Adding two linked list

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Here is code i have been trying which adds two linked list. Given a list for example like 3->4->5 and 1->6->3 ,this results into a new list 4->0->9 .
The most significant digit is the leftmost in the list. But the result is not expected . The printlist funtion prints the result linked list and it gives only one value 0 .Please help me out?

void add_list(struct list *node1,struct list *node2)
{
  if(node1==NULL && node2==NULL)

    {
      printf("empty list\n");
    }
  if(node1==NULL){
    return node2;
  }
  if(node2==NULL){
    return node1;
  }
  int i, carry=0;
  
  struct list *ptr1=node1,*ptr2=node2;
  
  struct list *result;
  
  result=malloc(sizeof(struct list));

  while(node1->next!=NULL && node2->next==NULL)

{
      result->data=((node1->data)+(node2->data)+(carry));
      node1=node1->next;
      node2=node2->next;
      result=result->next;
      if((result->data)>9){
         carry=(result->data)%10;
         
          }
  
    }
  return result;
}
 
   printlist(result);

 }</pre>

Posted 6-Sep-13 2:40am

Member 10200096

Updated 6-Sep-13 4:26am

v5

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Richard MacCutchan 6-Sep-13 9:25am

Most interesting, and do you have a question?

Member 10200096 6-Sep-13 9:51am

yes ! i updated it . I forgot to mention the problem

2 solutions

Solution 1

Well one problem that springs to mind is that you only ever allocate a single element in result, so you can't return more than a single element.
Second, you exit the function after you have processed the first element in both the lists!

A couple of suggestions for you:
1) sort out your indentation - as it is, it reflects a flow of control which does not happen.
2) Use a debugger, and step through your code - you would have spotted the return was in the loop almost immediately!

Posted 6-Sep-13 4:12am

OriginalGriff

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Fredrik Bornander · Accepted Answer · 2013-09-06T04:27:00

Your function makes little sense as you're trying to return values from a function with void return type;

Something like this might work for you:

C++

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


typedef struct list
{
    int data;
    list* next;
} list;

list* add_node(list* root, int data) {
    list* node = (list*)malloc(sizeof(node));
    node->data = data;
    node->next = NULL;
    if (root != NULL) {
        list* ptr = root;
        while(ptr->next != NULL) {
            ptr = ptr->next;
        }
        ptr->next = node;
    }

    return node;
}

void delete_list(list* root)
{
    list* ptr = root;
    while(ptr != NULL) {
        list* node = ptr;
        ptr = ptr->next;
        free(node);
    }
}

void print_list(list* root) {
    list* ptr = root;
    while(ptr != NULL) {
        printf("%d->", ptr->data);

        ptr = ptr->next;
    }
}

void add_list(list*l1, list* l2)
{
  list *l1Ptr = l1;
  list *l2Ptr = l2;


  int carry = 0;
  list *result = NULL;
  while(l1Ptr != NULL && l2Ptr != NULL) {
      int sum = l1Ptr->data + l2Ptr->data + carry;
      if (sum > 9) {
          sum %= 10;
          carry = 1;
      }

      if (result == NULL)
          result = add_node(result, sum);
      else
          add_node(result, sum);

      l1Ptr = l1Ptr->next;
      l2Ptr = l2Ptr->next;
  }
  printf("Result: ");
  print_list(result);
  printf("\r\n");
}

int main(int argc, char* argv[])
{
    list* l1 = add_node(NULL, 3);
    add_node(l1, 4);
    add_node(l1, 5);

    list* l2 = add_node(NULL, 1);
    add_node(l2, 6);
    add_node(l2, 3);

    printf("List 1: ");
    print_list(l1);
    printf("\r\n");
    printf("List 2: ");
    print_list(l2);
    printf("\r\n");


    add_list(l1, l2);


    return 0;
}

Hope this helps,
Fredrik