我遇到了双重链接列表的问题，所以我有两个问题。
首先，描述。
我用这种方式制作了一个struct：
typedef struct team{
char *name;
char *teamPlace;
}Team;
以这种方式创建了我的列表：
typedef struct nodeTeam{
int numberOfTeams;
Team team;
struct nodeTeam *next;
struct nodeTeam *prev;
}NodeTeam;
因此，我的列表将包含head和tail。当我在列表中添加一些Team时，我的head int numberOfTeams;将拥有我的列表中的团队数量。 tail将包含我的列表的最后一个元素，而int numberOfTeams;后的head将包含Team ID。
我的列表将以这种方式创建：
int createsListOfTeams(NodeTeam **head, NodeTeam **tail);
int addNodeTeamsSorted(NodeTeam *head, NodeTeam **tail, Team team);
void printListOfTeams(NodeTeam *listofTeams);
int main()
{
NodeTeam *headTeams,*tailTeams;
Team eq;
/*Creates the doubly linked list*/
if(createsListOfTeams(&headTeams,&tailTeams)){ /*See below this part of the code*/
printf("\nError\n");
return 0;
}
/*Teams are on a .txt file. The code for reading from a file is missing. It´s working ok so I believe it's not needed.
After reading one line after another it will do this
addNodeEquipasSorted(headTeams,tailTeams,eq);
where eq is a `struct` with the team data.
*/
/*Will print all the teams*/
printListOfTeams(headTeams);
return 0;
}
这是创建列表的代码：
/*Add the teams to the doubly linked list. At the end, all teams will be sorted by name*/
int createsListOfTeams(NodeTeam **head, NodeTeam **tail){
(*head) = (NodeTeam *)malloc(sizeof(NodeTeam));
if ((*head) == NULL){
return -1;
}
(*head)->numberOfTeams = 0;
strcpy((*head)->team.teamPlace,"");
strcpy((*head)->team.name,"");
(*head)->next = NULL;
(*head)->prev = NULL;
*tail = *head;
return 0;
}
将Team添加到我的列表中的代码（按团队名称排序）为：
/*Creates the doubly linked list*/
int addNodeTeamsSorted(NodeTeam *head, NodeTeam **tail, Team team){
NodeTeam *no, *aux;
/*Memory alloc for a new node*/
no = (NodeTeam*) malloc(sizeof(NodeTeam));
if (no == NULL){
return -1;
}
/*Updates the number of element of the list*/
head->numberOfTeams++;
/*Creates a copy of tail*/
aux = (*tail);
/*Puts team data on node*/
no->team = team;
/*to see if the list it's empty(no it's the first element of my list) or the last node as a name "smaler" then node*/
if(head == *tail || strcmp((*tail)->team.name,no->team.name) <= 0)
{
if (head == *tail){
no->numberOfTeams = 1;
}
else{
no->numberOfTeams = head->numberOfTeams;
(*tail)->numberOfTeams = no->numberOfTeams - 1;
}
no->next = (*tail)->next;
no->prev = *tail;
(no->prev)->next = no;
(*tail) = no;
aux = (*tail);
}
else{ /*If not the first element*/
head = head->next; /*To advance to the first item after my head*/
while(strcmp(head->team.name,no->team.name) < 0 && strcmp((*tail)->team.name,no->team.name) > 0 && head != *(tail)){
head = head->next;
(*tail) = (*tail)->prev;
}
if(strcmp(head->team.name,no->team.name) >= 0){
no->next = head;
no->prev = head->prev;
head->prev = no;
(no->prev)->next = no;
no->numberOfTeams = (no->next)->numberOfTeams;
(no->next)->numberOfTeams = no->numberOfTeams + 1;
if((no->prev)->prev != NULL){
(no->prev)->numberOfTeams = no->numberOfTeams - 1;
}
}
else{
no->next = (*tail)->next;
no->prev = (*tail);
no->numberOfTeams = (no->prev)->numberOfTeams + 1;
(no->prev)->next = no;
(no->next)->prev = no;
}
}
/*Puts `tail` pointing to the right position*/
if (aux != (*tail)){
(*tail) = aux;
}
return 0;
}
在我的.txt文件中，我有这些数据：
E team;E team place
J team;J team place
G team;G team place
F team;F team place
L team;L team place
A team;A team place
H team;H team place
O team;O team place
K team;K team place
P team;P team place
N team;N team place
B team;B team place
C team;C team place
M team;M team place
D team;D team place
I team;I team place
这是输出。
---------------------------------------------------------
| List of Teams |
---------------------------------------------------------
| Number of Teams | 16 | no 00740ff0 | prev 00000000 | next 00741240 |
--------------------------------------------------------
| ID | Team Name | Team Place |
--------------------------------------------------------
| 1 | A team | A team place | no 00741240 | prev 00740ff0 | next 00741450 |
| 2 | B team | B team place | no 00741450 | prev 00741240 | next 007436b0 |
| 3 | C team | C team place | no 007436b0 | prev 00741450 | next 00743760 |
| 4 | D team | D team place | no 00743760 | prev 007436b0 | next 00741088 |
| 5 | E team | E team place | no 00741088 | prev 00743760 | next 00741190 |
| 2 | F team | F team place | no 00741190 | prev 00741088 | next 00741138 |
| 3 | G team | G team place | no 00741138 | prev 00741190 | next 00741298 |
| 4 | H team | H team place | no 00741298 | prev 00741138 | next 007437b8 |
| 5 | I team | I team place | no 007437b8 | prev 00741298 | next 007410e0 |
| 4 | J team | J team place | no 007410e0 | prev 007437b8 | next 00741348 |
| 5 | K team | K team place | no 00741348 | prev 007410e0 | next 007411e8 |
| 7 | L team | L team place | no 007411e8 | prev 00741348 | next 00743708 |
| 8 | M team | M team place | no 00743708 | prev 007411e8 | next 007413f8 |
| 8 | N team | N team place | no 007413f8 | prev 00743708 | next 007412f0 |
| 9 | O team | O team place | no 007412f0 | prev 007413f8 | next 007413a0 |
| 10 | P team | P team place | no 007413a0 | prev 007412f0 | next 00000000 |
--------------------------------------------------------
通过此输出，我可以看到我的团队按名称添加到我的列表中。我打印内存地址的debbug显示一切正常。问题在于团队ID。它是int numberOfTeams;
所以，最后在所有这些文字之后，这些都是我的问题：
问题1 如何解决我的团队ID，即在我的列表中插入新的Team后，ID会更新为正确的ID。
问题2 尽管我的addNodeTeamsSorted正在使用ID执行，但我相信这是“chopy”的讽刺。你能推荐一些优化吗？
谢谢
答案 0 :(得分：2)
如何解决我的团队ID名称，即在将新团队插入我的列表后，ID会更新为正确的ID。
这是非法的记忆写：
(*head) = (NodeTeam *)malloc(sizeof(NodeTeam));
if ((*head) == NULL){
return -1;
}
(*head)->numberOfTeams = 0;
strcpy((*head)->team.teamPlace,""); <<------- here, and
strcpy((*head)->team.name,""); <<------- here
teamPlace和name被联合起来char*，因此strcpy()将写入不应该的地方。您需要为这些分配内存或将它们定义为固定大小的数组（如果可能）。