使用链接列表时出现EXC_BAD_ACCESS错误

Question

My program still compiles and prints from the inputted data files. However, an error still occurs after everything has printed and so the program doesn't end cleanly.

I get the error specifically in this part of my program

while (current->next != tail)

Basically this program is using a linked list to store information and outputting it into the screen. My particular error is with the clear() function that is supposed to clear the entire linked list with the pop_back() function.

//removes the last object from the linked list – deallocates memory
template <typename T>
void LL<T>::pop_back()
{

    if(count==0)
    {
        //Do nothing. Nothing to remove.
        return;
    }
    else{
        Node<T> *current;
        current=head;

        while (current->next != tail)
        {
            current=current->next;

        }

        delete tail;
        tail=current;

        count--;
    }
}

//Clears the linked list
template <typename T>
void LL<T>::clear()
{
    Node<T> *current= head;


    while (current != NULL)
    {
    pop_back();
    //current=tail;
}
current=tail;

head=tail=NULL;

}

Any help would be much appreciated.

Answer 1

Your pop_back method doesn't handle the case where the tail and the head are the same element (aka a linked list of 1 element). As a quick fix maybe an extra check for that case?

if(count==0)
{
    //Do nothing. Nothing to remove.
    return;
}
else if (count==1)
{
    head = tail = NULL;
    count--;
    return;
}

Also this loop is infinite as written:

while (current != NULL)
{
    pop_back();
    //current=tail;
}

maybe while (head != NULL) or while (count != 0) ?

Answer 2

You need to update node before the deleted one. Here is simplified version of your pop_back() method:

template <typename T>
void LL<T>::pop_back()
{
    Node<T> curr = head;
    Node<T> prev = NULL;

    while(curr != NULL && curr->next != NULL) // I'm assuming your tail's value is NULL
    {
        prev = curr;
        curr = curr->next;
    }

    if(curr != NULL)
    {
        if(prev != NULL)
            prev->next = NULL;
        delete curr;
        --count;
        if(count == 0)
            head = NULL;
    }
}

I didn't compiled the code, but I think idea is clear.

BTW you can improve performance of the clear() method:

Node<T> curr = head;
while(curr != NULL)
{
    Node<T> tmp = curr->next;
    delete curr;
    curr = tmp;
}

head = NULL;
tail = NULL;
count = 0;