Re: [Boost-users] [Proto] proto::function and constness

Subject: Re: [Boost-users] [Proto] proto::function and constness
From: Kim Kuen Tang (kuentang_at_[hidden])
Date: 2009-10-12 15:00:24

Hi Eric,

Eric Niebler schrieb:
> You're not missing anything. Proto handle operator() differently than 
> the other operators: operator() captures its arguments by const ref, 
> whereas the other operators have overloads for const and non-const 
> ref. Why the difference? Because operator() can have up to 
> BOOST_PROTO_MAX_FUNCTION_CALL_ARITY arguments, which defaults to 5. To 
> accomodate every permutation of const and non-const arguments would 
> require 2^5 + 2^4 + 2^3 + 2^2 + 2^1 + 2^0 overloads, which is ... a lot.
2^6-1=63 is very a lot.
>
> So what are your options. my suggestion is to first change this:
>
>     boost::result_of<Rhs(Expr)>::type & result = Rhs()(expr);
>
> to more correct:
>
>     boost::result_of<Rhs(Expr&)>::type result = Rhs()(expr);

Ok, this i understand.
> Now, if you are certain that the result is really a mutable lvalue, 
> you can const_cast the result to a std::vector<int>& and mutate it.
>
> The alternative would be to make _o_ an expression extension with 
> overloads of operator() to handle all the permutations.
Thanks for your advice.

Best regards,
Kim