带有指向下一个结构的指针的结构? [英] Structs with pointer to next struct?

问题描述

我想开个会话，在链表内的 size 变量中总共插入10个不同的整数.我相信我应该为此使用result_register()吗?

I want to have a session where I insert totally 10 different integers in the size variable that is inside the linked list. I believe I should use result_register() for this?

所有10个整数都存储完后，我想通过输入 result-> size 将它们打印出来.

When all 10 integers are stored I want to print them out by typing result->size.

我是链接列表的新手，所以请保持谦虚.

I am new with linked lists so please be gentle.

struct result {   
    int size;
};

struct result* result_next()
{
   // What do I type here?
}

void result_register(int size)
{
    // What do I type here?
}

主要

struct result* result;

while ((result = result_next()))
    printf("%d\n", result->size);

然后，我希望能够像上面一样打印出结果.

Then I want to be able to print out the results by doing like above.

推荐答案

结果是应该有一个next指针指向链接列表中的下一个节点:

You're result should have a next pointer to point to the next node in the linked list:

struct result {   
    int size;
    struct result *next;
};

然后，假设您有一个指针，您只需沿着列表走动即可；

Then, assuming you've got a pointer the head you just walk the list;

struct result *node = head;

while(node != null)
{
  printf("%d\n", node->size);
  node = node->next;
}

要在列表中创建新项目，您需要以下内容:

To create a new item in the list you need something like this:

struct result *Add(struct result *head, int size)
{
  struct result *newHead = (struct result*)malloc(sizeof(struct result));
  newHead->size = size;
  newHead->next = head;

  return newHead;
}

现在您可以说:

struct head *head=add(null, 0);
head *head=add(head, 1);
head *head=add(head, 2);
head *head=add(head, 3);

这将为您提供3 -> 2 -> 1 -> 0

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