Enqueue values at the back and dequeue them from the front in first-in, first-out order.

Algorithm

Basic Implementation

basic.f90
program stack_queue_lesson
    implicit none
    integer :: values(3) = [10, 20, 30]
    integer :: result(3)
    integer :: i, count
    count = 0
    do i = 1, 3
        count = count + 1
        result(count) = values(i)
    end do
    call print_values(result, count)
contains
    subroutine print_values(items, n)
        integer, intent(in) :: items(:)
        integer, intent(in) :: n
        integer :: j
        do j = 1, n
            if (j > 1) write(*, '(A)', advance='no') ' -> '
            write(*, '(I0)', advance='no') items(j)
        end do
        write(*, *)
    end subroutine print_values
end program stack_queue_lesson

The queue keeps the oldest value at the front and adds new values at the back.

Step 1 - Enqueue 10, 20, 30

New values join at the back. The oldest value, 10, stays at the front.

Queue after three enqueues: front 10, then 20, then back 30.nextnext10front2030back

Step 2 - Dequeue removes 10

Removing from the front returns 10 and makes 20 the new front.

After one dequeue: removed is 10; front moves to 20.next10removed20front30back

Complexity

  • Time: O(1) per operation with a real queue
  • Space: O(n)

Implementation notes

  • Keep the explicit stack/queue operations. Library shortcuts that only produce the final list hide the data-structure behavior this lesson is meant to replay.
  • The final output uses a deterministic a -> b -> c format for cross-language comparison.
front The front is the oldest value still waiting in the queue.
FIFO A queue removes values in first-in, first-out order.