我正在开发一个程序,该程序创建一个具有添加和显示新节点功能的结构。我有一个名为“add”的函数,我用它来创建新节点并将它们发送到struct-> next但是每当我尝试运行我的函数“displayData”时,该函数说我的结构是NULL /空。
这是代码。
#include <stdio.h>
#include <stdlib.h>
typedef struct node *nodePtr;
struct node {
int item;
nodePtr next;
};
typedef nodePtr Statistician;
int input();
Statistician newStatistician(); //allocates memory to the structure. Dynamic allocation
void add(Statistician s, int x); //Adds data to the rear
void displayData(Statistician s); //prints entire dataset
int main() {
int operation, data;
Statistician s = NULL;
data = input(); //get new input
add(s,data); //run add function
displayData(s); //run display function
}
int input(){
int x;
printf("Enter data: ");
if (scanf("%d", &x) != 1)
{
printf("\n\nInvalid Input!\n\n");
exit(0);
}
return x;
}
Statistician newStatistician(){
Statistician newStat;
newStat = malloc(sizeof(struct node));
return newStat;
}
void add(Statistician s, int x){
Statistician newNode = newStatistician();
newNode->item = x;
newNode->next = NULL;
if(s == NULL){
s = newNode;
return;
}
while (s != NULL) {
s = s->next;
}
s->next = newNode;
}
void displayData(Statistician s){
Statistician temp = s;
if(s==NULL){
printf("\n\nList is EMPTY.");
printf( "\n\nPress any key.\n" );
getch();
return;
}
printf( "\n\nThe List:\n" );
while (temp != NULL) {
printf(" %d", temp->item);
temp = temp->next;
}
printf( "\n\nPress any key.\n" );
getch();
return;
}
当我使用displayData时,输出为。
List is EMPTY
答案 0 :(得分:1)
您必须通过引用传递头节点。否则,更改列表的函数将处理头节点的副本,并且不会更改原始头节点。
例如
void add(Statistician *s, int x)
{
Statistician newNode = newStatistician();
newNode->item = x;
newNode->next = NULL;
while ( *s != NULL ) s = &( *s )->next;
*s = newNode;
}
该功能可以像
一样调用add( &s, data );