struct node{
int data; struct node *next;
};
void push(struct node* head, struct node* n){
if(n!= NULL){
if(head==NULL)
head = n;
else {
n->next = head;
head = n;
}
} else printf("Cannot insert a NULL node");
}
struct node* pop(struct node* head){
if(head!=NULL){
struct node *n = head;
head = head->next;
return n;
} else {
printf("The stack is empty");
return NULL;
}
}
int main(){
int i;
struct node *head = NULL, *n;
for(i=15;i>0;i--){
struct node *temp = malloc(sizeof(struct node));
temp -> data = i;
temp->next = NULL;
push(head,temp);
}
n = head;
while(n!=NULL){
printf("%d ",n->data);
n=n->next;
}
return 0;
}
You need to pass the address of the pointer head to the function push. I your case the head is not getting modified because you are only passing the value in the head.
void push(struct node** head, struct node* n){
if(n!= NULL){
if(*head==NULL)
*head = n;
else {
n->next = *head;
*head = n;
}
} else printf("Cannot insert a NULL node");}
int main(){
int i;
struct node *head = NULL, *n;
for(i=15;i>0;i--){
struct node *temp = (struct node *)malloc(sizeof(struct node));
temp -> data = i;
temp->next = NULL;
push(&head,temp);
}
n = head;
while(n!=NULL){
printf("%d ",n->data);
n=n->next;
}
return 0;}
You are passing the head
pointer by value to the function push(head,temp);
. The changes to head
done inside push
will not be reflected in the main()
function.
You should pass address of head
to push()
.
push(&head, temp);
and inside push()
:
*head = n;
Similar change will be required for pop()
. You can verify what I am saying by adding a printf
inside the loop in main()
as: printf("%p\\n", head);
. The value of head
will remain unchanged.
BTW, it is good practice to add a \\n
at the end of statement inside printf
, it flushes the stdout
stream immmediately hence your output is printed immediately on stdout
(your computer screen).
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