Thread: Epic System crash (quite funny)

Code:

temp2=temp1->next->next;

if temp1->next is null this will crash

PS. instead of i=0 you can just use break; statement

Further you have quite a few cases of bad logic. If front is the node you want to delete, your code obliterates the entire linked list.

Second, you update temp1, and then do temp2 = temp1->next->next. This means temp2 is always 2 steps ahead of temp1, not 1 like I believe you think it is.

Third, don't name variables temp unless they are holding a value for a very short time; see how I used (and didn't use) temp. When I declared current and prev, without even looking at the code where they are used, it should be immediately obvious what the function of each variable is -- current will hold the node we are currently examining, and prev will hold the node just previous to the current one. In fact in general I wouldn't even use temp for the special case (I'd call it newFront) but I did to illustrate a situation in which it's not too bad to use a poorly descriptive name.

Here's a simpler kind of code:

Code:

void delEntry()
{
        // Check the empty case
        if (front == NULL) return;

        cout<<"date: ";
        cin>>delDate;
        cout<<"time: ";
        cin>>delTime;

        // Check the special case where we delete from the front of the list
        if ( front->equals(delDate, delTime) ) {
                object * temp = front->next;
                delete front;
                front = temp;
                return;
        }

        object * prev = front;
        object * current = front->next;

        while(current != NULL){
                if (current->equals(delDate, delTime)){
                        prev->next = current->next;
                        delete current;
                        return;
                }
                prev = current;
                current = current->next;
        }
}