Help free structs and pointer.

Hey guys I'm having troubles understanding how to free a list of linked structs.

For simplicity reason we'll say that my struct has a value and a ptr to the next item. Here is waht I have but when I compile it, it doesn' t do anything.

Code:

void removePtr(LIST_ELEM *pHead, int loc){

	LIST_ELEM *ptr;
	LIST_ELEM *tmpPtr;
	int size = sizeOf(pHead);
	int i = 0;

	for(ptr = pHead; pHead != NULL; ptr++){
	   
	    if( loc == 0){
		tmpPtr = pHead;
		pHead = ptr->pNext;
		free(tmpPtr);
	     }
	    else if(loc = size-1){
		LIST_ELEM *ptrB4 = nodeB4(pHead,loc);
		ptrB4->pNext = NULL;
		free(ptr);
		}		
				
	    else if( i == loc){
		LIST_ELEM *ptrB4 = nodeB4(pHead,loc);
		tmpPtr = ptr;
		LIST_ELEM *ptrAfter = (tmpPtr->pNext);
		ptrB4->pNext = ptrAfter;
		free(tmpPtr);
		}
	i++;
	}
}

remove takes in the head of the list and the location/index where we want the item removed.

nodeB4 returns the item that come before the item we want to remove. Any ideas?

Originally Posted by TaiL

Code:

void removePtr(LIST_ELEM *pHead, int loc){

	LIST_ELEM *ptr;
	LIST_ELEM *tmpPtr;
	int size = sizeOf(pHead);
	int i = 0;

	for(ptr = pHead; pHead != NULL; ptr++){
	   
	    if( loc == 0){
		tmpPtr = pHead;
		pHead = ptr->pNext;
		free(tmpPtr);
	     } //if you pass in loc=0, won't you kill the entire list as you go through the loop?
	    else if(loc = size-1){ //this is an assignment, not a comparison (also, why is it a special case?
		LIST_ELEM *ptrB4 = nodeB4(pHead,loc);
		ptrB4->pNext = NULL;
		free(ptr);
		}		
				
	    else if( i == loc){ //because of the assignment above, this never happens
		LIST_ELEM *ptrB4 = nodeB4(pHead,loc);
		tmpPtr = ptr;
		LIST_ELEM *ptrAfter = (tmpPtr->pNext);
		ptrB4->pNext = ptrAfter;
		free(tmpPtr);
		}
	i++;
	}
}

See comments in code.

Hm if its equal to 0 that would mean we want to remove the head of the "list" right? So then we'd have to change where the head of the list is which it then next item.


checking to see if its the last item of the list allows us to change the 2nd to last item of the list to point to null.

I think thats what I was going for.

I agree that the head is a special case (although the tail isn't). But you can't do this inside the loop -- otherwise you're going to delete the head of the list; and then you're going to delete the head of the list; and then you're going to delete the head of the list; and then....

But I just now noticed the reason your for-loop never happens in the first place: pHead != NULL is never true.

I actually took a big red marker and drew == above the function keys.

Nothing to do with the iching tho.

above the function keys?

wouldn't I need pHead!=Null so it could iderate through it all to find the item at the location?

Originally Posted by TaiL

above the function keys?

wouldn't I need pHead!=Null so it could iderate through it all to find the item at the location?

Is pHead supposed to change through the loop?

HEY TABSTOP!! Sorry its suppose to be ptr not pHead. =)thnx

so after fixing that error. Here's what i have

Code:

void removePtr(LIST_ELEM *pHead, int loc){

	LIST_ELEM *ptr;
	LIST_ELEM *tmpPtr;
	int size = sizeOf(pHead);
	int i = 0;

	for(ptr = pHead; ptr != NULL; ptr++){
	   
	    if( loc == 0 && (size==0)){
		tmpPtr = pHead;
		pHead = ptr->pNext;
		free(tmpPtr);
		printf("The list is now empty.");
	     }		
	   else if( loc == 0){
		tmpPtr = pHead;
		pHead = ptr->pNext;
		free(tmpPtr);
	     }	
	    else if( i == loc){
		LIST_ELEM *ptrB4 = nodeB4(pHead,loc);
		tmpPtr = ptr;
		LIST_ELEM *ptrAfter = (tmpPtr->pNext);
		ptrB4->pNext = ptrAfter;
		free(tmpPtr);
		}
	i++;
	}
}

Checking the first two cases gives me a seg fault and checking the last error gives me a

Code:

 glibc detected *** a.out: free(): invalid pointer:

errror.

going home so I'll be back tomorrow =(

Man I really didn't look at your for-loop. "ptr++" makes no sense in a linked list. Use "ptr=ptr->next" instead.

lol you know I don't know why i did that I originally had pNext instead of p++ =\
but for some reason it still doesn't seem to work giving me the same glibc error.

So it executes now but it doesn't free anything.

Code:

if( loc == 0 && (size==1)){
		tmpPtr = ptr;  // points tmpPtr to where Ptr is pointing
		pHead = NULL;//since its our only item we'll set pHead to point to null
		free(tmpPtr);// we then free tmpPtr
		printf("The list is now empty.");//lets usr know item was deleted.
		
	     }

I don't know what you think size==1 is supposed to do, but size will always equal 4 (regardless of how many things there are on the list, et cetera).

sorry, size is equal to to a value returned by a method which returns the actual size of my list.

Okay, yes I see the capital O this time. So why do you say it doesn't free anything? If you mean the assigning of pHead to NULL isn't seen in the calling function, well, that's true. Changes made in the function stay in the function.

If you want to be able to change the value of the pointer, you need to pass in the address of the pointer (so that you would have a pointer-to-pointer-to-LIST_ELEM).