我正在尝试使用 Boost Spirit 来解析以下语法: 句子: 名词动词 句子连句
连词: “和”
名词: “鸟类” “猫”
动词: “飞” “喵”
当语法仅包含名词 >> 动词规则时,解析成功。 当语法修改为包括句子>>连接>>句子规则并且我提供了无效的输入(例如“birds Fly”而不是“birdsfly”)时,我在程序运行时收到未处理的异常。
这是根据 boost 文档中的示例修改的代码
#define BOOST_VARIANT_MINIMIZE_SIZE
#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/spirit/include/phoenix_statement.hpp>
#include <boost/spirit/include/phoenix_container.hpp>
#include <iostream>
#include <string>
using namespace boost::spirit;
using namespace boost::spirit::ascii;
template <typename Lexer>
struct token_list : lex::lexer<Lexer>
{
token_list()
{
noun = "birds|cats";
verb = "fly|meow";
conjunction = "and";
this->self.add
(noun)
(verb)
(conjunction)
;
}
lex::token_def<std::string> noun, verb, conjunction;
};
template <typename Iterator>
struct Grammar : qi::grammar<Iterator>
{
template <typename TokenDef>
Grammar(TokenDef const& tok)
: Grammar::base_type(sentence)
{
sentence = (tok.noun>>tok.verb)
|
(sentence>>tok.conjunction>>sentence)>>eoi
;
}
qi::rule<Iterator> sentence;
};
int main()
{
typedef lex::lexertl::token<char const*, boost::mpl::vector<std::string>> token_type;
typedef lex::lexertl::lexer<token_type> lexer_type;
typedef token_list<lexer_type>::iterator_type iterator_type;
token_list<lexer_type> word_count;
Grammar<iterator_type> g (word_count);
std::string str = "birdsfly";
//std::string str = "birds fly"; this input caused unhandled exception
char const* first = str.c_str();
char const* last = &first[str.size()];
bool r = lex::tokenize_and_parse(first, last, word_count, g);
if (r) {
std::cout << "Parsing passed"<< "\n";
}
else {
std::string rest(first, last);
std::cerr << "Parsing failed\n" << "stopped at: \""
<< rest << "\"\n";
}
system("PAUSE");
return 0;
}
最佳答案
您在句子
规则的第二个分支中具有左递归。
sentence = sentence >> ....
将始终在句子上递归,因此您会看到堆栈溢出。
我建议编写规则,例如:
sentence =
(tok.noun >> tok.verb)
>> *(tok.conjunction >> sentence)
>> qi::eoi
;
现在结果如下
g++ -Wall -pedantic -std=c++0x -g -O0 test.cpp -o test
Parsing failed
stopped at: " fly"
(当然还有不可避免的“sh:暂停:找不到命令”......)
PS。请不要使用命名空间
。相反:
namespace qi = boost::spirit::qi;
namespace lex = boost::spirit::lex;
这是一个清理版本,删除/修复了一些其他内容:http://coliru.stacked-crooked.com/view?id=1fb26ca3e8c207979eaaf4592c319316-e223fd4a885a77b520bbfe69dda8fb91
#define BOOST_VARIANT_MINIMIZE_SIZE
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/lex_lexertl.hpp>
// #include <boost/spirit/include/phoenix.hpp>
#include <iostream>
#include <string>
namespace qi = boost::spirit::qi;
namespace lex = boost::spirit::lex;
template <typename Lexer>
struct token_list : lex::lexer<Lexer>
{
token_list()
{
noun = "birds|cats";
verb = "fly|meow";
conjunction = "and";
this->self.add
(noun)
(verb)
(conjunction)
;
}
lex::token_def<std::string> noun, verb, conjunction;
};
template <typename Iterator>
struct Grammar : qi::grammar<Iterator>
{
template <typename TokenDef>
Grammar(TokenDef const& tok) : Grammar::base_type(sentence)
{
sentence =
(tok.noun >> tok.verb)
>> *(tok.conjunction >> sentence)
>> qi::eoi
;
}
qi::rule<Iterator> sentence;
};
int main()
{
typedef std::string::const_iterator It;
typedef lex::lexertl::token<It, boost::mpl::vector<std::string>> token_type;
typedef lex::lexertl::lexer<token_type> lexer_type;
typedef token_list<lexer_type>::iterator_type iterator_type;
token_list<lexer_type> word_count;
Grammar<iterator_type> g(word_count);
//std::string str = "birdsfly";
const std::string str = "birds fly";
It first = str.begin();
It last = str.end();
bool r = lex::tokenize_and_parse(first, last, word_count, g);
if (r) {
std::cout << "Parsing passed"<< "\n";
}
else {
std::string rest(first, last);
std::cerr << "Parsing failed\n" << "stopped at: \"" << rest << "\"\n";
}
}
关于c++ - 使用 Boost Spirit 解析语法的未处理异常,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17941355/