推词典？如何在Lua中实现这一目标？

说我在Lua有这本字典

places = {dest1 = 10, dest2 = 20, dest3 = 30}

在我的程序中,我检查字典是否符合我在这种情况下的大小限制3,如何从字典中推出最旧的键/值对并添加一个新的？

places["newdest"] = 50

--places should now look like this, dest3 pushed off and newdest added and dictionary has kept its size

places = {newdest = 50, dest1 = 10, dest2 = 20}

如果你真的需要它,这并不难做到,而且它也很容易重复使用.

local function ld_next(t, i) -- This is an ordered iterator, oldest first.
  if i <= #t then
    return i + 1, t[i], t[t[i]]
  end
end

local limited_dict = {__newindex = function(t,k,v)
  if #t == t[0] then -- Pop the last entry.
    t[table.remove(t, 1)] = nil
  end
  table.insert(t, k)
  rawset(t, k, v)
end, __pairs = function(t)
  return ld_next, t, 1
end}

local t = setmetatable({[0] = 3}, limited_dict)

t['dest1'] = 10
t['dest2'] = 20
t['dest3'] = 30
t['dest4'] = 50

for i, k, v in pairs(t) do print(k, v) end

dest2   20
dest3   30
dest4   50

订单存储在数字索引中,第0个索引指示表可以具有的唯一键的限制.