|
Alexander Gladysh wrote:
If you can afford to corrupt one of the tables, you can optimize out second loop (untested code): function compareTables(t1, t2) -- all key-value pairs in t1 must be in t2 for k, v in pairs(t1) do if t2[k] ~= v then return false end t2[k] = nil end return next(t2) == nil -- there must not be other keys in t2 end
Interesting idea. This time I have to do a non-destructive comparison, but it could be useful in other occasions.
(anyway, my question was more out of curiosity than for a real need of to-the-last-ns optimizazion).
Enrico