Compilation error by switching to higher Boost Version 1.6.1

Question

I switched my Boost version from 1.6.1 to >=1.6.2 and my boost::spirit parser code fails to compile. Actually, I thinking the problem has something to do with a bug fix in Boost Variant from version 1.6.1 to version 1.6.2.

Release notes of version 1.6.2 say:

Variant constructors and assignment operators now do not participate in overload resolutions if variant can not hold the input type #5871, #11602

Here is a stripped version of my failing code:

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;

}

The code compiles without problems in Version 1.6.1, but fails in Version 1.6.2 with the error:

.../proto/transform/default.hpp(154): error C2679: Binary operator "=": ...

I guess in Version 1.6.1 there was an implicit conversion from computationExpr to expr, which is no longer allowed.

How can I fix this code? I think something in _val = construct<Acc>(_1) must be changed, but I'm lacking the skills to do it.

1.6.1? 1.6.2? Can you clarify what Boost version you mean: http://www.boost.org/users/history/ — sehe, Jan 04 '18 at 22:36

score 3 · Accepted Answer · answered Jan 05 '18 at 01:18

Indeed, the recursive_wrapper restricts the options for implicit construction more since 1.62:

Wandbox on Boost 1.61

boost::variant<int, boost::recursive_wrapper<std::string> > x;
x = "worked before";
std::cout << boost::get<std::string>(x) << "\n";

Broken on Boost 1.62

boost::variant<int, boost::recursive_wrapper<std::string> > x;
x = "worked before";
std::cout << boost::get<std::string>(x) << "\n";

In this case, it's easy to fix: Fixed on Boost 1.62

x = std::string("Hello world");

Your Code

In your code the nested use of recursive wrappers complicates things. The good news is, you don't need to have two layers. Just drop one:

typedef boost::variant<
    int,
    computationExpr
> expr;

The instantiation is already sufficiently decoupled by the second recursive wrapper. Now, everything is fine again.

Demo Time

Note some style fixes/suggestions:

Also, I reordered the elements in the expr variant because they were triggering infinite recursion on default construction.

Live On Coliru

#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";
}

Prints

Parsed: Acc(3)

Many thanks. You really helped me out. Seemingly, there is still a lot to learn from your thorough answers. Now everything compiles. — Aleph0, Jan 05 '18 at 06:36

Compilation error by switching to higher Boost Version 1.6.1

1 Answers1

Your Code

Demo Time