Question

Trying not to lose it here. As you can see below I have assigned intFrontPtr to point to the first cell in the array. And intBackPtr to point to the last cell in the array…:

bool quack::popFront(int& nPopFront)
{   
    nPopFront = items[top+1].n;
    if ( count >= maxSize ) return false;
    else
    {
        items[0].n  = nPopFront;
        intFrontPtr = &items[0].n;
        intBackPtr  = &items[count-1].n;
    }
    for (int temp; intFrontPtr < intBackPtr ;)
    {
         ++intFrontPtr;
         temp = *intFrontPtr;
        *intFrontPtr = temp;
    }
    return true;
}

In the else statement I’m simply reassigning to ensure that my ptrs are where I want them. For some reason I’m popping off the back instead of off the front.

Anyone care to explain?

Answer 1

I’m not entirely sure I understand what you’re trying to do, but if I;m guessing right you’re trying to ‘pop’ the 1st element of the array (items[0]) into the nPopFront int reference, then move all the subsequent elements of the array over by one so that the 1st element is replaced by the 2nd, the 2nd by the 3rd, and so on. After this operation, the array will contain one less total number of elements.

Not having the full declaration of the quack class makes most of the following guesswork, but here goes:

I’m assuming that item[0] represents the ‘front’ of your array (so it’s the element you want ‘popped’).

I’m also assuming that ‘count` is the number of valid elements (so item[count-1] is the last valid element, or the ‘back’ of the array).

Given these assumptions, I’m honestly not sure what top is supposed to represent (so I might be entirely wrong on these guesses).

Problem #1: your nPopFront assignment is reversed, it should be:

nPopFront = items[0].n;

Problem #2; your for loop is a big no-op. It walks through the array assigning elements back to their original location. I think you want it to look more like:

for (int i = 1; i < count; ++i)
{
    items[i-1].n = items[i].n;  // move elements from back to front
}

Finally, you’ll want to adjust count (and probably top – if you need it at all) before you return to adjust the new number of elements in the data structure. The whole thing might look like:

bool quack::popFront(int& nPopFront)
{   
    if ( count >= maxSize ) return false;
    if ( count == 0 ) return false; // nothing to pop

    nPopFront = items[0].n;
    intFrontPtr = &items[0].n;      // do we really need to maintain these pointers?
    intBackPtr  = &items[count-1].n;

    for (int i = 1; i < count; ++i)
    {
        items[i-1].n = items[i].n;  // move elements from back to front
    }

    count -= 1;     // one less item in the array
    return true;
}