Rear = rear+1%n;where n = size of queue. The modulus returns the remainder of both of the operands (2 of them). For example, 4%2=0 since 4/2=2 with no remainder, while 4%3=1 since 4/3=1 with remainder 1. Since you can never have a remainder higher than the right operand, you have an effective "range" of answers for any modulus of 0 to (n-1). With that in mind, you can just plug in the numbers for each variable ((4+1)%5=? and (1+1)%4=?). Using long division you would then find the remainder. Therefore, the boundary condition achieved is when at the given position there is already an element that is present hence it will break the loop and element would not be enqueued until the position is free.