On Sep 10, 4:36 pm, Joel de Guzman <j...@boost-consulting.com> wrote:
On 9/11/10 4:37 AM, alfC wrote:
I noticed that in some contexts I have both the object class and the corresponding phoenix function
class impl; ...
impl f_impl; boost::phoenix::function<impl> f(f_impl);
now to avoid the copying I could use:
boost::phoenix::function<impl&> f(f_impl);
which works. Although boost::phoenix::function<impl const&> f(f_impl); //doesn't work. (error below)
Is there any problem with using references at all, in the first place?
I'm not sure. Could you provide a minimal cpp file I can try?
Yes, see below. Althougth now that I made this example I realize that the underlying object implementation is copied many times anyway (at least 4 times in the example), I guess during the building up of the expression template. [The final objective was to make an object function that not only evaluates but also is able to take expression and build expression templates]. #include<boost/spirit/home/phoenix.hpp> #include<iostream> struct F{ double a_; F(double a) : a_(a){std::cout << "F constructed" <<std::endl;} F(F const& other) : a_(other.a_){std::cout << "F copied" <<std::endl;} template <typename Arg> struct result{ typedef double type; }; result<double>::type operator()(double d) const{ return a_*d*d; } }; int main(){ using boost::phoenix::function; using namespace boost::phoenix::arg_names; { F f(4.); boost::phoenix::function<F> pf(f); std::cout << pf(arg1+1.)((double const&)5.) << std::endl; } { F f(4.); boost::phoenix::function<F&> pf(f); std::cout << pf(arg1+1.)((double const&)5.) << std::endl; } /* block doesn't work { F f(4.); boost::phoenix::function<F const&> pf(f); //does not compile std::cout << pf(arg1+1.)((double const&)5.) << std::endl; } */ } prints: F constructed F copied F copied F copied F copied F copied 144 F constructed F copied F copied F copied F copied 144