variable "tmp" used in loop condition not modified in loop body - c

Question

#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
$
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Answer 1

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 will be even better to write
for ( const node* tmp = list; tmp != NULL; tmp = tmp->next ){
because within the loop the list is not being changed.
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.

Answer 2

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
Sidenote: Your program is full of repetition which makes it hard to find errors and it's also what makes it harder to see that you forgot to add the node with the number 1. If you instead make functions of the things that you do repeatedly it'll be much clearer. You can also make functions with descriptive names to make the whole program easier to read and maintain.
If your program was rewritten to make use of functions, it could look like below - and then it'll be very obvious that the node with the number 1 is missing.
#include <stdio.h>
#include <stdlib.h>
typedef struct node node;
struct node {
int number;
node* next;
};
node *create_node(int number, node *next) {
node *nn = malloc(sizeof *nn);
if(!nn) exit(1);
// assign values to the new node:
*nn = (node){.number = number,
.next = next};
return nn;
}
void insert_first(node **list, int number) {
*list = create_node(number, *list);
}
void insert_last(node **list, int number) {
// find the last "next" pointer (the one pointing at NULL):
while(*list) list = &(*list)->next;
// make the pointer pointing at NULL now point at the new node:
*list = create_node(number, NULL);
}
void free_all(node **list) {
for(node *tmp; *list; *list = tmp) {
tmp = (*list)->next;
free(*list);
}
}
void print_all(const node *list) {
for (; list; list = list->next) {
printf("%d\n", list->number);
}
}
int main(void) {
node* list = NULL;
insert_last(&list, 2);
insert_last(&list, 3);
insert_last(&list, 4);
insert_first(&list, 0);
print_all(list);
free_all(&list); // list == NULL after this
}