {
         using namespace boost::fusion;
         std::wcerr << at_c<0>(t) << "\n";
      }
};


//! The definition of our grammar, as described above.
template <typename Iterator>
struct Grammar : boost::spirit::qi::grammar<Iterator> 
{
      template <typename TokenDef>
      Grammar(TokenDef const& tok, Echo const& e)
         : Grammar::base_type(path)
      {
         using boost::spirit::_val;
         path
            = 
            ((token >> tok.separator >> path)[boost::bind(&Echo::echo, e,::_1)]
             |
             (token)[boost::bind(&Echo::echo, &e, ::_1)]
             ) >> boost::spirit::eoi; // Look for end of input.

          token 
             = (tok.identifier) [_val=boost::spirit::qi::_1]
          ;

      }
      boost::spirit::qi::rule<Iterator> path;
      boost::spirit::qi::rule<Iterator, std::wstring()> token;
};


int main()
{
   // A set of typedefs to make things a little clearer. This stuff is
   // well described in the boost spirit documentation/examples.
   typedef std::wstring::iterator BaseIteratorType;
   typedef boost::spirit::lex::lexertl::token<BaseIteratorType, boost::mpl::vector<std::wstring> > TokenType;
   typedef boost::spirit::lex::lexertl::lexer<TokenType> LexerType;
   typedef Tokens<LexerType>::iterator_type TokensIterator;
   typedef LexerType::iterator_type LexerIterator;

   // Define some paths to parse.
   typedef std::vector<std::wstring> Tests;
   Tests paths;
   paths.push_back(L"abc");
   paths.push_back(L"abc:xyz");
   paths.push_back(L"abc:xyz:");
   paths.push_back(L":");

   // Parse 'em.
   for ( Tests::iterator iter = paths.begin(); iter != paths.end(); ++iter )
   {
      std::wstring str = *iter;
      std::wcerr << L"*****" << str << L"*****\n";

      Echo e;
      Tokens<LexerType> tokens;
      Grammar<TokensIterator> grammar(tokens, e);

      BaseIteratorType first = str.begin();
      BaseIteratorType last = str.end();

      // Have the lexer consume our string.
      LexerIterator lexFirst = tokens.begin(first, last);
      LexerIterator lexLast = tokens.end();

      // Have the parser consume the output of the lexer.
      bool r = boost::spirit::qi::parse(lexFirst, lexLast, grammar);

      // Print the status and whether or note all output of the lexer 
      // was processed.
      std::wcerr << r << L" " << (lexFirst==lexLast) << L"\n";
   }
}

尽管令牌不完整，但提升精神意味着成功解析

3 个答案: