Question

我正在研究链表并尝试理解指针，他们自己的物理地址和他们指向的地址。我编写用于打印列表的代码，并发现一个奇怪的现象，就像标题描述的那样。这是代码：

#include <iostream>
using namespace std;

struct Node {
    int num;
    struct Node *next;
};

void addNode(Node*&, int);

void printList(Node*);

int main() {
    Node *a = NULL;

    addNode(a, 1);
    addNode(a, 2);
    addNode(a, 3);

    cout << "After adding three nodes, a contains " << a << ", a locates at " << &a << endl;
    cout << "a->next contains " << a->next << endl;
    cout << "a->num = " << a->num << endl;
    cout << endl;

    printList(a);

    cout << "After printing, a contains " << a << ", a locates at " << &a << endl;
    cout << "a->next contains " << a->next << endl;
    cout << "a->num = " << a->num << endl;
    cout << endl;
}

void addNode(Node *&A, int i) {
    Node *temp;
    temp = new Node;
    temp->num = i;
    temp->next = A;
    A = temp;
}

void printList(Node *A) {
    cout << "1. printList: A contains " << A << ", A locates at " << &A << endl;
    cout << endl;

    Node *current;
    cout << "2. current contains " << current << " (uninitialized), current locates at " << &current << endl;
    //cout << current->next << endl;
    cout << endl;

    current = A;
    cout << "3. After being assigned by A, current contains " << current << ", current locates at " << &current << endl;
    cout << endl;

    while(true) {
        if(current == NULL) break;

        cout << "num is " << current->num << endl;
        current = current->next;
        cout << "After being assigned by next, current contains " << current << ", current locates at " << &current << endl;
        cout << endl;
    }
    cout << "End of the list" << endl;
    cout << endl;
}

结果如下：

添加三个节点后，a包含0x2822a98，a位于0x6bfebc   a->接下来包含0x2822a88 a-> num = 3



printList：A包含0x2822a98，A位于0x6bfea0



当前包含0x6bfebc（未初始化），当前位于0x6bfe8c



在被A分配后，当前包含0x2822a98，当前位于0x6bfe8c


num是3在被next分配后，current包含0x2822a88，   current位于0x6bfe8c

num是2在被next分配后，current包含0x2822a78，   current位于0x6bfe8c

num为1在被next分配后，current包含0，current   位于0x6bfe8c

列表的结尾

打印后，a包含0x2822a98，位于0x6bfebc a-＆gt; next   包含0x2822a88 a-> num = 3

注意，第2项：current包含0x6bfebc。 “0x6bfebc”与地址相同，地址显示在第一行。但我认为未初始化的指针变量应指向随机地址。为避免重合，我尝试重新打开程序并重新启动计算机，但每次只是锁定到a的地址。有人可以解释一下里面到底发生了什么。感谢。

为了更具可读性，我在最后添加了一个截图。希望它有所帮助。

screenshot

链表的本地未初始化指针变量包含与列表头

0 个答案: