Question

我正在尝试编写一个用于填充抽象语法树的算术表达式的精神解析器。如果我没有尝试填充AST，解析器会编译，但在当前版本中失败（有一个24K错误）。我使用clang ++ 3.5.0和-std = c ++ 11，并在Ubuntu 14.4上运行。

#include <string>
#include <vector>
#include <utility>

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

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

#include <boost/variant/variant.hpp>
#include <boost/variant/recursive_wrapper.hpp>

using std::string;
using std::vector;
using std::pair;

using boost::spirit::qi::grammar;
using boost::spirit::qi::space_type;
using boost::spirit::qi::rule;

struct Term; // forward dec
typedef boost::recursive_wrapper<Term> RWTerm;
typedef pair<char, RWTerm> OpAndRWTerm;
typedef pair<RWTerm, vector<OpAndRWTerm> > Expr;
typedef boost::variant<Expr, double> Factor;
typedef pair<char, Factor> OpAndFactor;
struct Term : public pair<Factor, vector<OpAndFactor> >{};

template<typename It>
struct formula_parser : grammar<It, Expr(), space_type> {
  formula_parser() : formula_parser::base_type(expr_rule) {
    using boost::spirit::qi::double_;
    using boost::spirit::ascii::char_;

    factor_rule %= double_ | parenthesis_rule;
    parenthesis_rule %= '(' >> expr_rule >> ')';
    op_and_factor_rule %= char_("/*") >> factor_rule;
    term_rule %= factor_rule >> *op_and_factor_rule;
    op_and_term_rule %= char_("+-") >> term_rule;
    expr_rule %= term_rule >> *op_and_term_rule;
  }
  rule<It, OpAndRWTerm(), space_type> op_and_term_rule;
  rule<It, Expr(), space_type> expr_rule;
  rule<It, OpAndFactor(), space_type> op_and_factor_rule;
  rule<It, RWTerm(), space_type> term_rule;
  rule<It, Expr(), space_type> parenthesis_rule;
  rule<It, Factor(), space_type> factor_rule;
};

int main() {
  formula_parser<string::const_iterator> grammar;
}

我从错误信息中理解的是，融合在规则term_rule中混合了类型因子和RWTerm。

我做错了什么？

Answer 1

如果我改变两件事，它会为我编译：

由于Term继承自std::pair，Term是新类型。因此，您需要将BOOST_FUSION_ADAPT_STRUCT应用于Term，无论std::pair中<boost/fusion/adapted/std_pair.hpp>是否已执行此操作：
```
BOOST_FUSION_ADAPT_STRUCT(
    Term,
    (Factor, first)
    (std::vector<OpAndFactor>, second)
)
```
或者，您可以使Term成为一个包含两个成员的独立结构，然后在其上应用BOOST_FUSION_ADAPT_STRUCT：
```
struct Term { Factor f; std::vector<OpAndFactor>  o;};

BOOST_FUSION_ADAPT_STRUCT(
    Term,
    (Factor, f)
    (std::vector<OpAndFactor>, o)
)
```
顺便说一句：您必须在此完全符合std::vector的条件，因为以下will not compile：
```
using std::vector;
BOOST_FUSION_ADAPT_STRUCT(
    Term,
    (Factor, f)
    (vector<OpAndFactor>, o)
)
```
在声明Term时使用RWTerm代替term_rule：
```
rule<It, Term(), space_type> term_rule;
```

完整代码：

#include <string>
#include <vector>
#include <utility>

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

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

#include <boost/variant/variant.hpp>
#include <boost/variant/recursive_wrapper.hpp>

using boost::spirit::qi::grammar;
using boost::spirit::qi::space_type;
using boost::spirit::qi::rule;

struct Term; // forward dec
typedef boost::recursive_wrapper<Term> RWTerm;
typedef std::pair<char, RWTerm> OpAndRWTerm;
typedef std::pair<RWTerm, std::vector<OpAndRWTerm> > Expr;
typedef boost::variant<Expr, double> Factor;
typedef std::pair<char, Factor> OpAndFactor;
struct Term : public std::pair<Factor, std::vector<OpAndFactor> >{};

BOOST_FUSION_ADAPT_STRUCT(
    Term,
    (Factor, first)
    (std::vector<OpAndFactor>, second)
)


template<typename It>
struct formula_parser : grammar<It, Expr(), space_type> {
  formula_parser() : formula_parser::base_type(expr_rule) {
    using boost::spirit::qi::double_;
    using boost::spirit::ascii::char_;

    factor_rule %= double_ | parenthesis_rule;
    parenthesis_rule %= '(' >> expr_rule >> ')';
    op_and_factor_rule %= char_("/*") >> factor_rule;
    term_rule %= factor_rule >> *op_and_factor_rule;
    op_and_term_rule %= char_("+-") >> term_rule;
    expr_rule %= term_rule >> *op_and_term_rule;
  }
  rule<It, OpAndRWTerm(), space_type> op_and_term_rule;
  rule<It, Expr(), space_type> expr_rule;
  rule<It, OpAndFactor(), space_type> op_and_factor_rule;
  rule<It, Term(), space_type> term_rule;
  rule<It, Expr(), space_type> parenthesis_rule;
  rule<It, Factor(), space_type> factor_rule;
};

int main() {
  formula_parser<std::string::const_iterator> grammar;
}

live example

boost :: spirit算术公式解析器无法编译

1 个答案: