Linked list in C is not printing - c

Question

I think it is not printing because I am not inserting into the list correctly. I see most people allocate space outside of the loop, but I am not sure that is the case.
#include <stdio.h>
#include <stdlib.h>
typedef struct intList {
int number;
struct intList *next;
} INT_LIST;
int main() {
INT_LIST *start = NULL, *temp;
INT_LIST *trvPtr = start;
/* Insert into list */
while (temp->number != 0) {
temp = malloc(sizeof(INT_LIST));
printf("Enter your integer: ");
scanf("%d", &(temp->number));
temp->next = start;
start = temp;
}
printf("List is: \n");
while (trvPtr != NULL) {
printf("%d", trvPtr->number);
trvPtr = trvPtr->next;
}
return 0;
}

Answer 1

temp is uninitialized and points to garbage. Checking temp->number != 0 is undefined behavior.
You can solve this with a do/while loop. The loop will always run once. Allocate temp, set it, then check.
INT_LIST *start = NULL, *temp;
do {
temp = malloc(sizeof(INT_LIST));
printf("Enter your integer: ");
scanf("%d", &(temp->number));
temp->next = start;
start = temp;
} while(temp->number != 0);
Note that your list will always start with a value of 0. To avoid this, check the input before doing anything else. Use while(1) for an infinite loop and use break to end the loop.
INT_LIST *start = NULL;
while(1) {
int input;
printf("Enter your integer: ");
scanf("%d", &input);
if(input == 0) {
break;
}
INT_LIST *temp = malloc(sizeof(INT_LIST));
temp->number = input;
temp->next = start;
start = temp;
}
When you print, trvPtr will always be NULL; it will not track changes to start. You have to assign it to the new value of start.
printf("List is: ");
INT_LIST *trvPtr = start;
while (trvPtr != NULL) {
printf("%d ", trvPtr->number);
trvPtr = trvPtr->next;
}
puts("");
Note that you're not limited to declaring all the variables at the start of the function.

Answer 2

I have made some modifications/corrections in your code and the following should work -
Correct Code -
#include <stdio.h>
#include <stdlib.h>
typedef struct intList {
int number;
struct intList *next;
} INT_LIST;
int main() {
INT_LIST *start = NULL, *temp;
temp = (INT_LIST*)malloc(sizeof(INT_LIST));
INT_LIST *trvPtr = temp;
/* Insert into list */
do {
printf("Enter your integer: ");
scanf("%d", &(temp->number));
start = temp;
temp->next = (INT_LIST*)malloc(sizeof(INT_LIST));
temp = temp->next;
}while (start->number != 0);
temp->next=NULL;
printf("List is: \n");
while (trvPtr->next != NULL) {
printf("%d", trvPtr->number);
trvPtr = trvPtr->next;
}
return 0;
}
SAMPLE OUTPUT :
Enter your integer: 1
Enter your integer: 2
Enter your integer: 3
Enter your integer: 0
List is:
1230
If you don't want the 0 to be printed, include temp = start, right after the first while loop ends.
Explanation :
Firstly, in your code, you hadn't assigned temp anything. It should have been allocated space and assigned something first or else the statement temp->number != 0 is just undefined behavior.
Nextly, I have used start to check for the last value entered by the user. If it is 0, the loop will break. Do note I have converted the loop into do-while loop, since start is NULL at the beginning.
Lastly, once the loop ends, I have set the next of temp to null so that the printing loop will execute correctly. In your code, this part also had a mistake since you had assigned NULL to trvPtr but never changed the value. So, in your case, the loop would never have executed.
NOTE : It's always better to mention the cast-type explicitly while using malloc. Some compiler would give warnings/errors. So, I have modified that as well wherever you have used malloc()