我正在编写一个双链表项目的语法问题，我是出于教育目的而编写的。我在头文件中创建了一个结构体，我的主程序似乎没问题，但在.cpp文件中实现我的功能给我带来了巨大的困难。我无法识别将记录插入列表的三种情况。具体来说，分配内存，初始化列表的头部和尾部，以及语句的顺序让我感到困惑，就像传递要添加到我的列表中的记录副本一样。
我的头文件如下：
struct rec
{
char * id;
char firstname[15];
char lastname[15];
struct rec* prev;
struct rec* next;
};
int AddItem ( rec r );
int DeleteItem ( char* delid );
void PrintList ( int order );
我的.cpp文件是困难所在，如下所示：
#include <iostream>
#include "list.h"
#include <string.h>
using namespace std;
// These pointers refer to the head and tail of the list.
rec* first = NULL;
rec* last = NULL;
int AddItem( Rec r )
{
rec* newRecEntry;
rec* current = NULL;
rec* previous = NULL;
// Check for duplicate id
current = first;
while (current)
{
if( strcmp(current -> id, r.id) == 0)
{
return 0;
}
else
// Create a new node
{
newRecEntry = new Rec;
newRecEntry->id = new char[strlen(r.id)+1];
strcpy(newRecEntry->id, r.id);
strcpy(newRecEntry->firstname,r.firstname);
strcpy(newRecEntry->lastname,r.lastname);
newRecEntry->next = NULL;
newRecEntry->prev = NULL;
}
// Find the appropriate position for the node and insert accordingly
// Check to see if the list is empty
if (first == NULL)
{
first = newRecEntry;
last = newRecEntry;
}
else if ( r.lastname>last.lastname)
{
else
{
return 0;
}
/*int DeleteItem(char* ID)
我应该能够在列表的开头，中间和结尾插入。根据ID从列表中删除项目，并根据用户输入按升序或降序打印列表，但我首先只想处理添加到列表中的项目。 我的函数定义如下，还包含一些错误
lists.cpp
#include <iostream>
#include "list.h"
#include <string.h>
using namespace std;
// These pointers refer to the head and tail of the list.
rec* first = NULL;
rec* last = NULL;
int AddItem( Rec r )
{
rec* newRecEntry;
rec* current = NULL;
rec* previous = NULL;
// Check for duplicate id
current = first;
while (current)
{
if( strcmp(current -> id, r.id) == 0)
{
return 0;
}
else
// Create a new node
{
newRecEntry = new Rec;
newRecEntry->id = new char[strlen(r.id)+1];
strcpy(newRecEntry->id, r.id);
strcpy(newRecEntry->firstname,r.firstname);
strcpy(newRecEntry->lastname,r.lastname);
newRecEntry->next = NULL;
newRecEntry->prev = NULL;
}
// Find the appropriate position for the node and insert accordingly
// Check to see if the list is empty
if (first == NULL)
{
first = newRecEntry;
last = newRecEntry;
}
else if ( r.lastname>last.lastname)
{
else
{
return 0;
}
/*int DeleteItem(char* ID)
{
rec
}
*/
/*void printList(int order)
{
loop
{
cout << ptr -> Id << " ";
cout << ptr -> firstname << " ";
cout << ptr -> lastname << " ";
cout << ptr -> prev << " "; // address of previous
cout << ptr << " "; // address of item
cout << ptr -> next << " "; // address of next item
}
}
主要内容如下：
#include <iostream>
#include "list.h"
#include <string.h> // <string>
using namespace std;
void main (void)
{
int choice, printorder;
char idbuffer[100];
rec r;
do
{
cout << "Enter your choice 1 Add, 2 Delete, 3 Print, 0 quit "<<endl;
cin >> choice;
switch ( choice )
{
case 1: //AddItem
cout << "\nEnter ID ";
cin >> idbuffer;
r.id = idbuffer;
cout << "\nFirst Name ";
cin >> r.firstname;
cout << "\nLast Name ";
cin >> r.lastname;
if ( AddItem ( r ) )
{
cout << "\nSuccess!\n";
}
else
{
cout << "\nItem failed to be added\n";
}
break;
case 2: //Delete
cout << "\nEnter id :";
cin >> idbuffer;
if ( DeleteItem ( idbuffer ) )
{
cout << "\nDelete OK\n";
}
else
{
cout << "\nDelete Failed for " << idbuffer;
}
break;
case 3: // Print
cout << "Enter order 0 - Ascending, 1 - Descending\n";
cin >> printorder;
PrintList (printorder);
break;
case 0: // quit
break;
default: // bad choice
break;
} // end switch
}
while ( choice != 0 );// end do while
} // end main
答案 0 :(得分：0)
尝试更改此内容：
int AddItem(Record entry);
对此：
Record* AddItem(Record entry, Record *insertion_point = NULL );
如果insertion_point为NULL，您可以假设Record是新列表的开头。
现在您有足够的信息来设置Next和Previous指针，并返回新创建的节点。
答案 1 :(得分：0)
首先，您必须定义第一个和最终项目。
可能性是
Previous分配给NULL的项目。Next分配给NULL的项目已结束。为简单起见，我删除了Record的一些字段，这不是最好的，请注意，你应该做很多错误处理并处理异常：
struct Record
{
int id;
Record* Next;
Record* Previous;
};
Record *CreateList(int id)
{
Record *rec = new Record;
rec->id = id;
rec->Next = NULL;
rec->Previous = NULL;
return rec;
}
void AddItemAfter(Record *item, int id)
{
if (!item)
return;
Record* newRecordPointer = new Record;
newRecordPointer->id = id;
newRecordPointer->Next = NULL;
newRecordPointer->Previous = NULL;
newRecordPointer->Next = item->Next;
item->Next = newRecordPointer;
newRecordPointer->Previous = item;
}
void ShowList(Record *list)
{
Record *ptr = list;
while (ptr)
{
cout << ptr->id << endl;
ptr = ptr->Next;
}
}
int main()
{
Record *list = CreateList(1);
AddItemAfter(list, 2);
AddItemAfter(list, 3);
ShowList(list);
}
答案 2 :(得分：0)
它可能看起来不像，但即便是这个功能
int AddItem(Record entry)
{
Record* newRecordPointer;
newRecordPointer=new Record;
strcpy(newRecordPointer->firstName,entry.firstName);
strcpy(newRecordPointer->lastName,entry.lastName);
newRecordPointer->ID=new char[strlen(entry.ID)+1];
strcpy(newRecordPointer->ID, entry.ID);
return 0;
}
正在尝试做太多事情。
让我们编写将项添加到列表中的伪代码描述：
我已经标记了动词和名词，你已经可以看到你的函数中缺少一个名词。您要求AddItem将项目添加到列表中......但是您没有给它一个列表来处理。
明确写出你的期望也很有用：
AddItem之前Record Record之后添加我们的新项目 AddItem后Record，其Next都应指向新节点Previous应指向传入的节点Record，其Next和Previous成员指向自己所以，让我们说可能有效的最小函数是：
void AddItem(Record *insert_after, Record value)
{
Record *new_node = CreateRecord();
CopyRecordValues(new_node, &value);
AttachAfter(insert_after, new_node);
}
请注意，如果我们编写真正的C ++，前两行可能只使用复制构造函数Record *new_node = new Record(value)，但是从我们开始的位置获取惯用的C ++代码需要更多的更改。
现在，鉴于此，你可以：
CreateRecord和CopyRecordValues已在您当前的代码中处理）