Thread: Stuck in simple while loop, can't break or return?

Code:

void printnode(Area*current){
	while(1){
	if(current->next==NULL){printf("[%p]", current->next); return;}
	if(current->next!=NULL){printnode(current->next);}	
	}
}

In this while loop, as I understand it, a structure Area*current is examined for the ->next. If that value is NULL, the value NULL should be printed and the break; return; should exit the function. Meanwhile, if *next is not null, then the function should recursively examine *next until it reaches a value of null. However, what I get is an infinite loop which just prints:

Code:

nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(nil)][(

Why is this? If it is printing nil, then the value == null, and if the value == null, then it should return and exit the loop?

Originally Posted by tabstop

Nothing happens after a break; that's the point of a break. So there is no way to reach the return statement. If you want to return just return.

break; was only there because return; wasn't working. I can take the break out but nothing changes.

Are you trying to implement printnode iteratively or recursively? It looks like you are trying to do both at the same time, which is probably the problem.

Also, is printnode supposed to print the current node in addition to all the following nodes? It looks like you only start printing from the next node.

I would have written it as:

Code:

void printnode(Area *current) {
    if (current) {
        printf("\n<%d||[%p]>: <%p|%p>\n",
               current->name, current, current->prev, current->next);
        printnode(current->next);
    }
}

Observe the tail recursion. It is easy to transform it to:

Code:

void printnode(Area *current) {
    while (current) {
        printf("\n<%d||[%p]>: <%p|%p>\n",
               current->name, current, current->prev, current->next);
        current = current->next;
    }
}

In which case you would no longer have the limitations of recursion depth.