Failed implementation of a Linked List in C

Question

I do not understand why this program is not working and the elements were not inserted into the List as intended.

Every time, when I am debugging, I see that when I go to the main method after the 'insert' method, the Linked List is still empty and I do not understand why because I think it should be well because I am using pointers (It seems like a 'Dangling Pointer' case, but if it is, I do not understand why).

Maybe should I use double star (**)? If yes, why in arrays it does not matter?

Here is the source code:

#include <stdio.h>
#include <stdlib.h>

struct A{
    int val;
    struct A* next;
} A;

void insert(struct A* L, int newVal) {

    if (L == NULL) {

        L = (struct A*) malloc(sizeof(struct A));
        L->val = newVal;
        L->next = NULL;
    }

    else {

        struct A* p = L;

        while (p->next != NULL) {

            p = p->next;
        }

        p->next = (struct A*) malloc(sizeof(struct A));
        p->next->val = newVal;
        p->next->next = NULL;
    }
}


void printA(struct A* printed) {

    struct A* p = printed;

    while (p != NULL) {
        printf("%d\n", p->val);
        p = p->next;
    }
}


int main() {

    struct A* L = NULL;

    insert(L, 1);
    printf("1 success\n");

    insert(L, 2);
    printf("2 success\n");

    insert(L, 3);
    printf("3 success\n");

    insert(L, 4);
    printf("4 success\n");

    insert(L, 5);
    printf("5 success\n");

    printf("\n\n\n");

    printA(L);

    return 0;
}

Thank You.

See https://stackoverflow.com/questions/2838038/c-programming-malloc-inside-another-function — Arkadiusz Drabczyk, Apr 29 '20 at 17:44
when passing a pointer, those contents you want to modify in a sub function, then you must pass the address of the pointer, not the contents of the pointer, just like you have to pass the address of any other variable that you want the sub function to modify — user3629249, Apr 30 '20 at 20:02

MrBens · Accepted Answer · 2020-04-30T18:45:09.107

insert function first argument is a pointer to struct. When you pass your struct, insert recieves the address, and creates a local pointer to the same place. In order to change what the actual struct (from main) is pointing at, you have to pass a double pointer.

Written below are the parts that need to be changed:

void insert(struct A** L, int newVal) {

    if (*L == NULL) {

        *L = (struct A*) malloc(sizeof(struct A));
        (*L)->val = newVal;
        (*L)->next = NULL;
    }

    else {

        struct A* p = *L;

        ...
        ...
        ...
    }
}    

int main() {

    struct A* L = NULL;

    insert(&L, 1);
    printf("1 success\n");

    ...
    ...
    ...

    printA(L);

    return 0;
}

A different approach would be to stay with a single pointer, but to change the return value of insert to struct A*. You'll just have to assign the return value to your main struct, like this:

struct A *insert(struct A* L, int newVal) {

    if (L == NULL) {

        L = (struct A*) malloc(sizeof(struct A));
        L->val = newVal;
        L->next = NULL;

        return L;
    }

    else {
        ...
    }

    return L;
}

int main() {

    struct A* L = NULL;

    L = insert(L, 1);
    ...
    return 0;
}

In addition, your print function isn't moving anywhere. Add the line p = p->next;

void printA(struct A* printed) {

    struct A* p = printed;

    while (p != NULL) {
        printf("%d\n", p->val);
        p = p->next;
    }
}

Failed implementation of a Linked List in C

1 Answers1