How to make a boost-proto function expression streamable?

Question

I'm going over the boost-proto tutorial, and ran into this problem with the lazy pow function example. This is the example code:

// Define a pow_fun function object
template<int Exp> // , typename Func>
struct pow_fun
{
    typedef double result_type;
    double operator()(double d) const
    {
        return pow(d, Exp);
    }
};

// Define a lazy pow() function for the calculator DSEL.
// Can be used as: pow< 2 >(_1)
template<int Exp, typename Arg>
typename proto::result_of::make_expr<
    proto::tag::function  // Tag type
  , pow_fun<Exp>          // First child (by value)
  , Arg const &           // Second child (by reference)
>::type const
mypow(Arg const &arg)
{
    return proto::make_expr<proto::tag::function>(
        pow_fun<Exp>()    // First child (by value)
      , boost::ref(arg)   // Second child (by reference)
    );    
}

Now, if I try to

proto::display_expr( mypow<2>(_1) );

the compiler complains that it doesn't have operator<< for the function expression. How do I define one?

Thanks.

The compiler error is:

/usr/include/boost/proto/debug.hpp:146: error: no match for ‘operator<<’ in ‘std::operator<< [with _Traits = std::char_traits](((std::basic_ostream >&)((std::basic_ostream >*)std::operator<< [with _Traits = std::char_traits](((std::basic_ostream >&)((std::basic_ostream >*)std::operator<< [with _Traits = std::char_traits](((std::basic_ostream >&)((std::basic_ostream >*)std::operator<< [with _CharT = char, _Traits = std::char_traits](((std::basic_ostream >&)((std::ostream*)((const boost::proto::functional::display_expr*)this)->boost::proto::functional::display_expr::sout_)), std::setw(((const boost::proto::functional::display_expr*)this)->boost::proto::functional::display_expr::depth_)))), (((const boost::proto::functional::display_expr*)this)->boost::proto::functional::display_expr::first_ ? ((const char*)"") : ((const char*)", "))))), boost::proto::tag::proto_tag_name((boost::proto::tag::terminal(), boost::proto::tag::terminal()))))), ((const char*)"(")) << boost::proto::value [with Expr = boost::proto::exprns_::expr >, 0l>](((const boost::proto::exprns_::expr >, 0l>&)((const boost::proto::exprns_::expr >, 0l>*)expr)))’

Joel Falcou Joel Falcou · Accepted Answer · 2011-11-25T06:50:43

Which proto version is this ? The latest don't require the << overload anymore and default to typeid to display name if needed. Could you post the actual error message ?

How to make a boost-proto function expression streamable?

1 Answers