Jody Hagins wrote:
The library, as used in Boost.Python, relies on calling the no-parameter template function, type_id, to get the type info object. Use of typeid() by itself is absent, because of the problem with compilers getting the typeid() implementation correct. For example:
struct foo { /* ... */ }; boost::python::type_info ti = boost::python::type_id< foo >();
Unfortunately, boost::python::type_id<>() has a couple major drawbacks (and a number of "minor" ones).
1. It only supports type-id calls, with a specific type passed as a template parameter. The ability to compute type_info based on the result of an expression is scant (a bit more functionality can possibly be obtained with more tricks like the ingenious typeof() stuff recently posted).
Please have a look at the article just published at the C++ Source (http://www.artima.com/cppsource/index.jsp) entitled "Conditional Love". If you want a BOOST_TYPEID macro that accepts an expression and returns the type_info without evaluating the expression, it's really quite simple: template<typename T> T const * encode_type( T const & ) { return 0; } template<typename T> type_info type_id_helper( T const * ) { // return type_info for type T } #define BOOST_TYPEID( expr ) \ type_id_helper( true? 0 : encode_type( expr ) ) With carefully selected overloads of encode_type and type_id_helper, you can correctly handle cv-qualifier and rvalues/lvalues. -- Eric Niebler Boost Consulting www.boost-consulting.com