如何使用 Boost Spirit 來解析具有運算子優先權的布林運算式？

布林表達式解析

給定一個以下形式的布林表達式：“a and b xor (c and d or a and b);”，目標是將其解析為一棵樹，遵循優先規則（非、與、異或或）。

Boost Spirit實作

Boost Spirit 可用來建立基於表達式範本的遞歸下降解析器。實作如下：

抽象資料型別

struct op_or  {}; // tag
struct op_and {}; // tag
struct op_xor {}; // tag
struct op_not {}; // tag

typedef std::string var;
template <typename tag> struct binop;
template <typename tag> struct unop;

typedef boost::variant<var, 
        boost::recursive_wrapper<unop <op_not> >, 
        boost::recursive_wrapper<binop<op_and> >,
        boost::recursive_wrapper<binop<op_xor> >,
        boost::recursive_wrapper<binop<op_or> >
        > expr;

語法規則

template <typename It, typename Skipper = qi::space_type>
    struct parser : qi::grammar<It, expr(), Skipper>
{
    parser() : parser::base_type(expr_)
    {
        using namespace qi;
        expr_  = or_.alias();

        not_ = ("not" > simple       ) [ _val = phx::construct<unop <op_not>>(_1)     ] | simple [ _val = _1 ];
#ifdef RIGHT_ASSOCIATIVE
        or_  = (xor_ >> "or"  >> or_ ) [ _val = phx::construct<binop<op_or >>(_1, _2) ] | xor_   [ _val = _1 ];
        xor_ = (and_ >> "xor" >> xor_) [ _val = phx::construct<binop<op_xor>>(_1, _2) ] | and_   [ _val = _1 ];
        and_ = (not_ >> "and" >> and_) [ _val = phx::construct<binop<op_and>>(_1, _2) ] | not_   [ _val = _1 ];
#else
        or_  = xor_ [ _val = _1 ] >> *("or"  >> xor_ [ _val = phx::construct<binop<op_or>> (_val, _1) ]);
        xor_ = and_ [ _val = _1 ] >> *("xor" >> and_ [ _val = phx::construct<binop<op_xor>>(_val, _1) ]);
        and_ = not_ [ _val = _1 ] >> *("and" >> not_ [ _val = phx::construct<binop<op_and>>(_val, _1) ]);
#endif

        simple = (('(' > expr_ > ')') | var_);
        var_ = qi::lexeme[ +alpha ];
    }

  private:
    qi::rule<It, var() , Skipper> var_;
    qi::rule<It, expr(), Skipper> not_, and_, xor_, or_, simple, expr_;
};

對語法樹進行運算

遍歷並列印語法樹：

struct printer : boost::static_visitor<void>
{
    printer(std::ostream&amp; os) : _os(os) {}
    std::ostream&amp; _os;

    //
    void operator()(const var&amp; v) const { _os << v; }

    void operator()(const binop<op_and>&amp; b) const { print(" &amp; ", b.oper1, b.oper2); }
    void operator()(const binop<op_or >&amp; b) const { print(" | ", b.oper1, b.oper2); }
    void operator()(const binop<op_xor>&amp; b) const { print(" ^ ", b.oper1, b.oper2); }

    void print(const std::string&amp; op, const expr&amp; l, const expr&amp; r) const
    {
        _os << "(";
            boost::apply_visitor(*this, l);
            _os << op;
            boost::apply_visitor(*this, r);
        _os << ")";
    }

    void operator()(const unop<op_not>&amp; u) const
    {
        _os << "(";
            _os << "!";
            boost::apply_visitor(*this, u.oper1);
        _os << ")";
    }
};

std::ostream&amp; operator<<(std::ostream&amp; os, const expr&amp; e)
{ boost::apply_visitor(printer(os), e); return os; }

測試輸出

解析器正確處理優先規則並輸出：

result: ((a & b) ^ ((c & d) | (a & b)))
result: ((a & b) ^ ((c & d) | (a & b)))
result: (a & b)
result: (a | b)
result: (a ^ b)
result: (!a)
result: ((!a) & b)
result: (!(a & b))
result: ((a | b) | c)

完整程式碼和實例（Coliru）

[點擊此處嘗試查看完整程式碼Coliru](https://coliru.stacked-crooked.com/a/a795dea9a310e79bb12db3fe86fa61cf)

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