At 19:02 29/01/2020, Glen Fernandes wrote:
You shouldn't expect that. For any unique_ptr
you can only expect that .get() gives you a D::pointer. For default_delete<T>
this might be T*. But it always depends on the Deleter.
That seems like a significantly more annoying interface than std::shared_ptr uses. shared_ptr can hold fancy pointers too, but it happens explicitly via using fancy<T> rather than plain T, so the explicit indirection on usage is obvious too. Having that be an implementation detail of the deleter changing the effective "type" of the first template argument (which is what people think of as the type of the pointer, even if the standard disagrees) seems quite annoying. Maybe we should have a type-erased unique ownership pointer as well. I know the reason why we didn't was to avoid heap allocation or wasted storage for the common case, but there ought to be some happy middle ground.
The correct way to get a raw pointer from any potentially-fancy-pointer x is: to_address(x)
This can be boost::to_address (C++03 or higher), or std::to_address (C++20 or higher)
Hmm. This is news to me; I don't think I've read about it anywhere in conjunction with unique_ptr. Perhaps since it doesn't actually exist yet (most compilers are still mostly C++17). But eg. https://en.cppreference.com/w/cpp/memory/unique_ptrhttps://en.cppreference.com/w/cpp/memory/unique_ptr appears entirely mute on the topic. (There's a passing reference to fancy pointers but even that doesn't talk about it.)