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2013/9/27 Rafis Ganeyev <>:

> Another gotcha (Lua so hard to use :( )

It becomes easier as you go along, particularly if you spend
some time reading the manual and "Programming in Lua".

> If you are deleting elements from collection using *pairs()*, you can't
> delete element with *table.remove()*, you must delete only element that
> *pairs()* pointed to you:
> local t = setmetatable({ 1, 2 }, mt)
> for i, obj in pairs(t) do
>   obj:teardown()
>   table.remove(t)
> end
> print(#t) -- prints 1
> This will not delete element t[1].

Of course it won't. You have programmed the following:

while true do
  if t[i] does not exist, break
  remove t[i] and move down the others

You can delete any element you like, but you must remember
that 'remove' changes the association between elements and