Question

当我运行这些方法时，我的列表永远不会增长，我不知道为什么。头部和尾部总是相等的。我的指针或结构设置有问题吗？我花了4个小时来切换这些东西，我似乎无法找到为什么事情不起作用，除非它是指针的东西（我正在学习并且不擅长）...

typedef struct EmployeeStruct
{
    char LastName[10];
    char FirstName[10];
    struct EmployeeStruct *next;  
} Employee;

struct EmployeeStruct *head,*tail, *temp;

Employee* hireEmployee(Employee* head, char LastName[10], char FirstName[10])
{
    // exit if max number of employees is reached
    if(eNum > MAX_EMPLOYEES)
    {
        printf("Sorry, but you can't hire any more employees right now\n");
        return 0;
    }
    // insert at tail
    newHire = head;
    while(newHire->next != NULL)
    {
        newHire = newHire->next;
        printf("nope\n");
    }
    // allocate memory
    newHire->next = malloc(sizeof(Employee));

    if(newHire == NULL)
    {
        printf("Memory allocation failed");
        return 0;
    }

    newHire->next = NULL;
    // insert values into this node
    strcpy(newHire->LastName, LastName );
    strcpy(newHire->FirstName, FirstName );
    newHire->EmployeeNumber = eNum;
    tail = newHire;
    //printf("%d\n",newHire->EmployeeNumber);
    eNum+=1;

    //printf("%d\n",newHire->EmployeeNumber);
    return newHire;
}

int main()
{  
    char choice;
    char first[20];
    char last[20];
    int i = 0;
    // allocate memory
    head = malloc(sizeof(Employee));

    if(head == NULL)
    {
        printf("Memory allocation failed");
        return 0;
    }
    head->next = tail;

    while(TRUE)
    {
        // prompt user for choice
        printf("Please choose from the following options:\n");
        printf("a: Hire new employee\nb: Promote employee\nc: Delete employee\nd: Display roster\ne: Exit\n");
            scanf("\n%c", &choice);

        switch(choice)
        {
            case 'a':
                printf("New employees first name:\n");
                    scanf("%s", first);
                printf("New employees last name:\n");
                    scanf("%s", last);
                tail = hireEmployee(head,last,first);
                temp = head;
                while(temp->next != NULL)
                {
                    temp = temp->next;
                    printf("nope\n");
                }
                temp->next = tail;
                tail = temp->next;
                printf("A%d: %s %s\n", tail->EmployeeNumber,tail->FirstName,tail->LastName);
                tail = tail->next;
                printf("A%d: %s %s\n", tail->EmployeeNumber,tail->FirstName,tail->LastName);
                tail->next = NULL;
                //printEmployees(head);
                break;
            case 'b':
                //TBD
                break;
        }
    }
}

**现在可以使用，就是这四个错误：

(newHire == NULL)    
(head == NULL)
newHire =newHire->next;
temp->next = tail;

顺便说一下，我总是做作业并上课，而我只是停留在一个较大的程序中。所以，谢谢那些没有侮辱我并实际提供有用建议的人。我真的很感激。

Answer 1

这是你的错误

newHire->next = malloc(sizeof(Employee));

malloc的返回类型为void *因此将其更改为

newHire->next = (EmployeeStruct *)malloc(sizeof(Employee));

if(newHire == NULL)

永远不会将NULL更改为

if(newHire->next == NULL)

也改变了

newHire->next = NULL;

你必须再遍历一次链接

newHire=newHire->next;
newHire->next = NULL;

Answer 2

我认为问题出在这里

temp->next = tail;
tail = temp->next;

在case 'b'

中

因为它将temp->next设为head

链表队列。坚持指针（我认为）

2 个答案: