Yuval Ronen wrote:
Hi. There's a shared_ptr constructor that accepts an auto_ptr<Y>&. Shouldn't it accept auto_ptr<Y> by value rather by non-const reference? It seems to me that the reference does not add anything, but only causes a VC warning (level 4):
--- warning C4239: nonstandard extension used : 'argument' : conversion from 'std::auto_ptr<_Ty>' to 'std::auto_ptr<_Ty> &' A reference that is not to 'const' cannot be bound to a non-lvalue ---
when calling this constructor with a temporary auto_ptr. This warning is a good thing and I don't want to disable it.
Thanks, Yuval
Don't call the constructor with a temporary. Anyway, there's a good reason for this: the semantics of auto_ptr. auto_ptr works with transfer of ownership: there can only be one auto_ptr at any time holding a specific pointer. When you assign or copy-construct auto_ptrs, ownership is transferred: the old pointer is set to NULL. Due to this, the auto_ptr will always destruct its pointee when it gets destructed. auto_ptr does not care about shared_ptr's reference count. To have shared_ptr not point to deleted memory, constructing a shared_ptr from an auto_ptr must prevent auto_ptr from deleting the pointee. And the only way to do that is to set it to NULL. This is why it needs to be passed as non-const reference; otherwise you couldn't modify it. Note that auto_ptr's own copy constructor and assignment operator also take non-const references. So even if you made shared_ptr's constructor take an auto_ptr by value, you'd still get the warning, this time from auto_ptr's constructor. Only this time you'd have an additional, useless temporary. Sebastian Redl