[Sasky]

Originally Posted by circonian Yes, since your always doing: result = result and <something else> if result is already false, its always going to return false, regardless of what <something else> is. Someone please correct me if I'm misunderstanding this part, but isn't the if statement unnecessary? for _,v in pairs(filters[filterType]) do    if(v) then       result = result and v(slot)    end end Doesn't v just represent functions in the table that tell if something should be displayed or not...since v is not a boolean (correct?) then wont the: if(v) always return true. If v did not exist it never would have made it to that part of the code, it would have broken out of the for loop before that. So then couldn't he just do: for _,v in pairs(filters[filterType]) do    result = result and v(slot)    if result == false then       return result    end end

The IF would return true as long as v isn't NIL or boolean false. If it's NIL, it shouldn't even be iterated over. If it's a function it can't be false.

As far as the short circuit, it could be simplified even more if v(slot) always returns a boolean:

function(slot)
    for _,v in pairs(filters[filterType]) do
        if not v(slot) then
            return false
        end
    end
    return true
end

If you hit false, it'll carry the false throughout, and the rest of the filters aren't necessary (as merlight noted) so you can immediately exit. Similarly, this also means a true return will only happen if and only if it iterates through the whole loop. And every time it processes the whole loop it'll return true, so the running assignment is unnecessary. Not as much of an optimization as early exit, but does save some assignment operations.