Boost Users :
From: Jens Theisen (jth01_at_[hidden])
Date: 2006-01-25 15:46:00
>> Why is there a need? Presumably you're potential surrogate for shared_ptr
>> will also support operator *, and that's all you use in your function.
> Why is there a need to do what? I'm not following you here.
You can always do
template< typename Pointer >
void print_handle(Pointer const& ptr)
std::cout << *ptr << std::endl;
which will work on any smart pointer.
The drawbacks are more weird error messages and less precise overload
resolution since the above matches on any type, not only pointers.
> Would the upcoming template typedef feature
> avoid the need to specify Type in the call?
I doubt it, though I haven't read the proposal. I'm pretty it's about an
abbreviation for traits classes so it's not going to introduce something
you don't have.
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