指向未初始化数据成员的指针 [英] Pointer to uninitialized data member

问题描述

不使用，但要存储一段时间.

Not use, but store at some moment of time.

我知道我必须使用智能指针.但是为了简化代码，我决定举一个拥有原始指针的示例.在实践中这不是很好，但是我认为足够好.

I know that I have to use smart pointers. But for code simplicity I decide to give an example with owning raw pointers. Which is not very good in practice but I think good enough for example.

这是代码示例:

struct Node {
    Node(Node*& node_ptr) : list_first_node_ptr{node_ptr} {}
    /* ... */
    Node*& list_first_node_ptr;
};

struct List {
    // Is this line - dangerous?
    List() : first_node_ptr{new Node{first_node_ptr}} {}
    /* ... */
    Node* first_node_ptr;

    ~List() {delete first_node_ptr;}
};

我用Node对象初始化了first_node_ptr，但是传入Node对象的构造函数仍然没有被未初始化的first_node_ptr.

I initialize first_node_ptr with Node object, but pass in constructor of Node object still not uninitialized first_node_ptr.

另一个问题:我通过它时是否已经为first_node_ptr分配了内存，那么Node构造函数中的引用地址是否有效?

One more question: is memory already allocated for first_node_ptr when I pass it, so will the reference address at Node constructor be valid?

我认为这个例子是相同的但简化的版本.我是对的吗?

I think that this example is about the same but simplified version. I'm right?

class Base {
 public:
    // var_ptr points to uninitialized va
    Base() : var_ptr{&var}, var{10} {}

    int* var_ptr;
    int var;
};

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PS 为什么当我写(Node*)&而不是Node*&时，我得到有关不完整类型的编译错误?

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P.S. Why when I write (Node*)& instead of Node*& I get compilation errors about incomplete type?

P.P.S. 使用智能指针可以改变情况吗?

P.P.S. Can the usage of smart pointers change the situation?

P.P.P.S. @JesperJuhl询问了用例. 这个想法是创建一个循环链表，其中每个节点在Head指针上都有指向第一个前哨节点的引用.我从CppCon视频2016 43:00中的 Herb Sutter中得到了这个想法.在视频中，他谈到了唯一的指针，但我举了一些原始指针的示例，这可能会变得有些混乱.

P.P.P.S. @JesperJuhl asked about use case. The idea is to create circular linked list where every Node has a reference on Head pointer to first sentinel Node. I get this idea from Herb Sutter in CppCon video 2016 43:00. In video he talked about unique pointers but I give example with raw pointers, it may be become a little bit confusing.

推荐答案

Node::list_first_node_ptr是参考.它并不关心被引用对象的 value ，只是被引用对象 itself 实际上存在.由于构造Node::list_first_node_ptr时肯定会存在List::first_node_ptr，因此这是合法代码.

Node::list_first_node_ptr is a reference. It doesn't care about the value of the referenced object, just that the referenced object itself actually exists. Since List::first_node_ptr is guaranteed to exist when Node::list_first_node_ptr is constructed, this is legal code.

用另一种方式表达；引用永远不能为空，但是它可以引用为空的指针.

To phrase it another way; a reference can never be null, but it can reference a pointer that is null.

但是，值得注意的是，在Node完成构造之前，List::first_node_ptr不会指向有效的对象.这意味着在Node构造函数完成之前，您不能取消引用List::first_node_ptr或Node::list_first_node_ptr.

It's worth noting, though, that List::first_node_ptr will not point to a valid object until Node has finished constructing. This means you can't dereference List::first_node_ptr or Node::list_first_node_ptr until the Node constructor has finished.

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