我将Boost版本从1.6.1切换到> = 1.6.2并且我的boost::spirit
解析器代码无法编译。实际上,我认为这个问题与Boost Variant中从版本1.6.1到版本1.6.2的错误修复有关。
版本1.6.2的发行说明说:
Variant constructors and assignment operators now do not participate in overload resolutions if variant can not hold the input type #5871, #11602
这是我失败的代码的剥离版本:
Parser.h
#pragma once
#include <string>
#include <boost/variant.hpp>
struct AccTag {};
template <typename tag> struct unop;
typedef unop<AccTag> Acc;
typedef boost::variant<
boost::recursive_wrapper<Acc>
> computationExpr;
typedef boost::variant<
boost::recursive_wrapper<computationExpr>,
int
> expr;
template <typename tag> struct unop
{
unop() : oper1() {
}
explicit unop(const expr& o) : oper1(o) { }
expr oper1;
};
expr parse(const std::string& expression, bool& ok);
Parser.cpp
#include "Parser.h"
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
using namespace boost;
template <typename Iterator = std::string::iterator, typename Skipper = spirit::qi::space_type>
class ParserImpl : public spirit::qi::grammar<Iterator, expr(), Skipper>
{
public:
ParserImpl() : ParserImpl::base_type(expr_)
{
using namespace boost::spirit::qi;
using namespace boost::phoenix;
expr_ = props_.alias();
props_ = (
(lit("Acc") >> "(" >> int_ >> ")")[_val = construct<Acc>(_1) /* Most likely the source of the error */]
);
}
spirit::qi::rule<Iterator, expr(), Skipper> props_;
spirit::qi::rule<Iterator, expr(), Skipper> expr_;
};
expr parse(const std::string& expression, bool& ok)
{
expr result;
std::string formula = expression;
ParserImpl<> parser;
auto b = formula.begin();
auto e = formula.end();
ok = spirit::qi::phrase_parse(b, e, parser, spirit::qi::space, result);
if (b != e) {
ok = false;
}
return result;
}
代码在版本1.6.1中编译没有问题,但在版本1.6.2中失败并出现错误:
.../proto/transform/default.hpp(154): error C2679: Binary operator "=": ...
我想在版本1.6.1中有一个从computationExpr
到expr
的隐式转换,这是不再允许的。
我该如何修复此代码?我认为_val = construct<Acc>(_1)
中的某些内容必须改变,但我缺乏这样做的技巧。
实际上,自1.62以来,recursive_wrapper更多地限制了隐式构造的选项:
boost::variant<int, boost::recursive_wrapper<std::string> > x;
x = "worked before";
std::cout << boost::get<std::string>(x) << "\n";
boost::variant<int, boost::recursive_wrapper<std::string> > x;
x = "worked before";
std::cout << boost::get<std::string>(x) << "\n";
在这种情况下,它很容易修复:Fixed on Boost 1.62
x = std::string("Hello world");
在您的代码中,嵌套使用递归包装器会使事情变得复杂。好消息是,你不需要有两层。只需放一个:
typedef boost::variant<
int,
computationExpr
> expr;
实例化已经被第二个递归包装器充分地解耦。现在,一切都很好。
请注意一些样式修复/建议:
此外,我重新排序了
expr
变体中的元素,因为它们在默认构造中触发无限递归。
#pragma once
#include <string>
#include <boost/variant.hpp>
struct AccTag {};
template <typename> struct unop;
typedef unop<AccTag> Acc;
typedef boost::variant<
boost::recursive_wrapper<Acc>
> computationExpr;
typedef boost::variant<
int,
computationExpr
> expr;
template <typename> struct unop {
unop() : oper1() { }
explicit unop(const expr& o) : oper1(o) { }
expr oper1;
};
expr parse(const std::string& expression, bool& ok);
#include "Parser.h"
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
namespace qi = boost::spirit::qi;
template <typename Iterator = std::string::const_iterator, typename Skipper = qi::space_type>
class ParserImpl : public qi::grammar<Iterator, expr(), Skipper>
{
public:
ParserImpl() : ParserImpl::base_type(expr_)
{
namespace phx = boost::phoenix;
using namespace qi;
expr_ = props_.alias();
props_ =
(lit("Acc") >> '(' >> int_ >> ')')[_val = phx::construct<Acc>(_1)]
;
}
private:
qi::rule<Iterator, expr(), Skipper> props_;
qi::rule<Iterator, expr(), Skipper> expr_;
};
expr parse(const std::string& formula, bool& ok)
{
expr result;
ParserImpl<> parser;
auto b = formula.begin();
auto e = formula.end();
ok = qi::phrase_parse(b, e, parser >> qi::eoi, qi::space, result);
return result;
}
static inline std::ostream& operator<<(std::ostream& os, Acc const& o) {
return os << "Acc(" << o.oper1 << ")";
}
int main() {
bool ok;
auto e = parse("Acc (3)", ok);
if (ok)
std::cout << "Parsed: " << e << "\n";
else
std::cout << "Parse failed\n";
}
打印
Parsed: Acc(3)