std :: enable_shared_from_this :: shared_from_this如何工作 [英] How std::enable_shared_from_this::shared_from_this works
问题描述
我只是不明白 std :: enable_shared_from_this :: shared_from_this
如何返回与现有指针共享所有权的共享pinter。换句话说,您此:
I just cannot understand how std::enable_shared_from_this::shared_from_this
returns a shared pinter that shared ownership with existing pointer. In other words you do this:
std::shared_ptr<Foo> getFoo() { return shared_from_this(); }
所以当您呼叫 getFoo
时,确切地说,它得到与共享所有权的另一个 shared_ptr
是什么,而不是创建拥有相同所有权的单独的 shared_ptr
这
。
So when you call getFoo
how does exactly it get what is the other shared_ptr
to share the ownership with and not to create a separate shared_ptr
that owns the same this
.
我需要了解这一点,以便能够理解如何从所有增加的对象中创建shared_ptr相同的引用计数,并且不初始化单独的 shared_ptr
s。
I need to understand this to be able to understand how to create shared_ptr from some object that all increase the same ref count and not initialize separate shared_ptr
s.
推荐答案
enable_shared_from_this< T>
具有 weak_ptr< T>
数据成员。 shared_ptr< T>
构造函数可以检测 T
是否源自 enable_shared_from_this< T>
。如果是,则 shared_ptr< T>
构造函数将分配 * this
(即 shared_ptr< T>
)到 enable_shared_from_this< T>
中的 weak_ptr
数据成员。 shared_from_this()
然后可以从 weak_ptr< T> <创建一个
。 shared_ptr< T>
enable_shared_from_this<T>
has a weak_ptr<T>
data member. The shared_ptr<T>
constructor can detect if T
is derived from enable_shared_from_this<T>
. If it is, the shared_ptr<T>
constructor will assign *this
(which is the shared_ptr<T>
) to the weak_ptr
data member in enable_shared_from_this<T>
. shared_from_this()
can then create a shared_ptr<T>
from the weak_ptr<T>
.
可能的实现示例:
template<class D>
class enable_shared_from_this {
protected:
constexpr enable_shared_from_this() { }
enable_shared_from_this(enable_shared_from_this const&) { }
enable_shared_from_this& operator=(enable_shared_from_this const&) {
return *this;
}
public:
shared_ptr<T> shared_from_this() { return self_.lock(); }
shared_ptr<T const> shared_from_this() const { return self_.lock(); }
private:
weak_ptr<D> self_;
friend shared_ptr<D>;
};
template<typename T>
shared_ptr<T>::shared_ptr(T* ptr) {
// ...
// Code that creates control block goes here.
// ...
// NOTE: This if check is pseudo-code. Won't compile. There's a few
// issues not being taken in to account that would make this example
// rather noisy.
if (is_base_of<enable_shared_from_this<T>, T>::value) {
enable_shared_from_this<T>& base = *ptr;
base.self_ = *this;
}
}
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