Default template parameters with forward declaration
Posted
by Seth Johnson
on Stack Overflow
See other posts from Stack Overflow
or by Seth Johnson
Published on 2009-11-24T19:09:02Z
Indexed on
2010/04/28
5:53 UTC
Read the original article
Hit count: 351
Is it possible to forward declare a class that uses default arguments without specifying or knowing those arguments?
For example, I would like to declare a boost::ptr_list< TYPE >
in a Traits class without dragging the entire Boost library into every file that includes the traits. I would like to declare
namespace boost { template<class T> class ptr_list< T >; }
, but that doesn't work because it doesn't exactly match the true class declaration:
template < class T,
class CloneAllocator = heap_clone_allocator,
class Allocator = std::allocator<void*>
>
class ptr_list { ... };
Are my options only to live with it or to specify boost::ptr_list< TYPE, boost::heap_clone_allocator, std::allocator<void*>
in my traits class? (If I use the latter, I'll also have to forward declare boost::heap_clone_allocator
and include <memory>
, I suppose.)
I've looked through Stroustrup's book, SO, and the rest of the internet and haven't found a solution. Usually people are concerned about not including STL, and the solution is "just include the STL headers." However, Boost is a much more massive and compiler-intensive library, so I'd prefer to leave it out unless I absolutely have to.
© Stack Overflow or respective owner