Can you please explain this?

Question

#include <iostream>
using namespace std;

class Queue {
int size;
int* queue;

public:
Queue() {
size = 0;
queue = new int[100];
}
void remove() {
if (size == 0) {
cout << "Queue is empty"<<endl;
return;
}
else {
for (int i = 0; i < size - 1; i++) {
queue[i] = queue[i + 1];
}
size--;
}
}
void print() {
if (size == 0) {
cout << "Queue is empty"<<endl;
return;
}
for (int i = 0; i < size; i++) {
cout<<queue[i]<<" <- ";
}
cout <<endl;
}
//your code goes here

void add(int x) {
queue[size++] = x;
}
};

int main() {
Queue q;
q.add(42); q.add(2); q.add(8); q.add(1);
q.print();
q.remove();
q.add(128);
q.print();
q.remove();
q.remove();
q.print();

return 0;
}

Specially, the remove and add method

Answer 1

So I guess you know that queue is a first in first out structure right? So just an example of the queue given in the codes, it'd be something like this
[42] add 42
[42, 2] add 2
[42, 2 ,8] add 8
[42, 2, 8, 1] add 1
[2, 8, 1] ---first remove
And so on
So what happen is that you want to always add to the end of the queue, and the end of the queue is precisely at index *size*, and after you add, you'd want to increment the size (because 1 more item), therefore you do
queue[size] = x; size++;
or combine together as shown in the codes.
For remove part, basically remove the first element by bringing every element at the back to the front, using the first remove above, it'll be like this
[42, 2, 8, 1]
[2, 2, 8, 1]
[2, 8, 8, 1]
[2, 8, 1, 1]
[2, 8, 1] - don't have to care for the last 1, it'll be overwritten if there's another add
So, basically remove is done by pushing every element one step forward, overwriting the element in front of it(therefore the first element is 'removed')

Answer 2

Hoh Shen Yien oOo, Ok, Thanks.

Answer 3

Hoh Shen Yien What if instead of replacing the 1st element with 2nd element, I set every element to NULL?

Answer 4

If setting every element to Null, you have reset the queue to an empty queue