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From: Howard Hinnant (hinnant_at_[hidden])
Date: 2001-10-11 08:33:30
On Thursday, October 11, 2001, at 07:13 AM, Peter Dimov wrote:
> Let's go back to the original question: is std::ostringstream a complete
> type after <sstream> has been included?
I'm not sure (and I would like to be).
I've read 27.7 and the only thing I can say for sure is that after
including <sstream> std::ostringstream is a typedef for
std::basic_ostringstream<char>. And std::basic_ostringstream<charT +
defaults> is a class _template_.
So perhaps the question can be reworded: Is a class template a complete
type?
template <class T> struct A;
template <class T>
struct B
: A<T>
{
};
B<T> is a class template now. Is it a complete type? Or is it just a
template for a type?
template <class T>
struct C
: T
{
};
Is C<T> a complete type now? I'm not positive that a class template can
constitute a complete type. However I think we both agree that a class
instantiated from a class template *is* a complete type.
I think the real question is: Can std::ostringstream be instantiated
after including <sstream>? I hope the answer is yes.
Before we get too far down the inheritance argument, there are other
examples which don't include inheritance. Is this portable code?
#include <string>
int
main()
{
std::string s;
}
I hope yes. But we've just instantiated and constructed a
std::allocator<char> (at least conceptually) which is defined over in
<memory>.
template<class charT, class traits = char_traits<charT>,
class Allocator = allocator<charT> >
class basic_string
{
...
explicit basic_string(const Allocator& a = Allocator());
...
};
Do we need to include <memory>? I hope not. But how do we know that we
don't need to include <memory>? Once we've decided that we don't need
to include <memory> in the above example, is this code legal:
#include <string>
int
main()
{
std::allocator<char> a;
}
I hope not. Though I'm guessing it would work on most of today's
platforms anyway.
-Howard
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