c++ - 灵气解析为嵌套函数的抽象语法树

Question

我正在尝试使用 boost 的 spirit qi 解析器创建一个解析器。它正在解析包含三种类型值的字符串。常量、变量或函数。这些函数可以相互嵌套。测试字符串为f(a, b) = f(g(z, x), g(x, h(x)), c)，其中a-e为常量，f-r是函数，s-z是变量。我成功地创建了一个可以正确解析表达式的规则。当我将解析规则的函数更改为 grammar 时出现了问题.我能够修复几个错误。我几乎得到语法来解析表达式并将其转换为我创建的抽象语法树。但是我得到了关于 boost 库中包含的文件的错误，我无法弄清楚它来自哪里，因为我不理解编译器消息。我正在按照网站上的示例使用员工示例将数据从解析器放入结构:http://www.boost.org/doc/libs/1_41_0/libs/spirit/example/qi/employee.cpp

main.cpp

#include "Parser.h"
#include "Term.h"

#include <boost/spirit/include/qi.hpp>
#include <string>
#include <iostream>
#include <list>

using std::string;
using std::cout;
using std::endl;

int main()
{
    cout << "Unification Algorithm" << endl << endl;

    string phrase = "f(a, b) = f(g(z, x), g(x, h(x)), c)";
    string::const_iterator itr = phrase.begin();
    string::const_iterator last = phrase.end();

    cout << phrase << endl;

    // Parser grammar
    Parser<string::const_iterator> g;

    // Output data
    Expression expression;

    if (phrase_parse(itr, last, g, boost::spirit::ascii::space, expression))
    {
        cout << "Expression parsed." << endl;
    }
    else
    {
        cout << "Could not parse expression." << endl;
    }
}

Parser.h

#ifndef _Parser_h_
#define _Parser_h_

#include "Term.h"

#include <boost/spirit/include/qi.hpp>
#include <vector>

namespace qi = boost::spirit::qi;
namespace ascii = boost::spirit::ascii;

template <typename Iterator>
struct Parser : qi::grammar<Iterator, Expression(), ascii::space_type>
{
    Parser() : Parser::base_type(expression)
    {
        using qi::char_;

        const_char = char_("a-eA-E");
        fn_char = char_("f-rF-R");
        var_char = char_("s-zS-Z");

        basic_fn = fn_char >> char_('(') >> (const_char | var_char) % char_(',') >> char_(')');
        first_fn_wrapper = fn_char >> char_('(') >> (basic_fn | const_char | var_char) % char_(',') >> char_(')');
        nested_fn = fn_char >> char_('(') >> (first_fn_wrapper | const_char | var_char) % char_(',') >> char_(')');

        expression = nested_fn >> char_("=") >> nested_fn;
    }
    // Constant character a - e
    qi::rule<Iterator, T_Cons, ascii::space_type> const_char;
    // Function character f - r
    qi::rule<Iterator, char(), ascii::space_type> fn_char;
    // Variable character s - z
    qi::rule<Iterator, T_Var, ascii::space_type> var_char;
    // Allows for basic function parsing eg. f(x, y, z)
    qi::rule<Iterator, T_Fn, ascii::space_type> basic_fn;
    // Allows for single nested functions eg. f(g(x), y, z)
    qi::rule<Iterator, T_Fn, ascii::space_type> first_fn_wrapper;
    // Allows for fully nested functions eg. f(g(x, h(y)), z) and so on
    qi::rule<Iterator, T_Fn, ascii::space_type> nested_fn;
    // Full rule for a nested function expression
    qi::rule<Iterator, Expression, ascii::space_type> expression;
};
#endif // _Parser_h_

Term.h

#ifndef _Term_h_
#define _Term_h_

#include <boost/fusion/include/adapt_struct.hpp>
#include <vector>

struct Term
{
    char name;
};

BOOST_FUSION_ADAPT_STRUCT(Term, (char, name))

struct T_Cons : Term
{

};

BOOST_FUSION_ADAPT_STRUCT(T_Cons, (char, name))

struct T_Var : Term
{

};

BOOST_FUSION_ADAPT_STRUCT(T_Var, (char, name))

struct T_Fn : Term
{
    std::vector<Term> * params;

    T_Fn() { params = new std::vector<Term>(); }

    ~T_Fn() { delete params; }
};

BOOST_FUSION_ADAPT_STRUCT(T_Fn, (std::vector<Term>*, params))

struct Expression
{
    Term lh_term;
    Term rh_term;
};

 BOOST_FUSION_ADAPT_STRUCT(Expression, (char, name) (Term, lh_term) (Term, rh_term))

#endif // _Term_h_

我无法链接来自编译器的整个错误消息，因为它非常长，但这里是最后几条。这些是它给出的编译错误:

boost_1_46_0\boost\mpl\assert.hpp|360|error: no matching function for call to 'assertion_failed(mpl_::failed************ (boost::spirit::qi::grammar<Iterator, T1, T2, T3, T4>::grammar(const boost::spirit::qi::rule<Iterator_, T1_, T2_, T3_, T4_>&, const string&) [with Iterator_ = __gnu_cxx::__normal_iterator<const char*, std::basic_string<char> >; T1_ = Expression; T2_ = boost::proto::exprns_::expr<boost::proto::tag::terminal, boost::proto::argsns_::term<boost::spirit::tag::char_code<boost::spirit::tag::space, boost::spirit::char_encoding::asci|
boost_1_46_0\boost\proto\extends.hpp|540|error: use of deleted function 'boost::proto::exprns_::expr<boost::proto::tag::terminal, boost::proto::argsns_::term<boost::spirit::qi::reference<const boost::spirit::qi::rule<__gnu_cxx::__normal_iterator<const char*, std::basic_string<char> >, Expression(), boost::proto::exprns_::expr<boost::proto::tag::terminal, boost::proto::argsns_::term<boost::spirit::tag::char_code<boost::spirit::tag::space, boost::spirit::char_encoding::ascii> >, 0l>, boost::spirit::unused_type, boost::spirit::unused_type> > >, 0l>:|
boost_1_46_0\boost\proto\detail\expr0.hpp|165|error: no matching function for call to 'boost::spirit::qi::reference<const boost::spirit::qi::rule<__gnu_cxx::__normal_iterator<const char*, std::basic_string<char> >, Expression(), boost::proto::exprns_::expr<boost::proto::tag::terminal, boost::proto::argsns_::term<boost::spirit::tag::char_code<boost::spirit::tag::space, boost::spirit::char_encoding::ascii> >, 0l>, boost::spirit::unused_type, boost::spirit::unused_type> >::reference()'|

Answer 1

UPDATE Showing a simplified parser with a a recursive ast parsing the sample expression shown

一如既往，断言消息恰恰导致了问题:

        // If you see the assertion below failing then the start rule
        // passed to the constructor of the grammar is not compatible with
        // the grammar (i.e. it uses different template parameters).
        BOOST_SPIRIT_ASSERT_MSG(
            (is_same<start_type, rule<Iterator_, T1_, T2_, T3_, T4_> >::value)
          , incompatible_start_rule, (rule<Iterator_, T1_, T2_, T3_, T4_>));

所以它告诉你应该将语法与开始规则相匹配:你有

struct Parser : qi::grammar<Iterator, Expression(), ascii::space_type>

但是

qi::rule<Iterator, Expression, ascii::space_type> expression;

显然你忘记了括号:

qi::rule<Iterator, Expression(), ascii::space_type> expression;

使用通用库时的指南:

其中一些“规则”是普遍适用的，除了没有。 2 与 Boost Spirit 具体相关:

1. 婴儿学步；从小处开始(空的，甚至)
2. 从 AST 开始以准确匹配语法
3. 逐步建立，
4. 编译沿途的每一步

更新

这是一个非常简化的语法。如前所述，在刚刚的“第一条 spirit 规则”中，从AST开始以精确匹配语法:

namespace ast {

    namespace tag {
        struct constant;
        struct variable;
        struct function;
    }

    template <typename Tag> struct Identifier { char name; };

    using Constant = Identifier<tag::constant>;
    using Variable = Identifier<tag::variable>;
    using Function = Identifier<tag::function>;

    struct FunctionCall;

    using Expression = boost::make_recursive_variant<
        Constant,
        Variable,
        boost::recursive_wrapper<FunctionCall>
    >::type;

    struct FunctionCall {
        Function function;
        std::vector<Expression> params;
    };

    struct Equation {
        Expression lhs, rhs;
    };
}

当然，这可能简单得多，因为所有标识符都只是 char 并且您可以动态切换 ( impression ) .

现在，语法必须遵循。 1. 保持简单2. 仔细格式化3. 直接匹配ast，4. 添加调试宏:

template <typename It, typename Skipper = ascii::space_type>
    struct Parser : qi::grammar<It, ast::Equation(), Skipper>
{
    Parser() : Parser::base_type(equation_)
    {
        using namespace qi;

        constant_ = qi::eps >> char_("a-eA-E");
        function_ = qi::eps >> char_("f-rF-R");
        variable_ = qi::eps >> char_("s-zS-Z");

        function_call = function_ >> '(' >> -(expression_ % ',') >> ')';
        expression_   = constant_ | variable_ | function_call;
        equation_     = expression_ >> '=' >> expression_;

        BOOST_SPIRIT_DEBUG_NODES((constant_)(function_)(variable_)(function_call)(expression_)(equation_))
    }
    qi::rule<It, ast::Constant()> constant_;
    qi::rule<It, ast::Function()> function_;
    qi::rule<It, ast::Variable()> variable_;

    qi::rule<It, ast::FunctionCall(), Skipper> function_call;
    qi::rule<It, ast::Expression(),   Skipper> expression_;
    qi::rule<It, ast::Equation(),     Skipper> equation_;
};

Note how the comments have become completely unneeded. Also note how recursively using expression_ solved your biggest headache!

完整程序

Live On Coliru

//#define BOOST_SPIRIT_DEBUG
#include <boost/spirit/include/qi.hpp>

namespace qi    = boost::spirit::qi;
namespace ascii = boost::spirit::ascii;

namespace ast {

    namespace tag {
        struct constant;
        struct variable;
        struct function;
    }

    template <typename Tag> struct Identifier { char name; };

    using Constant = Identifier<tag::constant>;
    using Variable = Identifier<tag::variable>;
    using Function = Identifier<tag::function>;

    struct FunctionCall;

    using Expression = boost::make_recursive_variant<
        Constant,
        Variable,
        boost::recursive_wrapper<FunctionCall>
    >::type;

    struct FunctionCall {
        Function function;
        std::vector<Expression> params;
    };

    struct Equation {
        Expression lhs, rhs;
    };
}

BOOST_FUSION_ADAPT_STRUCT(ast::Constant, (char, name))
BOOST_FUSION_ADAPT_STRUCT(ast::Variable, (char, name))
BOOST_FUSION_ADAPT_STRUCT(ast::Function, (char, name))
BOOST_FUSION_ADAPT_STRUCT(ast::FunctionCall, (ast::Function, function)(std::vector<ast::Expression>, params))
BOOST_FUSION_ADAPT_STRUCT(ast::Equation, (ast::Expression, lhs)(ast::Expression, rhs))

namespace qi = boost::spirit::qi;
namespace ascii = boost::spirit::ascii;

template <typename It, typename Skipper = ascii::space_type>
struct Parser : qi::grammar<It, ast::Equation(), Skipper>
{
    Parser() : Parser::base_type(equation_)
    {
        using namespace qi;

        constant_ = qi::eps >> char_("a-eA-E");
        function_ = qi::eps >> char_("f-rF-R");
        variable_ = qi::eps >> char_("s-zS-Z");

        function_call = function_ >> '(' >> -(expression_ % ',') >> ')';
        expression_   = constant_ | variable_ | function_call;
        equation_     = expression_ >> '=' >> expression_;

        BOOST_SPIRIT_DEBUG_NODES((constant_)(function_)(variable_)(function_call)(expression_)(equation_))
    }
    qi::rule<It, ast::Constant()> constant_;
    qi::rule<It, ast::Function()> function_;
    qi::rule<It, ast::Variable()> variable_;

    qi::rule<It, ast::FunctionCall(), Skipper> function_call;
    qi::rule<It, ast::Expression(),   Skipper> expression_;
    qi::rule<It, ast::Equation(),     Skipper> equation_;
};

int main() {
    std::cout << "Unification Algorithm\n\n";

    std::string const phrase = "f(a, b) = f(g(z, x), g(x, h(x)), c)";
    using It = std::string::const_iterator;
    It itr = phrase.begin(), last = phrase.end();

    std::cout << phrase << std::endl;

    Parser<It> g;
    ast::Equation parsed;

    if (phrase_parse(itr, last, g, ascii::space, parsed)) {
        std::cout << "Expression parsed.\n";
    } else {
        std::cout << "Could not parse equation.\n";
    }

    if (itr != last) {
        std::cout << "Remaining unparsed input: '" << std::string(itr,last) << "'\n";
    }
}