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Question

I am working on problem 2 in leetcode (Two Sum) and I keep getting this error. I don't understand how this reflects what I wrote: here is my code that I wrote, I think it has something to do with the dynamic allocations. any help would be appreciated.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
int GetLength(struct ListNode* list){
    int count = 0;
    while (list != NULL){
        count ++;
        list = list->next;
    }
    return count;
}


struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){
    int count1 = GetLength(l1);
    int count2 = GetLength(l2);
    struct ListNode* newVal = malloc(sizeof(struct ListNode));
    struct ListNode* start;
    if (count1 < count2){
        int* tmp = l1;
        l1 = l2;
        l2 = tmp;
        int tmpcount = count1;
        count1 = count2;
        count2 = tmpcount;
    }
    // updated l1 is longer
    int carryflag = 0;
    int iteration = 0;
    while (l2 != NULL){
        iteration ++;
        int sum = l1->val + l2->val + carryflag;
        printf("%d\n", iteration);
        if (iteration = 1){
         start = newVal;
        }
        newVal->val = sum % 10;
        if (sum >= 10) carryflag = 1;
        else carryflag = 0;
        l1 = l1->next;
        l2 = l2->next;
        newVal->next = malloc(sizeof(struct ListNode));
        newVal = newVal->next;
    }
    while(l1 != NULL){
        newVal->next = malloc(sizeof(struct ListNode));
        newVal->val = l1->val;
        l1 = l1->next;
        newVal = newVal->next;
    }
    return start;
}

Answer 1

If not take into account typos as for example in this statement

if (iteration = 1){

where there is used the assignment operator = instead of the comparison operator == the program contains serious bugs.

In this code snippet

if (count1 < count2){
    int* tmp = l1;
    l1 = l2;
    l2 = tmp;
    int tmpcount = count1;
    count1 = count2;
    count2 = tmpcount;
}

you are assigning a pointer of the type int * to a pointer of the type ListNode *

    l2 = tmp;

So the compiler should issue a message that there are used pointers of incompatible types.

Also the function can produce a memory leak if empty lists are passed and can be a reason of undefined behavior because for example the last node can be leaved unitialized.

    newVal->next = malloc(sizeof(struct ListNode));
    newVal = newVal->next;

And if the second list is empty then the pointer start will not be even itialized.

The second while loop ignores the value of carryFlag .

In any case the approach is inefficient. There is no any need to count nodes in the two lists.

The function can be declared and defined the following way

struct ListNode * addTwoNumbers( const struct ListNode *l1, const struct ListNode *l2 )
{
    const int Base = 10;

    struct ListNode *head = NULL;
    struct ListNode **current = &head;

    int carryFlag = 0;

    while ( carryFlag != 0 || l1 != NULL || l2 != NULL )
    {
        *current = malloc( sizeof( struct ListNode ) );

        int value = ( l1 == NULL ? 0 : l1->val ) + 
                    ( l2 == NULL ? 0 : l2->val ) +
                    carryFlag;

        ( *current )->val = value % Base;
        carryFlag = value / Base;

        ( *current )->next =  NULL;
        current = &( *current )->next;

        if ( l1 != NULL ) l1 = l1->next;
        if ( l2 != NULL ) l2 = l2->next;
    }   

    return head;
}

Answer 2

Variation on @Vlad from Moscow suggested code:

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

struct ListNode * addTwoNumbers(const struct ListNode *l1, const struct ListNode *l2) {
  struct ListNode head = {.next = NULL};
  struct ListNode *previous = &head;

  const int Base = 10;
  int carry = 0;

  while (l1 || l2 || carry) {
    int value = carry;
    if (l1) {
      value += l1->val;
      l1 = l1->next;
    }
    if (l2) {
      value += l2->val;
      l2 = l2->next;
    }

    struct ListNode *current = malloc(sizeof current[0]);
    if (current == NULL) {
      // TBD code to handle error
      return NULL;
    }

    current->next = NULL;
    current->val = value % Base;
    carry = value / Base;
    previous->next = current;
    previous = current;
  }

  return head.next;
}