我有一个解析器，它解析布尔表达式。我应该如何修改它以支持＆＃34;隐含 - 和＆＃34;例如＆＃34; A1 A2＆＃34;必须解析为A1和A2？我试图改变＆＃34;和_＆＃34;规则支持它，但它开始治疗＆＃34; xor＆＃34;即使它被用作操作也是一个变量。不包括＆＃34; xor＆＃34;来自var的规则与（！（＆＃34; xor＆＃34;）＆gt;＆gt;）对于简单表达式仍然失败。
源代码：
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/variant/recursive_wrapper.hpp>
namespace qi = boost::spirit::qi;
namespace phx = boost::phoenix;
struct op_or {};
struct op_and {};
struct op_xor {};
struct op_not {};
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;
template <typename tag> struct binop
{
explicit binop(const expr& l, const expr& r) : oper1(l), oper2(r) { }
expr oper1, oper2;
};
template <typename tag> struct unop
{
explicit unop(const expr& o) : oper1(o) { }
expr oper1;
};
struct printer : boost::static_visitor<void>
{
printer(std::ostream& os) : _os(os) {}
std::ostream& _os;
void operator()(const var& v) const { _os << v; }
void operator()(const binop<op_and>& b) const { print(" & ", b.oper1, b.oper2); }
void operator()(const binop<op_or >& b) const { print(" | ", b.oper1, b.oper2); }
void operator()(const binop<op_xor>& b) const { print(" ^ ", b.oper1, b.oper2); }
void print(const std::string& op, const expr& l, const expr& r) const
{
_os << "(";
boost::apply_visitor(*this, l);
_os << op;
boost::apply_visitor(*this, r);
_os << ")";
}
void operator()(const unop<op_not>& u) const
{
_os << "(";
_os << "!";
boost::apply_visitor(*this, u.oper1);
_os << ")";
}
};
std::ostream& operator<<(std::ostream& os, const expr& e)
{ boost::apply_visitor(printer(os), e); return os; }
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();
or_ = xor_ [_val = _1] >> *(("or" >> xor_ ) [ _val = phx::construct<binop<op_or >>(_val, _1) ]) | xor_ [ _val = _1 ];
xor_ = and_ [_val = _1] >> *(("xor" >> and_) [ _val = phx::construct<binop<op_xor>>(_val, _1) ]) | and_ [ _val = _1 ];
and_ = not_ [_val = _1] >> *(("and" >> not_) [ _val = phx::construct<binop<op_and>>(_val, _1) ]) | not_ [ _val = _1 ];
not_ = ("not" > simple ) [ _val = phx::construct<unop <op_not>>(_1) ] | simple [ _val = _1 ];
simple = (('(' > expr_ > ')') | var_);
var_ = qi::lexeme[ +alpha ];
BOOST_SPIRIT_DEBUG_NODE(expr_);
BOOST_SPIRIT_DEBUG_NODE(or_);
BOOST_SPIRIT_DEBUG_NODE(xor_);
BOOST_SPIRIT_DEBUG_NODE(and_);
BOOST_SPIRIT_DEBUG_NODE(not_);
BOOST_SPIRIT_DEBUG_NODE(simple);
BOOST_SPIRIT_DEBUG_NODE(var_);
}
private:
qi::rule<It, var() , Skipper> var_;
qi::rule<It, expr(), Skipper> not_, and_, xor_, or_, simple, expr_;
};
int main()
{
for (auto& input : std::list<std::string> {
"(a and b) xor ((c and d) or (a and b));",
"a and b xor c and d or a and b;",
/// Simpler tests:
"a and b;",
"a or b;",
"xor_in1 xor xor_in2;",
"xorin1 xor xorin2;",
"and1 xor xor;",
"not a;",
"not a and b;",
"not (a and b);",
"a or b or c;",
})
{
auto f(std::begin(input)), l(std::end(input));
parser<decltype(f)> p;
try
{
expr result;
bool ok = qi::phrase_parse(f,l,p > ';',qi::space,result);
if (!ok)
std::cerr << "invalid input\n";
else
std::cout << "result: " << result << "\n";
} catch (const qi::expectation_failure<decltype(f)>& e)
{
std::cerr << "expectation_failure at '" << std::string(e.first, e.last) << "'\n";
}
if (f!=l) std::cerr << "unparsed: '" << std::string(f,l) << "'\n";
}
return 0;
}
答案 0 :(得分：1)
我首先使用你明显开始的Boolean expression (grammar) parser in c++中的原始规则恢复了解析器规则：
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 ];
not_ = ("not" > simple ) [ _val = phx::construct<unop <op_not>>(_1) ] | simple [ _val = _1 ];
现在像"xorin1 xor xorin2;"这样的失败案例与您提出的问题（隐式和）没有任何关系。事实上，您刚刚发现需要做一些事情来正确地解析关键字：
以下是使用distinct directive from the Spirit Repository：
or_ = (xor_ >> qr::distinct(alnum|'_')[ "or" ] >> or_ ) [ _val = phx::construct<binop<op_or >>(_1, _2) ] | xor_ [ _val = _1 ];
xor_ = (and_ >> qr::distinct(alnum|'_')[ "xor" ] >> xor_) [ _val = phx::construct<binop<op_xor>>(_1, _2) ] | and_ [ _val = _1 ];
and_ = (not_ >> qr::distinct(alnum|'_')[ "and" ] >> and_) [ _val = phx::construct<binop<op_and>>(_1, _2) ] | not_ [ _val = _1 ];
not_ = (qr::distinct(alnum|'_')[ "not" ] > simple ) [ _val = phx::construct<unop <op_not>>(_1) ] | simple [ _val = _1 ];
下一步，也无关，您显然希望将“xor_in1”等视为有效标识符。相应地更改规则：
var_ = qi::lexeme[ alpha >> *(alnum | char_("_")) ];
现在所有案件都通过了：
<强> Live On Coliru
result: ((a & b) ^ ((c & d) | (a & b)))
result: (((a & b) ^ (c & d)) | (a & b))
result: (a & b)
result: (a | b)
result: (xor_in1 ^ xor_in2)
result: (xorin1 ^ xorin2)
result: (and1 ^ xor)
result: (!a)
result: ((!a) & b)
result: (!(a & b))
result: (a | (b | c))
您提到了隐含的内容和？
and_ = (not_ >> -qr::distinct(alnum|'_')[ "and" ] >> and_) [ _val = phx::construct<binop<op_and>>(_1, _2) ] | not_ [ _val = _1 ];
一些不错的测试用例：
<强> Live On Coliru
//#define BOOST_SPIRIT_DEBUG
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/spirit/repository/include/qi_distinct.hpp>
#include <boost/variant/recursive_wrapper.hpp>
namespace qi = boost::spirit::qi;
namespace qr = boost::spirit::repository::qi;
namespace phx = boost::phoenix;
struct op_or {};
struct op_and {};
struct op_xor {};
struct op_not {};
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;
template <typename tag> struct binop
{
explicit binop(const expr& l, const expr& r) : oper1(l), oper2(r) { }
expr oper1, oper2;
};
template <typename tag> struct unop
{
explicit unop(const expr& o) : oper1(o) { }
expr oper1;
};
struct printer : boost::static_visitor<void>
{
printer(std::ostream& os) : _os(os) {}
std::ostream& _os;
void operator()(const var& v) const { _os << v; }
void operator()(const binop<op_and>& b) const { print(" & ", b.oper1, b.oper2); }
void operator()(const binop<op_or >& b) const { print(" | ", b.oper1, b.oper2); }
void operator()(const binop<op_xor>& b) const { print(" ^ ", b.oper1, b.oper2); }
void print(const std::string& op, const expr& l, const expr& r) const
{
_os << "(";
boost::apply_visitor(*this, l);
_os << op;
boost::apply_visitor(*this, r);
_os << ")";
}
void operator()(const unop<op_not>& u) const
{
_os << "(";
_os << "!";
boost::apply_visitor(*this, u.oper1);
_os << ")";
}
};
std::ostream& operator<<(std::ostream& os, const expr& e)
{ boost::apply_visitor(printer(os), e); return os; }
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();
or_ = (xor_ >> qr::distinct(alnum|'_')[ "or" ] >> or_ ) [ _val = phx::construct<binop<op_or >>(_1, _2) ] | xor_ [ _val = _1 ];
xor_ = (and_ >> qr::distinct(alnum|'_')[ "xor" ] >> xor_) [ _val = phx::construct<binop<op_xor>>(_1, _2) ] | and_ [ _val = _1 ];
and_ = (not_ >> -qr::distinct(alnum|'_')[ "and" ] >> and_) [ _val = phx::construct<binop<op_and>>(_1, _2) ] | not_ [ _val = _1 ];
not_ = (qr::distinct(alnum|'_')[ "not" ] > simple ) [ _val = phx::construct<unop <op_not>>(_1) ] | simple [ _val = _1 ];
simple = (('(' > expr_ > ')') | var_);
var_ =
!qr::distinct(alnum|'_') [ lit("or")|"xor"|"and"|"not" ] >>
qr::distinct(alnum|'_') [ alpha >> *(alnum | char_("_")) ]
;
BOOST_SPIRIT_DEBUG_NODES((expr_) (or_) (xor_) (and_) (not_) (simple) (var_))
}
private:
qi::rule<It, var()> var_;
qi::rule<It, expr(), Skipper> not_, and_, xor_, or_, simple, expr_;
};
int main()
{
for (auto& input : std::list<std::string> {
#if 0
"a or b or c;",
"(a and b);",
"a xor b;",
"a or b;",
"(a) or (b);",
"((c and d) or (a and b));",
#endif
"(a and b) xor ((c and d) or (a and b));",
"(a b) xor ((c d) or (a b));",
"a and b xor c and d or a and b;",
"a b xor c d or a b;",
/// Simpler tests:
"a and b;",
"a b;",
"not a and b;",
"not a b;",
"not (a and b);",
"not (a b);",
})
{
auto f(std::begin(input)), l(std::end(input));
parser<decltype(f)> p;
try
{
expr result;
bool ok = qi::phrase_parse(f,l,p > ';',qi::space,result);
std::cout << "\n======= input '" << input << "'\n";
if (!ok)
std::cerr << "invalid input\n";
else
std::cout << "result: " << result << "\n";
} catch (const qi::expectation_failure<decltype(f)>& e)
{
std::cerr << "expectation_failure at '" << std::string(e.first, e.last) << "'\n";
}
if (f!=l) std::cerr << "unparsed: '" << std::string(f,l) << "'\n";
}
return 0;
}
打印
======= input '(a and b) xor ((c and d) or (a and b));'
result: ((a & b) ^ ((c & d) | (a & b)))
======= input '(a b) xor ((c d) or (a b));'
result: ((a & b) ^ ((c & d) | (a & b)))
======= input 'a and b xor c and d or a and b;'
result: (((a & b) ^ (c & d)) | (a & b))
======= input 'a b xor c d or a b;'
result: (((a & b) ^ (c & d)) | (a & b))
======= input 'a and b;'
result: (a & b)
======= input 'a b;'
result: (a & b)
======= input 'not a and b;'
result: ((!a) & b)
======= input 'not a b;'
result: ((!a) & b)
======= input 'not (a and b);'
result: (!(a & b))
======= input 'not (a b);'
result: (!(a & b))