PostgreSQL数据库事务系统——Push\Pop快照

快照使用快照链栈来保存获取的多个快照。栈的元素是ActiveSnapshotElt结构体，ActiveSnapshot指针指向栈顶，OldestActiveSnapshot指向栈底。

/* Elements of the active snapshot stack.
 * Each element here accounts for exactly one active_count on SnapshotData.
 * NB: the code assumes that elements in this list are in non-increasing
 * order of as_level; also, the list must be NULL-terminated. */
typedef struct ActiveSnapshotElt {
  Snapshot  as_snap;
  int      as_level;
  struct ActiveSnapshotElt *as_next;
} ActiveSnapshotElt;
/* Top of the stack of active snapshots */
static ActiveSnapshotElt *ActiveSnapshot = NULL;
/* Bottom of the stack of active snapshots */
static ActiveSnapshotElt *OldestActiveSnapshot = NULL;

PushActiveSnapshot

PushActiveSnapshot第一步先申请ActiveSnapshotElt内存，第二步判定形参snap是否是CurrentSnapshot或者SecondarySnapshot，或者没有拷贝过，则需要将该快照拷贝到ActiveSnapshotElt的as_snap快照成员，否则直接将snap赋值给ActiveSnapshotElt的as_snap快照成员。

/* PushActiveSnapshot Set the given snapshot as the current active snapshot
 * If the passed snapshot is a statically-allocated one, or it is possibly
 * subject to a future command counter update, create a new long-lived copy
 * with active refcount=1.  Otherwise, only increment the refcount. */
void PushActiveSnapshot(Snapshot snap) {
  ActiveSnapshotElt *newactive;
  Assert(snap != InvalidSnapshot);
  newactive = MemoryContextAlloc(TopTransactionContext, sizeof(ActiveSnapshotElt));
  /* Checking SecondarySnapshot is probably useless here, but it seems better to be sure. */
  if (snap == CurrentSnapshot || snap == SecondarySnapshot || !snap->copied)
    newactive->as_snap = CopySnapshot(snap);
  else
    newactive->as_snap = snap;

  newactive->as_next = ActiveSnapshot;
  newactive->as_level = GetCurrentTransactionNestLevel();
  newactive->as_snap->active_count++;

  ActiveSnapshot = newactive;
  if (OldestActiveSnapshot == NULL)
    OldestActiveSnapshot = ActiveSnapshot;
}

第三步将ActiveSnapshotElt的as_next指向当前栈顶，ActiveSnapshotElt作为当前快照的栈顶，为as_level赋值为GetCurrentTransactionNestLevel()，将快照的active_count自增。（If the passed snapshot is a statically-allocated one, or it is possibly subject to a future command counter update, create a new long-lived copy with active refcount=1. Otherwise, only increment the refcount.）

PopActiveSnapshot

从快照链栈中删除栈顶的元素，并减少快照引用计数，如果引用计数为零，需要释放该栈顶快照。

/* PopActiveSnapshot
 * Remove the topmost snapshot from the active snapshot stack, decrementing the
 * reference count, and free it if this was the last reference. */
void PopActiveSnapshot(void) {
  ActiveSnapshotElt *newstack;
  newstack = ActiveSnapshot->as_next;
  Assert(ActiveSnapshot->as_snap->active_count > 0);
  ActiveSnapshot->as_snap->active_count--;
  if (ActiveSnapshot->as_snap->active_count == 0 && ActiveSnapshot->as_snap->regd_count == 0)
    FreeSnapshot(ActiveSnapshot->as_snap);
  pfree(ActiveSnapshot);
  ActiveSnapshot = newstack;
  if (ActiveSnapshot == NULL) OldestActiveSnapshot = NULL;
  SnapshotResetXmin();
}