package genstack // GenStack implements a generational stack. // // GenStack works as common stack except for the fact that all elements in the // older generation are guaranteed to be popped before any element in the newer // generation. New elements are always pushed to the current (newest) // generation. // // We could also say that GenStack behaves as a stack in case of a single // generation, but it behaves as a queue of individual generation stacks. type GenStack[T any] struct { // We can represent arbitrary number of generations using 2 stacks. The // new stack stores all new pushes and the old stack serves all reads. // Old stack can represent multiple generations. If old == new, then all // elements pushed in previous (not current) generations have already // been popped. old *stack[T] new *stack[T] } // NewGenStack creates a new empty GenStack. func NewGenStack[T any]() *GenStack[T] { s := &stack[T]{} return &GenStack[T]{ old: s, new: s, } } func (s *GenStack[T]) Pop() (T, bool) { // Pushes always append to the new stack, so if the old once becomes // empty, it will remail empty forever. if s.old.len() == 0 && s.old != s.new { s.old = s.new } if s.old.len() == 0 { var zero T return zero, false } return s.old.pop(), true } // Push pushes a new element at the top of the stack. func (s *GenStack[T]) Push(v T) { s.new.push(v) } // NextGen starts a new stack generation. func (s *GenStack[T]) NextGen() { if s.old == s.new { s.new = &stack[T]{} return } // We need to pop from the old stack to the top of the new stack. Let's // have an example: // // Old: 4 3 2 1 // New: 8 7 6 5 // PopOrder: 1 2 3 4 5 6 7 8 // // // To preserve pop order, we have to take all elements from the old // stack and push them to the top of new stack: // // New: 8 7 6 5 4 3 2 1 // s.new.push(s.old.takeAll()...) // We have the old stack allocated and empty, so why not to reuse it as // new new stack. s.old, s.new = s.new, s.old } // Len returns number of elements in the stack. func (s *GenStack[T]) Len() int { l := s.old.len() if s.old != s.new { l += s.new.len() } return l }