使用结构的 C 函数，为什么它不起作用？

Question

我在这个实现中遇到了很多错误。

typedef struct EmployeeStruct
{
    char lastName[MAX_LENGTH];
    char firstName[MAX_LENGTH];
    int employeeNumber;  // Holds the employee's ID. This value is
                             // equal to the number of employees
    struct EmployeeStruct *Next;  // Pointer to the next most recently hired Employee
}Employee;

当尝试创建一个将返回指向该结构的指针的函数时，问题就来了。该错误出现在 malloc 调用中，导致“new”未被正确声明，因此此函数中的所有行都存在错误。

Employee* hireEmployee(Employee tail, char lastName[MAX_LENGTH], char firstName[MAX_LENGTH])
{
    struct Employee *new = (Employee*)malloc(sizeof(Employee));
    new.lastName = lastName;
    new.firstName = firstName;
    new.next = tail;
    tail.next = new;
    new.employeeNumber = employeeCount;

    return tail;
}

这是错误列表。感谢您的帮助!

lab6.c:19: warning: initialization from incompatible pointer type
lab6.c:20: error: request for member ‘lastName’ in something not a structure or union
lab6.c:21: error: request for member ‘firstName’ in something not a structure or union
lab6.c:22: error: request for member ‘next’ in something not a structure or union
lab6.c:23: error: ‘Employee’ has no member named ‘next’
lab6.c:24: error: request for member ‘employeeNumber’ in something not a structure or union
lab6.c:26: error: incompatible types in return

Answer 1

这里有几个不同的问题:

您需要使用指针取消引用运算符 -> 来访问指向您的结构的指针的成员。
然后你需要使用strcpy分配给您的 char 数组。
你需要避免链表中的循环(你设置了 new 和 tail 以作为 next 指向彼此) .显而易见的解决方法是将 new 设置为新的 tail。调用代码可能需要更新以反射(reflect)这一点。
最后，你不应该从 malloc 中转换返回值
真的最后，next应该是Next。或者您可以更改结构定义中的大小写。

Employee *new = malloc(sizeof(Employee));
strcpy(new->lastName, lastName);
strcpy(new->firstName, firstName);
new->Next = NULL;
tail->Next = new;
new->employeeNumber = employeeCount;