c - malloc 和 free 具有动态变化的结构

Question

我在动态变化的结构中移动指针时遇到问题。 我已经创建了我的代码，您可以在其中分配更多内存，这似乎有效。 我遇到的问题是如何添加到结构中，如何释放内存以及如何从一个结构移动到另一个结构并打印所有项目。

我正在尝试测试添加和打印(那里的删除功能似乎不起作用，段错误)

当我添加到结构然后打印结构时，我从添加的值中得到了段错误。我不知道我是否正确地从第一个结构移动到下一个结构。

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

/********************************************
Creates more memory for size (strut * rec+1)
*********************************************/

employee *create(int record){
employee *new_employee = malloc(sizeof(employee) * (record+1));

return new_employee;    
}

/********************************************
Copies the data from one structure to a new structure with 
size "structure" multipled by rec+1
***********************************************/
employee *copy(employee *data, int record){
    employee *new_employee = create(record);
int i;
    for(i = 0; i<record;i++){
        new_employee->first = data->first;
        new_employee->last = data->last;
        new_employee->start_date = data->start_date;
        new_employee->sal = data->sal;
        data++;
    }
    /********************
    Needs to free the old struct
    *********************/
    //deleteData(data, record);

return new_employee;
}
/********************************************
Function prints everything in the struct
*********************************************/
void printStruct(employee *data, int record){
int i;

    for(i = 0; i<record; i++){
        printf("\nEntry: %d\n", i+1);           
        printf("The employee's name is %s %s\n", data->first, data->last);
        printf("The employee was hired on: %s\n", data->start_date);
        printf("The employee make $%f\n\n", data->sal); 
        data++;     
    }
}
/******************************************
Function frees the old data base
*******************************************/
void deleteData(employee *data, int record){
int i;
    for(i = 0; i<record; i++){
        free(data->first);
        free(data->last);
        free(data->start_date);
        data++;
    }
    free(data);
}
/******************************************
Adds an employee to the new structure
*******************************************/
employee *add(employee *data,char *fname, char *lname, char *date, float salary, int record){
employee *employeeDB = create(record);
employeeDB = copy(data, record);
int i;
    employeeDB++;
    employeeDB->first = fname;
    employeeDB->last = lname;
    employeeDB->start_date = date;
    employeeDB->sal = salary;

return employeeDB;
}




/**************************
Starts of the main function
***************************/

int main(void){
    //Keeps track of the number of records that are in the structure
int rec = 0;
    //Keeps the number of accesses to the structure. Even with the one entry   the structure has not been accessed. 
int acc = 0;
    //Holds the input information for the menu
int input;
    //holds the information for inputing user first name
char *fname;
    //holds the information for inputing user last name
char *lname;
    //holds the information for for the startdate
char *start;
    //holds the information for the salary;
float sal;
/*********************************
This next section adds an employee to the record

************************************/
//This creates the first entry to the dynamic structure.
employee *first_employee = create(rec);
first_employee->first = "FIRST";
first_employee->last = "LAST";
first_employee->start_date = "June-20th-2006";
first_employee->sal = 55555.55;
//increase the number of records    
rec = rec+1;

employee *new_employeeDB = add(first_employee, "fname", "lname", "JUNE-20th-2010", 55555.55, rec);
rec = rec + 1;
printStruct(new_employeeDB, rec);


printf("%d\n", (sizeof(employee)* rec));


}

Answer 1

第一个问题:好的...您没有包含员工类型的声明。 我猜第一个和最后一个被声明为字符指针，这是一个错误。 您需要它们是固定大小的字符数组，否则这将永远不起作用。 选择字符串的最大长度并声明如下结构。

typedef struct
{
    char name[50];
    char surname[50];
    ....
} employee;

上一篇文章:

好吧，我必须承认我不太喜欢这种实现方式。 我会使用更稳定的方法。

首先，由于它只是一个项目列表，因此您可以使用双向链表。 这将允许您以 O(1) 复杂度添加和删除元素。确实不错。

这还允许您迭代从第一个到最后一个的所有项目。

为此，我将使用更面向 OOP 的方法:) 我知道，我们使用 C，但请听听这个想法。

typedef struct
{
    MyList* OwnerList;
    Employee* Previous;
    Employee* Next;

    char name[50];
    int age;
} Employee;

typedef struct
{
    Employee* First;
    Employee* Last;
    int Count;
} MyList;

MyList* AllocList() { return calloc(sizeof(MyList), 1); }

void DeleteList(MyList* list)
{
    Employee* current;
    Employee* next;
    for (current = list->First; current != NULL; current = next)
    {
        next = current->Next;
        free(current);
    }
    free(list);
}

int GetCount(const MyList* list)
{
    return list->Count;
}

Employee* AddAmployee(MyList* list)
{
    Employee* result = calloc(sizeof(Employee), 1);
    Employee* last = list->Last;
    if (last != null)
        last->Next = result;
    else
        list->First = result;
    result->Previous = last;
    list->Last = result;
    ++list->Count;
    result->OwnerList = list;
    return result;
}

void RemoveEmployee(Employee* employee)
{
    /* i leave removal for you as exercise :) look for doubly linked list */
    free(employee);
}

现在，迭代所有项目很简单:

Employee* current;
for (current = list->First; current != null; current = current->Next)
    printf("%s %d\n", current->name, current->age);