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c++ - C++中的 bool 表达式(语法)解析器

转载 作者:IT老高 更新时间:2023-10-28 12:38:00 25 4
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我想解析一个 bool 表达式(在 C++ 中)。输入形式:

a and b xor (c and d or a and b);

我只是想把这个表达式解析成一棵树,知道优先规则(not,and,xor,or)。所以上面的表达式应该是这样的:

(a and b) xor ((c and d) or (a and b));

到解析器。

树的形式是:

                        a
and
b
or
c
and
d
xor
a
and
b

输入将通过命令行或字符串的形式。我只需要解析器。

是否有任何资源可以帮助我做到这一点?

最佳答案

这是一个基于 Boost Spirit 的实现。

因为 Boost Spirit 会根据 表达式模板 生成 递归下降 解析器,因此遵守“特殊”(原文如此)优先规则(正如其他人所提到的)是相当乏味的。因此语法缺乏一定的优雅。

抽象数据类型

我使用Boost Variant的递归变体支持定义了一个树形数据结构,注意expr的定义:

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;

(以下为完整来源)

语法规则

如前所述,以下是(略显乏味的)语法定义。

虽然我不认为这种语法是最佳的,但它的可读性很强,而且我们自己有一个静态编译的解析器具有强类型的 AST 数据类型,大约 50 行代码。情况可能会更糟。

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_;
};

Note: I've left the old rule definitions that would result in right-associativity in the binary operators for reference, but left-associativity is the more natural, and therefore taken by default

在语法树上操作

显然,您想要评估表达式。现在,我决定停止打印,所以我不必为命名变量做查找表:)

遍历递归变体一开始可能看起来很神秘,但 boost::static_visitor<>一旦掌握了窍门,它就非常简单:

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; }

测试输出:

对于代码中的测试用例,输出如下,通过添加(冗余)括号来演示正确处理优先规则:

Live On Coliru

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)

Note, compare to Live On Coliru, with -DRIGHT_ASSOCIATIVE

完整代码:

Live On Coliru

#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();

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 ];

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> {
// From the OP:
"(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;",
"a xor b;",
"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;
}

奖金:

对于奖励积分,获得一棵与 OP 中显示的完全相同的树:

Live On Coliru

static const char indentstep[] = "    ";

struct tree_print : boost::static_visitor<void>
{
tree_print(std::ostream& os, const std::string& indent=indentstep) : _os(os), _indent(indent) {}
std::ostream& _os;
std::string _indent;

void operator()(const var& v) const { _os << _indent << v << std::endl; }

void operator()(const binop<op_and>& b) const { print("and ", b.oper1, b.oper2); }
void operator()(const binop<op_or >& b) const { print("or ", b.oper2, b.oper1); }
void operator()(const binop<op_xor>& b) const { print("xor ", b.oper2, b.oper1); }

void print(const std::string& op, const expr& l, const expr& r) const
{
boost::apply_visitor(tree_print(_os, _indent+indentstep), l);
_os << _indent << op << std::endl;
boost::apply_visitor(tree_print(_os, _indent+indentstep), r);
}

void operator()(const unop<op_not>& u) const
{
_os << _indent << "!";
boost::apply_visitor(tree_print(_os, _indent+indentstep), u.oper1);
}
};

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

结果:

            a
and
b
or
c
and
d
xor
a
and
b

关于c++ - C++中的 bool 表达式(语法)解析器,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/8706356/

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