我正在做一个双向链表。据我所知，它正在工作，但来到这里是为了确保并查看我是否以正确的方式进行操作。
另一方面，当我做这个时，我遇到了其他与双向链表无关但与 C 文件之间的结构和“可见性”有关的问题。如果您了解我应该对这两个其他疑问提出其他问题，请告诉。否则请随时启发我。
在我的 file1.c 我有这个：
代码
#include <stdio.h>
#include <stdlib.h>
typedef struct team{
char *name;
char *teamPlace;
}Team;
typedef struct nodeTeam{
int numberOfTeams;
Team team;
struct nodeTeam *next;
struct nodeTeam *prev;
}NodeTeam;
int createsListOfTeams(NodeTeam **head, NodeTeam **tail);
void printListOfTeams(NodeTeam *listofTeams);
int addNodeTeamsSorted(NodeTeam *head, NodeTeam **tail, Team team);
int main()
{
NodeTeam *headEquipas,*tailEquipas;
Team eq;
/*Creates the doubly linked list*/
if(createsListOfTeams(&headEquipas,&tailEquipas)){
printf("\nError\n");
return 0;
}
/*Add the teams to the doubly linked list. At the end, all teams will be sorted by name*/
eq.name = "D team";
eq.teamPlace = "D team place";
if (addNodeTeamsSorted(headEquipas,&tailEquipas,eq)){
printf("\nError\n");
return 0;
}
eq.name = "A team";
eq.teamPlace = "A team place";
if (addNodeTeamsSorted(headEquipas,&tailEquipas,eq)){
printf("\nError\n");
return 0;
}
eq.name = "C team";
eq.teamPlace = "C team place";
if (addNodeTeamsSorted(headEquipas,&tailEquipas,eq)){
printf("\nError\n");
return 0;
}
eq.name = "B team";
eq.teamPlace = "B team place";
if (addNodeTeamsSorted(headEquipas,&tailEquipas,eq)){
printf("\nError\n");
return 0;
}
/*Will print all the teams*/
printListOfTeams(headEquipas);
return 0;
}
在我的 file2.c 我有这个
代码
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct team{
char *name;
char *teamPlace;
}Team;
typedef struct nodeTeam{
int numberOfTeams;
Team team;
struct nodeTeam *next;
struct nodeTeam *prev;
}NodeTeam;
/*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;
(*head)->team.teamPlace = "";
(*head)->team.name = "";
(*head)->next = NULL;
(*head)->prev = NULL;
*tail = *head;
return 0;
}
/*Creates the doubly linked list*/
int addNodeTeamsSorted(NodeTeam *head, NodeTeam **tail, Team team){
NodeTeam *no,*listIni;
no = (NodeTeam*) malloc(sizeof(NodeTeam));
if (no == NULL){
return -1;
}
/*copy of my list*/
listIni = head;
no->team = team;
/*to see is it's the first element of my list*/
if(head->numberOfTeams == 0)
{
no->next = head->next;
no->prev = head;
head->next = no;
*tail = no;
}
else{ /*If not the first element*/
head = head->next;
while(head->prev != *tail && strcmp(head->team.name,no->team.name) < 0 && strcmp((*tail)->team.name,no->team.name)>0){
head = head->next;
(*tail) = (*tail)->prev;
}
if(strcmp(head->team.name,no->team.name) >= 0 || head->prev == *tail){
no->next = head;
no->prev = head->prev;
(head->prev)->next = no;
head->prev = no;
}
else if(strcmp((*tail)->team.name,no->team.name) <= 0){
no->next = (*tail)->next;
no->prev = (*tail);
(*tail)->next = no;
*tail = no;
}
}
/*Updates the number of element of the list*/
head = listIni;
head->numberOfTeams++;
return 0;
}
/*Prints my lists*/
void printListOfTeams(NodeTeam *listofTeams){
printf("| number of teams %22d |\n",listofTeams->numberOfTeams);
printf("| team name | team place |\n");
printf("--------------------------------------------------\n");
listofTeams = listofTeams->next;
while (listofTeams != NULL){
printf("| %-21s | %-22s |\n",listofTeams->team.name,listofTeams->team.teamPlace);
listofTeams = listofTeams->next;
}
printf("--------------------------------------------------\n\n");
}
所以这是我的树问题：
Q1 - 这是实现一个头和尾分别指向列表开头和结尾的双向链表的正确方法吗？
Q2 - 为什么要在我的两个文件上声明struct team和struct nodeTeam？既然它们都在同一个项目中，那么声明不应该对我项目中的所有文件“可见”吗？
Q3 -struct team为什么我必须申报char *name而不是char name[31]？