链接列表抛出无限循环，但最新指针设置为null

I've written a linked list and when I'm doing the append to end, it seems to be going into an infinite loop.

我写了一个链表，当我正在做追加结束时，它似乎进入了无限循环。

// this function will make a node and give it a value at the end of the list
void add_at_end(struct Node* begin, int v) {
    // new node
    struct Node* temporary = begin; // pointer to start of list
    temporary->vv = v;
    temporary->nx = NULL; // latest node
    if (begin == NULL) { // this will make a new list if it is empty
        begin = temporary;
        return;
    }
    struct Node* iterate = begin;
    while (iterate->nx != NULL) { iterate = iterate->nx; }
    iterate->nx = temporary;
    return;
}

I call it using:

我用它来称呼它：

struct Node alpha;
add_at_end(&alpha, 1);

Why is this throwing an infinite loop?

为什么这会抛出无限循环？

1 个解决方案

#1

• alpha is not initalized, it contains garbage, including the next pointer. Access the content is undefined behaviour, which could result in an infinite loop... or crash.. or whatever.
• alpha不是初始化的，它包含垃圾，包括下一个指针。访问内容是未定义的行为，这可能导致无限循环...或崩溃..或其他。
• begin = temporary; this will set the variable in the local scope, and won't have any effect on what you've passed to the function. You could either pass a pointer to the pointer (**), or pass an existing node to the function (in which case there's no need to check). You are doing a mix of these which doesn't make sense.
• begin = temporary;这将在本地范围内设置变量，并且不会对传递给函数的内容产生任何影响。您可以将指针传递给指针（**），也可以将现有节点传递给函数（在这种情况下无需检查）。你正在混合使用这些没有意义的东西。
• It's very inefficient to traverse the whole list just to add an element. You should maintain a link to the last node, so that it becomes an O(1) operation.
• 遍历整个列表只是为了添加一个元素是非常低效的。您应保持指向最后一个节点的链接，以使其成为O（1）操作。