On Wed, 10 May 2006, Rush Manbert wrote:
I have a template class that is designed to contain built in types or std::string types. I always need to be able to cast an instance of the class as a std::string, and I also need to be able to cast it as the parameter type with which it was instantiated.
In my class definition, I have defined two cast operators, like so: (Extraneous stuff omitted) [...] I thought that if I could conditionally define the cast as T operator (or the cast as std::string) I would be okay, but that means testing against the value of T with the preprocessor. I don't see how to do that, but I know that some of you folks are really good at bending the preprocessor to your will.
Can anyone show me a way out of this predicament?
No need for the preprocessor. It can all be done with templates. I could imagine something like this: template < typename Derived , typename T > struct convert_operator { operator const T& () { // ... your code, you can access the subclass with: // static_cast< Derived& >( * this ) // Add a const before Derived if the operator is const } } ; template < typename Derived > struct convert_operator< Derived , ::std::string > { // nothing } ; template < typename T > class MyDataObject : public MyDataObjectBase , public convert_operator< MyDadaObject< T > , T > { // Cast as std::string operator const std::string& () { // ...some code here } // ... } ; -- François Duranleau LIGUM, Université de Montréal "As the caterpillar chooses the fairest leaves to lay her eggs on, so the priest lays his curse on the fairest joys." - William Blake, _Marriage of Heaven & Hell_