C for 循环与推出循环的效果不同

Question

我正在使用一个跟踪最后一个元素的链表在 c 中编写一个简单的队列。

这是我的代码：

#include <stdlib.h>
#include <stdio.h>
#include <stdbool.h>
#include <stdint.h>   

typedef struct node_s {
    uint32_t x, y;
} node;


// very cheap linked list queue that lives as long as its scope
typedef struct queue_s {
    node* node;
    struct queue_s* next;
} queue;

bool isEmpty(queue** queue)
{
    return !*queue;
}

node* dequeue(queue** head, queue** tail)
{
    if (isEmpty(head)) {
        printf("attempting to dequeue from an empty queue\n");
        return NULL;
    }
    node* ret = (**head).node;
    *head = (*head)->next;
    if (isEmpty(head)) *tail = NULL;
    printf("next %p\n", (void*) *head);
    // no need to free anything, the queue lives as long as its scope
    return ret;
}

void enqueue(queue** head, queue** tail, queue* new)
{
    if (isEmpty(head)) {
        // has no head -> make new head
        *head = new;
        *tail = new;
        printf("empty\n");
    } else {
        // has head -> make new tail // TODO does this work?
        (*tail)->next = new;
        *tail = (*tail)->next;
        printf("not empty\n");
    }
}

int main()
{
    queue* head = NULL;
    queue* tail = NULL;

    enqueue(&head, &tail, &(queue) {.node=&(node) {1, 1}, NULL});
    printf("h %i\n", head->node->x);
    printf("t %i\n", tail->node->x);

    enqueue(&head, &tail, &(queue) {.node=&(node) {2, 2}, NULL});
    printf("h %i\n", head->node->x);
    printf("t %i\n", tail->node->x);

    enqueue(&head, &tail, &(queue) {.node=&(node) {3, 3}, NULL});
    printf("h %i\n", head->node->x);
    printf("t %i\n", tail->node->x);

    // for (int i = 1; i <= 3; ++i) {
    //  enqueue(&head, &tail, &(queue) {.node=&(node) {i, i}, NULL});
    //  printf("h %i\n", head->node->x);
    //  printf("t %i\n", tail->node->x);
    // }

    return 0;
}

现在 main() 中推出的 for 循环的行为正确，因为头元素保持 1 而新元素排队：

empty
h 1
t 1
not empty
h 1
t 2
not empty
h 1
t 3

...但是 for 循环本身给出了错误的输出，因为头元素总是等于尾元素：

empty
h 1
t 1
not empty
h 2
t 2
not empty
h 3
t 3

我是一个 c 初学者，这让我发疯，有人可以帮帮我吗？

Answer 1

这段代码：

for (int i = 1; i <= 3; ++i) {
      enqueue(&head, &tail, &(queue) {.node=&(node) {i, i}, NULL});
      printf("h %i\n", head->node->x);
      printf("t %i\n", tail->node->x);
}

等价于：

for (int i = 1; i <= 3; ++i) 
      queue next = {.node=&(node) {i, i}, NULL};
      enqueue(&head, &tail, &next);
      printf("h %i\n", head->node->x);
      printf("t %i\n", tail->node->x);
} // End of scope

新元素next 将在作用域结束时不复存在（见注释）。要让它循环工作，您需要做的是动态内存分配。