标签：boost-spirit-qi c parsing boost-spirit expression-trees
首先,如果使用Boost Variant或Utree更容易,那么我将与他们达成和解,我将尝试在另一个主题中解决我们的问题.但是,我非常希望能够像下面一样构建一棵树.
背景,如果你想直接讨论这个问题,请忽略：我希望能够构建一个解析像
"({a} == 0) && ({b} > 5)"
或标准的数学表达式
"(2 * a) + b"
然后我将在评估树之前定义a和b是什么,如下所示：
a = 10;
double val = myExpression->Evaluate();
我的问题来自于当我尝试构建try以将字符串解析为我的表达式树时.我正在使用一个抽象类“Expression”,然后导出“变量”,“常量”和“二进制”表达式(它也会做一元,但它不应该影响我的问题.我一直有使用我的规则添加到树的问题所以我显然做错了什么.我很难绕着属性缠头.
我的树如下(Tree.h)：
class BinaryExpression;
typedef double (*func)(double, double);
class Expression
{
public:
virtual double Evaluate() = 0;
};
class BinaryExpression : public Expression
{
private:
Expression* lhs;
Expression* rhs;
func method;
double Evaluate();
public:
BinaryExpression(void);
BinaryExpression(char op, Expression* lhs, Expression* rhs);
BinaryExpression(char op);
void operator()(Expression* lhs, Expression* rhs);
};
class ConstantExpression : public Expression
{
private:
double value;
public:
ConstantExpression(void);
ConstantExpression(char op);
ConstantExpression(double val);
double Evaluate();
};
// Require as many types as there are fields in expression?
static double a;
static double b;
class VariableExpression : public Expression
{
private:
char op;
public:
VariableExpression(char op);
double Evaluate();
};
BOOST_FUSION_ADAPT_STRUCT(
BinaryExpression,
(Expression*, lhs)
(Expression*, rhs)
(func, method)
)
BOOST_FUSION_ADAPT_STRUCT(
VariableExpression,
(char, op)
)
BOOST_FUSION_ADAPT_STRUCT(
ConstantExpression,
(double, op)
)
Tree.cpp
typedef double (*func)(double, double);
/////////////////////////////////////////////////////////////////////////////
// BINARY EXPRESSION
////////////////////////////////////////////////////////////////////////////
BinaryExpression::BinaryExpression(void) {}
BinaryExpression::BinaryExpression(char op, Expression* lhs, Expression* rhs)
{
this->lhs = lhs;
this->rhs = rhs;
// Example, methods are held in another header
if (op == '+')
method = Add;
else if (op == '-')
method = Subtract;
}
double BinaryExpression::Evaluate()
{
return method(lhs->Evaluate(), rhs->Evaluate());
}
BinaryExpression::BinaryExpression(char op)
{
if (op == '+')
method = Add;
else if (op == '-')
method = Subtract;
}
void BinaryExpression::operator()(Expression* lhs, Expression* rhs)
{
this->lhs = lhs;
this->rhs = rhs;
}
/////////////////////////////////////////////////////////////////////////////
// CONSTANT EXPRESSION
////////////////////////////////////////////////////////////////////////////
ConstantExpression::ConstantExpression() {}
ConstantExpression::ConstantExpression(char op)
{
this->value = op - 48;
}
ConstantExpression::ConstantExpression(double val)
{
value = val;
}
double ConstantExpression::Evaluate()
{
return value;
}
/////////////////////////////////////////////////////////////////////////////
// VARIABLE EXPRESSION
////////////////////////////////////////////////////////////////////////////
VariableExpression::VariableExpression(char op)
{
this->op = op;
}
double VariableExpression::Evaluate()
{
// a and b are defined in the header, and are used to fill in the variables we want to evaluate
if (op == 'a')
return a;
if (op == 'b')
return b;
return 0;
}
现在,如果我手动构建树,它一切正常,所以我不认为它的结构方式存在问题.
这是Grammar.h(很多评论来自我尝试各种各样的东西,我可以删除它们,但我可能值得展示我尝试过的东西/我想去的地方)
#include "Tree.h"
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix_function.hpp>
namespace qi = boost::spirit::qi;
namespace ascii = boost::spirit::ascii;
qi::_1_type _1;
qi::_2_type _2;
// Pass functions to boost
boost::phoenix::function<BinaryExpression> plus = BinaryExpression('+');
boost::phoenix::function<BinaryExpression> minus = BinaryExpression('-');
template <typename Iterator>
struct ExpressionParser : qi::grammar<Iterator, BinaryExpression(), ascii::space_type>
{
ExpressionParser() : ExpressionParser::base_type(expression)
{
qi::_3_type _3;
qi::_4_type _4;
qi::char_type char_;
qi::uint_type uint_;
qi::_val_type _val;
qi::raw_type raw;
qi::lexeme_type lexeme;
qi::alpha_type alpha;
qi::alnum_type alnum;
qi::bool_type bool_;
qi::double_type double_;
expression = //?
additive_expr [_val = _1]
;
//equality_expr =
// relational_expr >>
// *(lit("==") > relational_expr) [/*Semantice action to add to tree*/]
// ;
additive_expr =
primary_expr >>
( '+' > primary_expr) [plus(_val, _1)]
| ( '-' > primary_expr) [minus(_val, _1)]
;
// Also tried "_val = plus(_1, _2)"
primary_expr =
constant [_val = _1]
| variable [_val = _1]
//| '(' > expression > ')' [_val = _1]
;
string %=
'{' >> *(char_ - '}') >> '}'
;
// Returns ConstantExpression
constant =
double_ [_val = _1];
// Returns VariableExpression
variable =
char_ [_val = _1]
;
}
// constant expression = double
// variable expression = string
qi::rule<Iterator, BinaryExpression(), ascii::space_type>
expression;
qi::rule<Iterator, BinaryExpression(), ascii::space_type>
// eventually will deal with all these rules
equality_expr,
relational_expr,
logical_expr,
additive_expr,
multiplicative_expr,
primary_expr
;
qi::rule<Iterator, ConstantExpression(), ascii::space_type>
constant
;
qi::rule<Iterator, VariableExpression(), ascii::space_type>
variable
;
qi::rule<Iterator, std::string(), ascii::space_type>
string
;
};
所以这是一个真正被黑客攻击,但希望它会显示我想要实现的目标.任何建议或提示将非常感激.有没有一个例子,有人建立了这样的树而不使用变体或utree.
也很抱歉,如果我违反惯例,并为我的格式,我试图使其尽可能可读.
解决方法:
我不清楚你对(递归)变体的抱怨是什么,但是这里有一个变化,你希望使用动态分配的节点来使用’老式’树构建：
> http://liveworkspace.org/code/3VS77n$0
我有目的地回避了你的语法中运算符优先级的问题,因为
>你的语法不完整
>我不知道所需的语义(毕竟,你似乎也支持布尔评估,但我不知道如何)
>您可以在其他答案中了解这些：
> Boolean expression (grammar) parser in c++
> Boost::Spirit Expression Parser
> Compilation error with a boost::spirit parser显示了另一种方法
请注意我
>使用shared_ptr删除无处不在的内存泄漏(如果没有TR1库,可以使用Boost)
>我删除了错误的重用特定BinaryExpression实例作为凤凰懒惰的演员.相反,我现在做了一个当地的二元演员.
>注意现在如何支持操作符链(1 2 5 6-10)：
additive_expr =
primary_expr [ _val = _1 ]
>> *(char_("-+*/") >> primary_expr) [ _val = makebinary(_1, _val, _2)]
;
>我添加了{var},/,*和(expr)支持
>为显示添加了序列化(打印虚拟方法,运算符<<)(为了便于显示,BinaryExpression现在存储运算符而不是结果方法)
>因此现在你可以使用BOOST_SPIRIT_DEBUG(取消注释第一行)
>我已将Expression重命名为AbstractExpression(并使de构造函数受保护)
>我已将PrimaryExpression重命名为Expression(现在这是您的主要表达式数据类型)
>我将展示如何在静态地图中存储简单变量
>一定要看看qi ::符号和
>例如How to add qi::symbols in grammar<Iterator,double()>?
>使用更少的融合结构适应(仅适用于变量)
>使用模板化的构造函数技巧,可以很容易地从不同的解析类型构造表达式：
struct Expression : AbstractExpression {
template <typename E>
Expression(E const& e) : _e(make_from(e)) { } // cloning the expression
// ...
};
足以有效支持例如：
primary_expr =
( '(' > expression > ')' ) [ _val = _1 ]
| constant [ _val = _1 ]
| variable [ _val = _1 ]
;
>为了好玩,还包括了一些测试用例：
Input: 3*8 + 6
Expression: Expression(BinaryExpression(BinaryExpression(ConstantExpression(3) * ConstantExpression(8)) + ConstantExpression(6)))
Parse success: true
Remaining unparsed: ''
(a, b): 0, 0
Evaluation result: 30
----------------------------------------
Input: 3*(8+6)
Expression: Expression(BinaryExpression(ConstantExpression(3) * BinaryExpression(ConstantExpression(8) + ConstantExpression(6))))
Parse success: true
Remaining unparsed: ''
(a, b): 0, 0
Evaluation result: 42
----------------------------------------
Input: 0x1b
Expression: Expression(ConstantExpression(27))
Parse success: true
Remaining unparsed: ''
(a, b): 0, 0
Evaluation result: 27
----------------------------------------
Input: 1/3
Expression: Expression(BinaryExpression(ConstantExpression(1) / ConstantExpression(3)))
Parse success: true
Remaining unparsed: ''
(a, b): 0, 0
Evaluation result: 0.333333
----------------------------------------
Input: .3333 * 8e12
Expression: Expression(BinaryExpression(ConstantExpression(0.3333) * ConstantExpression(8e+12)))
Parse success: true
Remaining unparsed: ''
(a, b): 0, 0
Evaluation result: 2.6664e+12
----------------------------------------
Input: (2 * a) + b
Expression: Expression(BinaryExpression(BinaryExpression(ConstantExpression(2) * VariableExpression('a')) + VariableExpression('b')))
Parse success: true
Remaining unparsed: ''
(a, b): 10, 7
Evaluation result: 27
----------------------------------------
Input: (2 * a) + b
Expression: Expression(BinaryExpression(BinaryExpression(ConstantExpression(2) * VariableExpression('a')) + VariableExpression('b')))
Parse success: true
Remaining unparsed: ''
(a, b): -10, 800
Evaluation result: 780
----------------------------------------
Input: (2 * {a}) + b
Expression: Expression(BinaryExpression(BinaryExpression(ConstantExpression(2) * VariableExpression('a')) + VariableExpression('b')))
Parse success: true
Remaining unparsed: ''
(a, b): -10, 800
Evaluation result: 780
----------------------------------------
Input: {names with spaces}
Expression: Expression(VariableExpression('names with spaces'))
Parse success: true
Remaining unparsed: ''
(a, b): 0, 0
Evaluation result: 0
----------------------------------------
// #define BOOST_SPIRIT_DEBUG
// #define BOOST_RESULT_OF_USE_DECLTYPE
// #define BOOST_SPIRIT_USE_PHOENIX_V3
#include <cassert>
#include <memory>
#include <iostream>
#include <map>
struct AbstractExpression;
typedef std::shared_ptr<AbstractExpression> Ptr;
struct AbstractExpression {
virtual ~AbstractExpression() {}
virtual double Evaluate() const = 0;
virtual std::ostream& Print(std::ostream& os) const = 0;
friend std::ostream& operator<<(std::ostream& os, AbstractExpression const& e)
{ return e.Print(os); }
protected: AbstractExpression() {}
};
template <typename Expr> // general purpose, static Expression cloner
static Ptr make_from(Expr const& t) { return std::make_shared<Expr>(t); }
struct BinaryExpression : AbstractExpression
{
BinaryExpression() {}
template<typename L, typename R>
BinaryExpression(char op, L const& l, R const& r)
: _op(op), _lhs(make_from(l)), _rhs(make_from(r))
{}
double Evaluate() const {
func f = Method(_op);
assert(f && _lhs && _rhs);
return f(_lhs->Evaluate(), _rhs->Evaluate());
}
private:
char _op;
Ptr _lhs, _rhs;
typedef double(*func)(double, double);
static double Add(double a, double b) { return a+b; }
static double Subtract(double a, double b) { return a-b; }
static double Multuply(double a, double b) { return a*b; }
static double Divide(double a, double b) { return a/b; }
static BinaryExpression::func Method(char op)
{
switch(op) {
case '+': return Add;
case '-': return Subtract;
case '*': return Multuply;
case '/': return Divide;
default: return nullptr;
}
}
std::ostream& Print(std::ostream& os) const
{ return os << "BinaryExpression(" << *_lhs << " " << _op << " " << *_rhs << ")"; }
};
struct ConstantExpression : AbstractExpression {
double value;
ConstantExpression(double v = 0) : value(v) {}
double Evaluate() const { return value; }
virtual std::ostream& Print(std::ostream& os) const
{ return os << "ConstantExpression(" << value << ")"; }
};
struct VariableExpression : AbstractExpression {
std::string _name;
static double& get(std::string const& name) {
static std::map<std::string, double> _symbols;
return _symbols[name];
/*switch(name) {
* case 'a': static double a; return a;
* case 'b': static double b; return b;
* default: throw "undefined variable";
*}
*/
}
double Evaluate() const { return get(_name); }
virtual std::ostream& Print(std::ostream& os) const
{ return os << "VariableExpression('" << _name << "')"; }
};
struct Expression : AbstractExpression
{
Expression() { }
template <typename E>
Expression(E const& e) : _e(make_from(e)) { } // cloning the expression
double Evaluate() const { assert(_e); return _e->Evaluate(); }
// special purpose overload to avoid unnecessary wrapping
friend Ptr make_from(Expression const& t) { return t._e; }
private:
Ptr _e;
virtual std::ostream& Print(std::ostream& os) const
{ return os << "Expression(" << *_e << ")"; }
};
//Tree.cpp
/////////////////////////////////////////////////////////////////////////////
// BINARY EXPRESSION
////////////////////////////////////////////////////////////////////////////
//#include "Tree.h"
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/fusion/adapted.hpp>
BOOST_FUSION_ADAPT_STRUCT(VariableExpression, (std::string, _name))
namespace qi = boost::spirit::qi;
namespace ascii = boost::spirit::ascii;
namespace phx = boost::phoenix;
// Pass functions to boost
template <typename Iterator>
struct ExpressionParser : qi::grammar<Iterator, Expression(), ascii::space_type>
{
struct MakeBinaryExpression {
template<typename,typename,typename> struct result { typedef BinaryExpression type; };
template<typename C, typename L, typename R>
BinaryExpression operator()(C op, L const& lhs, R const& rhs) const
{ return BinaryExpression(op, lhs, rhs); }
};
phx::function<MakeBinaryExpression> makebinary;
ExpressionParser() : ExpressionParser::base_type(expression)
{
using namespace qi;
expression =
additive_expr [ _val = _1]
;
additive_expr =
primary_expr [ _val = _1 ]
>> *(char_("-+*/") >> primary_expr) [ _val = makebinary(_1, _val, _2)]
;
primary_expr =
( '(' > expression > ')' ) [ _val = _1 ]
| constant [ _val = _1 ]
| variable [ _val = _1 ]
;
constant = lexeme ["0x" >> hex] | double_ | int_;
string = '{' >> lexeme [ *~char_("}") ] > '}';
variable = string | as_string [ alpha ];
BOOST_SPIRIT_DEBUG_NODE(expression);
BOOST_SPIRIT_DEBUG_NODE(additive_expr);
BOOST_SPIRIT_DEBUG_NODE(primary_expr);
BOOST_SPIRIT_DEBUG_NODE(constant);
BOOST_SPIRIT_DEBUG_NODE(variable);
BOOST_SPIRIT_DEBUG_NODE(string);
}
qi::rule<Iterator, Expression() , ascii::space_type> expression;
qi::rule<Iterator, Expression() , ascii::space_type> additive_expr;
qi::rule<Iterator, Expression() , ascii::space_type> primary_expr;
qi::rule<Iterator, ConstantExpression(), ascii::space_type> constant;
qi::rule<Iterator, VariableExpression(), ascii::space_type> variable;
qi::rule<Iterator, std::string() , ascii::space_type> string;
};
void test(const std::string input, double a=0, double b=0)
{
typedef std::string::const_iterator It;
ExpressionParser<It> p;
Expression e;
It f(input.begin()), l(input.end());
bool ok = qi::phrase_parse(f,l,p,ascii::space,e);
std::cout << "Input: " << input << "\n";
std::cout << "Expression: " << e << "\n";
std::cout << "Parse success: " << std::boolalpha << ok << "\n";
std::cout << "Remaining unparsed: '" << std::string(f,l) << "'\n";
std::cout << "(a, b): " << a << ", " << b << "\n";
VariableExpression::get("a") = a;
VariableExpression::get("b") = b;
std::cout << "Evaluation result: " << e.Evaluate() << "\n";
std::cout << "----------------------------------------\n";
}
int main()
{
test("3*8 + 6");
test("3*(8+6)");
test("0x1b");
test("1/3");
test(".3333 * 8e12");
test("(2 * a) + b", 10, 7);
test("(2 * a) + b", -10, 800);
test("(2 * {a}) + b", -10, 800);
test("{names with spaces}");
}
标签：boost-spirit-qi,c,parsing,boost-spirit,expression-trees 来源： https://codeday.me/bug/20190929/1830036.html
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