11

可以说我有这样的代码(行号供参考):

1:
2:function FuncName_1 {
3:    var Var_1 = 3;
4:    var  Var_2 = 4;
5:    ...

我想编写一个解析此类文本的语法,将所有标识符(函数和变量名)信息放入树(utree?)中。每个节点应保留:line_num、column_num 和符号值。例子:

root: FuncName_1 (line:2,col:10)
  children[0]: Var_1 (line:3, col:8)
  children[1]: Var_1 (line:4, col:9)

我想将它放入树中,因为我计划遍历该树并且对于每个节点,我必须知道“上下文”:(当前节点的所有父节点)。

例如,在使用 Var_1 处理节点时,我必须知道这是函数 FuncName_1 的局部变量的名称(当前正在作为节点处理,但更早一级)

我无法弄清楚一些事情

  1. 这可以通过语义动作和 utree 在 Spirit 中完成吗?或者我应该使用 variant<> 树?
  2. 如何同时将这三个信息(列、行​​、符号名)传递给节点?我知道我必须使用 pos_iterator 作为语法的迭代器类型,但是如何在语义动作中访问这些信息?

我是 Boost 的新手,所以我一遍又一遍地阅读 Spirit 文档,我尝试用谷歌搜索我的问题,但我无法将所有部分放在一起找到解决方案。好像以前没有我这样的用例(或者我只是找不到它)看起来像位置迭代器的唯一解决方案是解析错误处理的解决方案,但我不是这种情况我感兴趣。仅解析我正在使用的代码的代码如下,但我不知道如何继续前进。

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

  namespace qi = boost::spirit::qi;
  typedef boost::spirit::line_pos_iterator<std::string::const_iterator> pos_iterator_t;

  template<typename Iterator=pos_iterator_t, typename Skipper=qi::space_type>
  struct ParseGrammar: public qi::grammar<Iterator, Skipper>
  {
        ParseGrammar():ParseGrammar::base_type(SourceCode)
        {
           using namespace qi;
           KeywordFunction = lit("function");
           KeywordVar    = lit("var");
           SemiColon     = lit(';');

           Identifier = lexeme [alpha >> *(alnum | '_')];
           VarAssignemnt = KeywordVar >> Identifier >> char_('=') >> int_ >> SemiColon;
           SourceCode = KeywordFunction >> Identifier >> '{' >> *VarAssignemnt >> '}';
        }

        qi::rule<Iterator, Skipper> SourceCode;
        qi::rule<Iterator > KeywordFunction;
        qi::rule<Iterator,  Skipper> VarAssignemnt;
        qi::rule<Iterator> KeywordVar;
        qi::rule<Iterator> SemiColon;
        qi::rule<Iterator > Identifier;
  };

  int main()
  {
     std::string const content = "function FuncName_1 {\n var Var_1 = 3;\n var  Var_2 = 4; }";

     pos_iterator_t first(content.begin()), iter = first, last(content.end());
     ParseGrammar<pos_iterator_t> resolver;    //  Our parser
     bool ok = phrase_parse(iter,
                            last,
                            resolver,
                            qi::space);

     std::cout << std::boolalpha;
     std::cout << "\nok : " << ok << std::endl;
     std::cout << "full   : " << (iter == last) << std::endl;
     if(ok && iter == last)
     {
        std::cout << "OK: Parsing fully succeeded\n\n";
     }
     else
     {
        int line   = get_line(iter);
        int column = get_column(first, iter);
        std::cout << "-------------------------\n";
        std::cout << "ERROR: Parsing failed or not complete\n";
        std::cout << "stopped at: " << line  << ":" << column << "\n";
        std::cout << "remaining: '" << std::string(iter, last) << "'\n";
        std::cout << "-------------------------\n";
     }
     return 0;
  }
4

1 回答 1

15

这是一个有趣的练习,我最终整理了一个on_success[1]的工作演示 来注释 AST 节点。

假设我们想要一个像这样的 AST:

namespace ast
{
struct LocationInfo {
    unsigned line, column, length;
};

struct Identifier     : LocationInfo {
    std::string name;
};

struct VarAssignment  : LocationInfo {
    Identifier id;
    int value;
};

struct SourceCode     : LocationInfo {
    Identifier function;
    std::vector<VarAssignment> assignments;
};
}

我知道,“位置信息”对于SourceCode节点来说可能是多余的,但你知道......无论如何,为了方便为这些节点分配属性,而不需要语义操作或大量专门制作的构造函数:

#include <boost/fusion/adapted/struct.hpp>
BOOST_FUSION_ADAPT_STRUCT(ast::Identifier,    (std::string, name))
BOOST_FUSION_ADAPT_STRUCT(ast::VarAssignment, (ast::Identifier, id)(int, value))
BOOST_FUSION_ADAPT_STRUCT(ast::SourceCode,    (ast::Identifier, function)(std::vector<ast::VarAssignment>, assignments))

那里。现在我们可以声明规则来公开这些属性:

qi::rule<Iterator, ast::SourceCode(),    Skipper> SourceCode;
qi::rule<Iterator, ast::VarAssignment(), Skipper> VarAssignment;
qi::rule<Iterator, ast::Identifier()>         Identifier;
// no skipper, no attributes:
qi::rule<Iterator> KeywordFunction, KeywordVar, SemiColon;

我们根本不(本质上)修改语法:属性传播是“自动的” [2]

KeywordFunction = lit("function");
KeywordVar      = lit("var");
SemiColon       = lit(';');

Identifier      = as_string [ alpha >> *(alnum | char_("_")) ];
VarAssignment   = KeywordVar >> Identifier >> '=' >> int_ >> SemiColon; 
SourceCode      = KeywordFunction >> Identifier >> '{' >> *VarAssignment >> '}';

魔法

我们如何获取附加到节点的源位置信息?

auto set_location_info = annotate(_val, _1, _3);
on_success(Identifier,    set_location_info);
on_success(VarAssignment, set_location_info);
on_success(SourceCode,    set_location_info);

现在,annotate它只是一个可调用对象的惰性版本,定义为:

template<typename It>
struct annotation_f {
    typedef void result_type;

    annotation_f(It first) : first(first) {}
    It const first;

    template<typename Val, typename First, typename Last>
    void operator()(Val& v, First f, Last l) const {
        do_annotate(v, f, l, first);
    }
  private:
    void static do_annotate(ast::LocationInfo& li, It f, It l, It first) {
        using std::distance;
        li.line   = get_line(f);
        li.column = get_column(first, f);
        li.length = distance(f, l);
    }
    static void do_annotate(...) { }
};

由于get_column工作方式的原因,函子是有状态的(因为它记住了开始迭代器)[3]。如您所见do_annotate,只接受来自LocationInfo.

现在,布丁的证明:

std::string const content = "function FuncName_1 {\n var Var_1 = 3;\n var  Var_2 = 4; }";

pos_iterator_t first(content.begin()), iter = first, last(content.end());
ParseGrammar<pos_iterator_t> resolver(first);    //  Our parser

ast::SourceCode program;
bool ok = phrase_parse(iter,
        last,
        resolver,
        qi::space,
        program);

std::cout << std::boolalpha;
std::cout << "ok  : " << ok << std::endl;
std::cout << "full: " << (iter == last) << std::endl;
if(ok && iter == last)
{
    std::cout << "OK: Parsing fully succeeded\n\n";

    std::cout << "Function name: " << program.function.name << " (see L" << program.printLoc() << ")\n";
    for (auto const& va : program.assignments)
        std::cout << "variable " << va.id.name << " assigned value " << va.value << " at L" << va.printLoc() << "\n";
}
else
{
    int line   = get_line(iter);
    int column = get_column(first, iter);
    std::cout << "-------------------------\n";
    std::cout << "ERROR: Parsing failed or not complete\n";
    std::cout << "stopped at: " << line  << ":" << column << "\n";
    std::cout << "remaining: '" << std::string(iter, last) << "'\n";
    std::cout << "-------------------------\n";
}

这打印:

ok  : true
full: true
OK: Parsing fully succeeded

Function name: FuncName_1 (see L1:1:56)
variable Var_1 assigned value 3 at L2:3:14
variable Var_2 assigned value 4 at L3:3:15

完整的演示程序

在Coliru现场观看

还显示:

  • 错误处理,例如:

    Error: expecting "=" in line 3: 
    
    var  Var_2 - 4; }
               ^---- here
    ok  : false
    full: false
    -------------------------
    ERROR: Parsing failed or not complete
    stopped at: 1:1
    remaining: 'function FuncName_1 {
    var Var_1 = 3;
    var  Var_2 - 4; }'
    -------------------------
    
  • BOOST_SPIRIT_DEBUG 宏

  • LocationInfo一种方便地流式传输任何 AST 节点的部分的 hacky 方式,抱歉 :)
//#define BOOST_SPIRIT_DEBUG
#define BOOST_SPIRIT_USE_PHOENIX_V3
#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/spirit/include/support_line_pos_iterator.hpp>
#include <iomanip>

namespace qi = boost::spirit::qi;
namespace phx= boost::phoenix;

typedef boost::spirit::line_pos_iterator<std::string::const_iterator> pos_iterator_t;

namespace ast
{
    namespace manip { struct LocationInfoPrinter; }

    struct LocationInfo {
        unsigned line, column, length;
        manip::LocationInfoPrinter printLoc() const;
    };

    struct Identifier     : LocationInfo {
        std::string name;
    };

    struct VarAssignment  : LocationInfo {
        Identifier id;
        int value;
    };

    struct SourceCode     : LocationInfo {
        Identifier function;
        std::vector<VarAssignment> assignments;
    };

    ///////////////////////////////////////////////////////////////////////////
    // Completely unnecessary tweak to get a "poor man's" io manipulator going
    // so we can do `std::cout << x.printLoc()` on types of `x` deriving from
    // LocationInfo
    namespace manip {
        struct LocationInfoPrinter {
            LocationInfoPrinter(LocationInfo const& ref) : ref(ref) {}
            LocationInfo const& ref;
            friend std::ostream& operator<<(std::ostream& os, LocationInfoPrinter const& lip) {
                return os << lip.ref.line << ':' << lip.ref.column << ':' << lip.ref.length;
            }
        };
    }

    manip::LocationInfoPrinter LocationInfo::printLoc() const { return { *this }; }
    // feel free to disregard this hack
    ///////////////////////////////////////////////////////////////////////////
}

BOOST_FUSION_ADAPT_STRUCT(ast::Identifier,    (std::string, name))
BOOST_FUSION_ADAPT_STRUCT(ast::VarAssignment, (ast::Identifier, id)(int, value))
BOOST_FUSION_ADAPT_STRUCT(ast::SourceCode,    (ast::Identifier, function)(std::vector<ast::VarAssignment>, assignments))

struct error_handler_f {
    typedef qi::error_handler_result result_type;
    template<typename T1, typename T2, typename T3, typename T4>
        qi::error_handler_result operator()(T1 b, T2 e, T3 where, T4 const& what) const {
            std::cerr << "Error: expecting " << what << " in line " << get_line(where) << ": \n" 
                << std::string(b,e) << "\n"
                << std::setw(std::distance(b, where)) << '^' << "---- here\n";
            return qi::fail;
        }
};

template<typename It>
struct annotation_f {
    typedef void result_type;

    annotation_f(It first) : first(first) {}
    It const first;

    template<typename Val, typename First, typename Last>
    void operator()(Val& v, First f, Last l) const {
        do_annotate(v, f, l, first);
    }
  private:
    void static do_annotate(ast::LocationInfo& li, It f, It l, It first) {
        using std::distance;
        li.line   = get_line(f);
        li.column = get_column(first, f);
        li.length = distance(f, l);
    }
    static void do_annotate(...) {}
};

template<typename Iterator=pos_iterator_t, typename Skipper=qi::space_type>
struct ParseGrammar: public qi::grammar<Iterator, ast::SourceCode(), Skipper>
{
    ParseGrammar(Iterator first) : 
        ParseGrammar::base_type(SourceCode),
        annotate(first)
    {
        using namespace qi;
        KeywordFunction = lit("function");
        KeywordVar      = lit("var");
        SemiColon       = lit(';');

        Identifier      = as_string [ alpha >> *(alnum | char_("_")) ];
        VarAssignment   = KeywordVar > Identifier > '=' > int_ > SemiColon; // note: expectation points
        SourceCode      = KeywordFunction >> Identifier >> '{' >> *VarAssignment >> '}';

        on_error<fail>(VarAssignment, handler(_1, _2, _3, _4));
        on_error<fail>(SourceCode, handler(_1, _2, _3, _4));

        auto set_location_info = annotate(_val, _1, _3);
        on_success(Identifier,    set_location_info);
        on_success(VarAssignment, set_location_info);
        on_success(SourceCode,    set_location_info);

        BOOST_SPIRIT_DEBUG_NODES((KeywordFunction)(KeywordVar)(SemiColon)(Identifier)(VarAssignment)(SourceCode))
    }

    phx::function<error_handler_f> handler;
    phx::function<annotation_f<Iterator>> annotate;

    qi::rule<Iterator, ast::SourceCode(),    Skipper> SourceCode;
    qi::rule<Iterator, ast::VarAssignment(), Skipper> VarAssignment;
    qi::rule<Iterator, ast::Identifier()>             Identifier;
    // no skipper, no attributes:
    qi::rule<Iterator> KeywordFunction, KeywordVar, SemiColon;
};

int main()
{
    std::string const content = "function FuncName_1 {\n var Var_1 = 3;\n var  Var_2 - 4; }";

    pos_iterator_t first(content.begin()), iter = first, last(content.end());
    ParseGrammar<pos_iterator_t> resolver(first);    //  Our parser

    ast::SourceCode program;
    bool ok = phrase_parse(iter,
            last,
            resolver,
            qi::space,
            program);

    std::cout << std::boolalpha;
    std::cout << "ok  : " << ok << std::endl;
    std::cout << "full: " << (iter == last) << std::endl;
    if(ok && iter == last)
    {
        std::cout << "OK: Parsing fully succeeded\n\n";

        std::cout << "Function name: " << program.function.name << " (see L" << program.printLoc() << ")\n";
        for (auto const& va : program.assignments)
            std::cout << "variable " << va.id.name << " assigned value " << va.value << " at L" << va.printLoc() << "\n";
    }
    else
    {
        int line   = get_line(iter);
        int column = get_column(first, iter);
        std::cout << "-------------------------\n";
        std::cout << "ERROR: Parsing failed or not complete\n";
        std::cout << "stopped at: " << line  << ":" << column << "\n";
        std::cout << "remaining: '" << std::string(iter, last) << "'\n";
        std::cout << "-------------------------\n";
    }
    return 0;
}

[1]可悲的是没有(der)记录,除了召唤样本

[2]好吧,我过去常常不用太多工作as_string就能得到适当的分配Identifier

[3]在性能方面可能有更聪明的方法,但现在,让我们保持简单

于 2013-10-27T01:48:10.607 回答