我找到了一个关于布尔转换器的非常好的例子， * Boolean expression (grammar) parser in c++
我现在想的是再做一步，将(!T|F)&T翻译成F或0，这样计算很长的布尔表达式非常方便。
是否有一些使用精神的例子？我所做的是首先制作一个计算器，然后让它计算'（T +！F * T）'，它等于(T||!F&&T)但是当我输入()时，就会出现错误。如何修改？非常感谢！
#include <iostream>
#include <stack>
#include <boost/lexical_cast.hpp>
#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
using namespace std;
namespace phoenix = boost::phoenix;
namespace qi = boost::spirit::qi;
namespace ascii = boost::spirit::ascii;
struct calculator
{
bool interpret(const string& s);
void do_neg();
void do_add();
void do_sub();
void do_mul();
void do_div();
void do_number(const char* first, const char* last);
int val() const;
private:
stack<int> values_;
int *pn1_, n2_;
void pop_1();
void pop_2();
};
template <typename Iterator>
struct calc_grammar : qi::grammar<Iterator, ascii::space_type>
{
calc_grammar(calculator& calc)
: calc_grammar::base_type(add_sub_expr)
, calc_(calc)
{
using namespace qi;
using boost::iterator_range;
#define LAZY_FUN0(f) phoenix::bind(&calculator::f, calc_)
#define LAZY_FUN2(f) phoenix::bind(&calculator::f, calc_, phoenix::bind(&iterator_range<Iterator>::begin, qi::_1), phoenix::bind(&iterator_range<Iterator>::end, qi::_1))
add_sub_expr =
(
-lit('+') >> mul_div_expr |
(lit('-') >> mul_div_expr)[LAZY_FUN0(do_neg)]
) >>
*(
lit('+') >> mul_div_expr[LAZY_FUN0(do_add)] |
lit('-') >> mul_div_expr[LAZY_FUN0(do_sub)]
) >> eoi;
mul_div_expr =
basic_expr >>
*(
lit('*') >> basic_expr[LAZY_FUN0(do_mul)] |
lit('/') >> basic_expr[LAZY_FUN0(do_div)]
);
basic_expr =
raw[number][LAZY_FUN2(do_number)] |
lit('(') >> add_sub_expr >> lit(')');
number = lexeme[+digit];
}
qi::rule<Iterator, ascii::space_type> add_sub_expr, mul_div_expr, basic_expr, number;
calculator& calc_;
};
bool calculator::interpret(const string& s)
{
calc_grammar<const char*> g(*this);
const char* p = s.c_str();
return qi::phrase_parse(p, p + s.length(), g, ascii::space);
}
void calculator::pop_1()
{
pn1_ = &values_.top();
}
void calculator::pop_2()
{
n2_ = values_.top();
values_.pop();
pop_1();
}
void calculator::do_number(const char* first, const char* last)
{
string str(first, last);
int n = boost::lexical_cast<int>(str);
values_.push(n);
}
void calculator::do_neg()
{
pop_1();
*pn1_ = -*pn1_;
}
void calculator::do_add()
{
pop_2();
*pn1_ += n2_;
}
void calculator::do_sub()
{
pop_2();
*pn1_ -= n2_;
}
void calculator::do_mul()
{
pop_2();
*pn1_ *= n2_;
}
void calculator::do_div()
{
pop_2();
*pn1_ /= n2_;
}
int calculator::val() const
{
assert(values_.size() == 1);
return values_.top();
}
int main()
{
for(;;){
cout << ">>> ";
string s;
getline(cin, s);
if(s.empty()) break;
calculator calc;
if(calc.interpret(s))
cout << calc.val() << endl;
else
cout << "syntax error" << endl;
}
return 0;
}
答案 0 :(得分：11)
根据旧的Boolean Parser答案，这是一个快速而又肮脏的演示。这是一个评估您传递的AST的访问者：
struct eval : boost::static_visitor<bool>
{
eval() {}
//
bool operator()(const var& v) const
{
if (v=="T" || v=="t" || v=="true" || v=="True")
return true;
else if (v=="F" || v=="f" || v=="false" || v=="False")
return false;
return boost::lexical_cast<bool>(v);
}
bool operator()(const binop<op_and>& b) const
{
return recurse(b.oper1) && recurse(b.oper2);
}
bool operator()(const binop<op_or>& b) const
{
return recurse(b.oper1) || recurse(b.oper2);
}
bool operator()(const unop<op_not>& u) const
{
return !recurse(u.oper1);
}
private:
template<typename T>
bool recurse(T const& v) const
{ return boost::apply_visitor(*this, v); }
};
bool evaluate(const expr& e)
{ return boost::apply_visitor(eval(), e); }
我希望以后可以找一些时间来解释。请注意，_var现在是用词不当，因为您希望将所有操作数视为文字。另请注意，文字的评价现在有点......快速和肮脏：）
<强> 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>
#include <boost/lexical_cast.hpp>
namespace qi = boost::spirit::qi;
namespace phx = boost::phoenix;
struct op_or {};
struct op_and {};
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_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 eval : boost::static_visitor<bool>
{
eval() {}
//
bool operator()(const var& v) const
{
if (v=="T" || v=="t" || v=="true" || v=="True")
return true;
else if (v=="F" || v=="f" || v=="false" || v=="False")
return false;
return boost::lexical_cast<bool>(v);
}
bool operator()(const binop<op_and>& b) const
{
return recurse(b.oper1) && recurse(b.oper2);
}
bool operator()(const binop<op_or>& b) const
{
return recurse(b.oper1) || recurse(b.oper2);
}
bool operator()(const unop<op_not>& u) const
{
return !recurse(u.oper1);
}
private:
template<typename T>
bool recurse(T const& v) const
{ return boost::apply_visitor(*this, v); }
};
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 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 << ")";
}
};
bool evaluate(const expr& e)
{ return boost::apply_visitor(eval(), e); }
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_ = (and_ >> '|' >> or_ ) [ _val = phx::construct<binop<op_or > >(_1, _2) ] | and_ [ _val = _1 ];
and_ = (not_ >> '&' >> and_) [ _val = phx::construct<binop<op_and> >(_1, _2) ] | not_ [ _val = _1 ];
not_ = ('!' > simple ) [ _val = phx::construct<unop <op_not> >(_1) ] | simple [ _val = _1 ];
simple = (('(' > expr_ > ')') | var_);
var_ = qi::lexeme[ +(alpha|digit) ];
BOOST_SPIRIT_DEBUG_NODE(expr_);
BOOST_SPIRIT_DEBUG_NODE(or_);
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_, or_, simple, expr_;
};
int main()
{
const std::string inputs[] = {
std::string("true & false;"),
std::string("true & !false;"),
std::string("!true & false;"),
std::string("true | false;"),
std::string("true | !false;"),
std::string("!true | false;"),
std::string("T&F;"),
std::string("T&!F;"),
std::string("!T&F;"),
std::string("T|F;"),
std::string("T|!F;"),
std::string("!T|F;"),
std::string("") // marker
};
for (const std::string *i = inputs; !i->empty(); ++i)
{
typedef std::string::const_iterator It;
It f(i->begin()), l(i->end());
parser<It> 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:\t" << result << "\n";
std::cout << "evaluated:\t" << evaluate(result) << "\n";
}
} catch (const qi::expectation_failure<It>& 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;
}
输出：
result: (true & false)
evaluated: 0
result: (true & (!false))
evaluated: 1
result: ((!true) & false)
evaluated: 0
result: (true | false)
evaluated: 1
result: (true | (!false))
evaluated: 1
result: ((!true) | false)
evaluated: 0
result: (T & F)
evaluated: 0
result: (T & (!F))
evaluated: 1
result: ((!T) & F)
evaluated: 0
result: (T | F)
evaluated: 1
result: (T | (!F))
evaluated: 1
result: ((!T) | F)
evaluated: 0