#include <stdio.h>
#include <stdlib.h>
typedef struct node{
int number;
struct node* next;
}
node;
int main(void){
node* list = NULL;
node *n = malloc(sizeof(node));
if(n==NULL){
return 1;
}
n->number = 2;
n-> next = NULL;
list = n;
n = malloc(sizeof(node));
if(n == NULL){
free(list);
return 1;
}
n->number = 3;
n->next = NULL;
list->next = n;
n = malloc(sizeof(node));
if(n == NULL){
free(list->next);
free(list);
}
n->number = 4;
n->next = NULL;
list->next->next =n;
n = malloc(sizeof(node));
if(n!=NULL){
n->number = 0;
n->next = NULL;
n->next = list;
list = n;
}
for( node* tmp = list; tmp != NULL; tmp->next){
printf("%i\n" , tmp->number);
}
while(list!=NULL){
node*tmp = list->next;
free(list);
list=tmp;
}
}
was trying linked list. expected when running the code: 0 1 2 3 4 $ //asdoihasidashiofdhiohdfgdiwheifiopioioiophfaifjasklfhafiashfauiosfhwuiohwefuiowhfaslfidasdaskdasjdlaksdjqwfiqpweiojfkldfjsdfklwhefiowefweopfiosfkosid;fjwdfp;fdasiopfjew[0fowejfwepfojmofejmiwrfgj;wdfjewio;fijwefjsdp;jfkl;wjw
CodePudding user response:
Actually you are not changing the pointer
for( node* tmp = list; tmp != NULL; tmp->next){
You need to write
for( node* tmp = list; tmp != NULL; tmp = tmp->next){
It even be better to write
for ( const node* tmp = list; tmp != NULL; tmp = tmp->next ){
Also in this code snippet
if(n!=NULL){
n->number = 0;
n->next = NULL;
n->next = list;
list = n;
}
the statement
n->next = NULL;
is redundant.
CodePudding user response:
In this loop, tmp->next
has no effect because you don't assign it to anything.
for (node* tmp = list; tmp != NULL; tmp->next) {
printf("%i\n", tmp->number);
}
You must do tmp = tmp->next
:
for (node* tmp = list; tmp != NULL; tmp = tmp->next) {
// ^^^^^
printf("%i\n", tmp->number);
}
Also, you can't expect 1
to be in the output because you never create a node with that number. With the above change the program will therefore output:
0
2
3
4