As you can see in the code size=0 ; then how later code size-1 will lead to -1, can you explain me the code !!!

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){
size+=1;
queue[size-1]=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;
}

Answer 1

Question is not clear..?
do you mean about queue[size-1]=x?

Answer 2

It's first incrementing , then using current position..
When size = 0
size+=1 causes size=1 but current position is 0 index so queue[1-1]=queue[0] = x
It won't go to -1

But better way to write is :

void add(int x) {
queue[size] = x;
size = size + 1;
}
//you can put overflow condition as well in add.

Hope it helps..

Answer 3

Yes