[Boost-users] Proto: extending the example mixed.cpp

Subject: [Boost-users] Proto: extending the example mixed.cpp
From: Kim Kuen Tang (kuentang_at_[hidden])
Date: 2008-12-11 16:09:31

Hi all,

i wnat to extend the example in the file mixed.cpp with an operator[]:

double operator []( std::ptrdiff_t i ) const
    {
        DereferenceCtx const deref = {};
        AdvanceCtx adctx(i); //just shift the iterator with std::advance
        typename boost::result_of<Begin(Expr const &)>::type expr2 = 
Begin()(*this);
        proto::eval(expr2,adctx);
        return proto::eval(expr2, deref);
    }
This works well.

However is it possible to let proto figure out what the result_type of 
this function is?

I've tried to program in this way.

typedef typename boost::result_of<Begin(Expr const &)>::type Expr2;
    typedef typename proto::result_of::eval<Expr,DereferenceCtx>::type 
result_type;
    result_type operator []( std::ptrdiff_t i ) const
    {
        DereferenceCtx const deref = {};
        AdvanceCtx adctx(i);
        Expr2 expr2 = Begin()(*this);
        proto::eval(expr2,adctx);
        return proto::eval(expr2, deref);
    }

But the compiler complained that the nested struct eval in the struct 
struct DereferenceCtx
{
  
    template<typename Expr, typename EnableIf = void>
    struct eval
        : proto::default_eval<Expr, DereferenceCtx const>
    {};

cannot be instantiated by this particulat template. So I guess I ' ve to 
write a grammar for this issue. Is this true?