C 中的链表实现失败

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.

Answer 1

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;
    }
}