C 中的链表不打印

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:

1. 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.

2. 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.

3. 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()