CODEWITHSUNDEEP
c
Jim
Jim

Reputation: 3

return char* within a while loop

I am trying to print the contents of a linked list which is a char* but the while loops are messing up the code:

the function to get the next item in a linked list:

char * list_next(list *l)
{
    list *currentPosition = NULL;
    currentPosition = l->next; //since the first node is a dummy value in the singly linked list

    while (currentPosition != NULL)
    {
        return currentPosition->charValue;
        currentPosition = currentPosition->next;
    }

    return NULL;
}

in my main:

char * item;
while(item = list_next(list))
    printf("%s ",item);

can someone please help me I am pretty sure the problem is the returns inside the while loop but i cant seem to fix it

Upvotes: 0

Views: 176

Answers (4)

ComFreek
ComFreek

Reputation: 29424

You've mentioned in a comment that you just need to return all values in the linked list.

function iterate(someNode)
   if someNode ≠ null
     node := someNode
     do
       do something with node.value [1]
       node := node.next
     while node ≠ someNode

From this Wikipedia article, CC BY-SA.

Now you could simply printf() at position [1] in the code. However it seems that you have to return all values. Therefore you have to create an (probably dynamic) array.

Upvotes: 0

Michael Dorgan
Michael Dorgan

Reputation: 12515

Your program, even with the return position swapped is still only going to print the "2nd" item in the list every time through - if it exists. You either need to work with a double pointer to update the base value, or come up with some better way to iterate your list.

Upvotes: 2

Eddie
Eddie

Reputation: 427

You pass the original list to list_next(). I believe you will always print the second item in a forever loop.

I suggest you can simplify it as below:

char *item;
for (item=list->next; // Skip the first item as you said the first node is dummy.
     item != NULL; item=item->next) {
    printf("%s ",item->charValue);
}

Upvotes: 0

user529758
user529758

Reputation:

Swap the two lines. return immediately exits from the function. It should read

currentPosition = currentPosition->next;
return currentPosition->charValue;

instead.

(Not to mention numerous other errors that others pointed out as well - the lack of ability to actually update the next pointer because of confusion about scopes, the missing check for NULL before dereferencing, etc.)

Upvotes: 3

Related Questions