带有链表的c中的堆栈-将元素与头部错误中的元素进行比较

Posted 2023-02-18

【中文标题】带有链表的c中的堆栈-将元素与头部错误中的元素进行比较

【英文标题】：stack in c with linked list - comparing an element with element in head error

【发布时间】：2021-05-19 10:35:06

【问题描述】：

我正在尝试实现一个应用程序，该应用程序使用单链接列表使用一堆employee 结构，其中每个员工都由他/她的增值税号、姓名和工资来描述。

我正在尝试扩展代码，以便当用户选择“1”时，它将从控制台读取新员工的数据，检查他的薪水是否高于堆栈顶部员工的薪水，并打印比较结果。

新元素在栈中的push正常，但是if检查salary是否高于栈顶salary却不起作用。

同样，我想知道要添加什么，以便用户可以点击“3”来打印堆栈中每个唯一员工的数据。

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

struct employee 
int am;
char name[50];
float salary;
struct employee *next;
;

void init(struct employee **head);
void show(struct employee *head);
int push(struct employee s, struct employee **head);
int pop(struct employee *s, struct employee **head);

int main()

struct employee data;
struct employee *head;
int ch;
init(&head);
do 
    printf("\nMenu ");
    printf("\n1.Push ");
    printf("\n2.Pop ");
    printf("\n3.Show contents ");
    printf("\n0.Exit ");
    printf("\nChoice: ");
    scanf("%d",&ch);

    switch (ch) 
    case 1:
        printf("Enter employee data:");
        scanf("%d %s %f", &data.am, data.name, &data.salary);
        if (push(data,&head))
            printf("Successfully pushed to stack.\n");
            printf("%f\n", data.next->salary);
            if (data.salary > head->salary)
                 printf("salary on the head is bigger\n");
            else
                printf("salary on the head is less\n");
        
        else
            printf("No memory available, push failed.\n");
    break;

    case 2:
        if (pop(&data,&head))
            printf("Popped: %d %s %.1f\n",data.am, data.name, data.salary);
        else
            printf("Empty stack, no data retrieved.\n");
        break;
     case 3:
        show(head);
        break;
     case 0:
        printf("Bye!\n");
        break;
    default: 
        printf("Try again.\n");
        break;
     
 while (ch!=0);
exit(0);


void init(struct employee **head)
    *head = NULL;


int push(struct employee d, struct employee **head)

    int done=0;
    struct employee *s;

    s = (struct employee *)malloc(sizeof(struct employee));
    if (s!=NULL)
    
        *s = d;
        s->next = *head;  
        *head = s;
         done = 1;
    

    return done;


int pop(struct employee *s, struct employee **head)

    int done=0;
    struct employee *temp;

    if (*head != NULL)
   
        temp = *head;
        *s = **head;
        *head = (*head)->next;
        free(temp);
        done = 1;
    

    return done;


void show(struct employee *head)

    struct employee *i;
    printf("--- Top of stack  ---\n");
    for (i=head; i!=NULL; i=i->next) //maybe here edit for the odd positions
        printf("%d %s %.2f\n",i->am,i->name,i->salary);
    printf("---Bottom of stack---\n");  

【参考方案1】：

对于您的第一个问题，您将数据推送到堆栈顶部（因此您的 data 成为负责人），然后检查与负责人的工资差异。否则，你是在和自己比较。

您可能希望在推入堆栈之前检查，或将if 行替换为

if (head->salary > data.next->salary)

请注意，在检查其薪水之前，您还需要检查next 节点是否存在。