CODEWITHSUNDEEP
c++boostboost-spiritboost-spirit-qiboost-phoenix
rickba
rickba

Reputation: 53

How to access boost::variant members from Spirit::Qi rule?

I can't find a proper way how to access members of boost::variant using boost::phoenix in my Spirit-Qi grammar. Here is simple an example what I’m trying to achieve. (my whole grammar is much more complex, this is simple fragment where I'm testing mentioned problem).

namespace ph = boost::phoenix;
typedef boost::variant<std::string,int> VariantType;
typedef std::list<VariantType> TlstVariants;

rule<Iterator, void(TlstVariants&), Skipper>    rule1;

rule1 = 
   qi::eps [ ph::push_back(qi::_r1, ph::construct<int>(2)) ]
>> qi::eps [ ph::get<int>(ph::back(qi::_r1)) = ph::val(3) ] //THIS IS EXAMPLE OF WHAT I NEED
;

TlstVariants lstVals;
ExecuteParser("5",rule1( ph::ref(lstVals) ));   

BOOST_FOREACH( VariantType &val, lstVals )
{
    std::cout << val.which() << " - " << val;
}

But I can't find any phoenix::get<> or any similar method to access boost::variant using Phoenix. Reason why I need phoenix::get<> is because I need to insert variant to list with specific type and then pass this specific type as reference to the child rule as inherited attribute:

qi::rule<Iterator, void(structTest&), Skipper> rule_child;

rule = 
  qi::lit("test") [ph::push_back(sp::_r1, ph::construct<structTest>())]
> qi::lit('(') 
> rule_child( ph::get<structTest>(ph::back(sp::_r1)) ) 
> qi::lit(')') 
...

Is there any way how to achieve behavior like this?

Thank for any reply

Rick

Upvotes: 4

Views: 1567

Answers (2)

rickba
rickba

Reputation: 53

I think I found a way how to do it. (Don't know if this is best way, but it works ;-) ). The problem was in int& type, because boost::variant holds int and not int&. So I update your template to accept two types, one for variant getter and one for return type.

I updated get_impl template in this way:

template <typename Result, typename Inner>
struct get_impl
{
    template <typename T>
    struct result
    {
        typedef Result type;
    };

    template <BOOST_VARIANT_ENUM_PARAMS(typename T)>
    Result operator()(boost::variant<BOOST_VARIANT_ENUM_PARAMS(T)> & v) const
    {
        return boost::get<Inner>(v);
    }
};

And my grammar now looks like this:

typedef boost::variant<std::string,int> VariantType;
qi::rule<DG_Iterator, void(VariantType&), DG_Skipper>   rule1;

ph::function<get_impl<int,int> >  const get_int = get_impl<int, int>();
ph::function<get_impl<int&,int> > const get_int_ref = get_impl<int&,int>();

rule1 = 
    qi::eps [ std::cout << ph::val("variant=") << qi::_r1 << ph::val("\n") ]
>> qi::eps [ std::cout << ph::val("before=") << get_int(qi::_r1) << ph::val("\n") ]
>> qi::eps [ get_int_ref(qi::_r1) = ph::val(7) ]
>> qi::eps [ std::cout << ph::val("after=") << get_int(qi::_r1) << ph::val("\n") ]
;

VariantType val(2134);
TestSimpleRuleValidity("x",rule1( ph::ref(val) ), true);    
std::cout << val << "\n";

And everything seems to works Ok. Thank you again hkaiser for your initial reply which helps me a lot.

Upvotes: 1

hkaiser
hkaiser

Reputation: 11521

It's fairly easy to write your own 'lazy' Phoenix function. Here is one for boost::variant.

#include <boost/variant.hpp>
#include <boost/spirit/include/phoenix.hpp>

template <typename Result>
struct get_impl
{
    template <typename T>
    struct result
    {
        typedef Result type;
    };

    template <BOOST_VARIANT_ENUM_PARAMS(typename T)>
    Result operator()(boost::variant<BOOST_VARIANT_ENUM_PARAMS(T)> const& v) const
    {
        return boost::get<Result>(v);
    }
};

ph::function<get_impl<int> > const get_int = get_impl<int>();

Now, this can be used in a semantic action:

... qi::eps [ get_int(ph::back(qi::_r1)) = ph::val(3) ]

Upvotes: 3

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