main函数去掉注释就会出现总线错误：bus error，是因为复制function对象的原因吗？

Question

class DebugDelete {
public:
	DebugDelete(const std::string &msg = "unique_ptr", std::ostream &s = std::cerr) : message(msg), os(s) {}
	template <typename T> void operator()(T *p) const
		{ os << "deleting " << message << std::endl; delete p; }
private:
	std::ostream &os;
	std::string message;
};

template <typename T>
class MySharedPtr {
public:
	typedef std::function<void(T*)> Delete;
	explicit MySharedPtr(T *t, Delete d = nullptr)
		: ptr(t), use(new size_t(1)), del(d) {}
	MySharedPtr(const MySharedPtr &rhs)
		: ptr(rhs.ptr), del(rhs.del) { ++*use; }
	MySharedPtr& operator=(const MySharedPtr&);
	T& operator*() const { return *ptr; }
	T* operator->() const { return ptr; }
	~MySharedPtr();
private:
	T *ptr;
	size_t *use;
	Delete del;
};

template <typename T>
MySharedPtr<T>& MySharedPtr<T>::operator=(const MySharedPtr &rhs)
{
	++*rhs.use; //递增右侧对象的引用计数
	if (--*use == 0) {
		del ? del(ptr) : delete ptr;
		delete use;
	}
	ptr = rhs.ptr;
	use = rhs.use;
	del = rhs.del;
}

template <typename T>
MySharedPtr<T>::~MySharedPtr()
{
	if (--*use == 0) {
		del ? del(ptr) : delete ptr;
		delete use;
	}
}

template <typename T>
MySharedPtr<T> make_myshared(T t = T(), typename MySharedPtr<T>::Delete del = nullptr)
{ return MySharedPtr<T>(new T(t), del); }

int main()
{
	auto iptr = make_myshared<int>(10, DebugDelete("my_shared_ptr"));
	//auto iptr2 = iptr;
	std::cout << *iptr << std::endl;
}

Answer 1

就算注释了也有问题。

	MySharedPtr(const MySharedPtr &rhs)
		: ptr(rhs.ptr), del(rhs.del) { ++*use; }

当调用复制构造函数的时候，use指针未被初始化，然后就使用了。好像应该先use = rhs.use;