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Table.insert() and Table.remove()? [closed]

Asked by 9 years ago

Please provide more explanation in your question. If you explain exactly what you are trying to accomplish, it will be much easier to answer your question correctly.

In a localscript.

local plr = game.Players.LocalPlayer
repeat wait() until plr.Character
local char = plr.Character

table.insert(_G.PlayersInQueue,plr.Name)

game.Players.PlayerRemoving:connect(function(player)
if player.Name == plr.Name then
table.remove(_G.PlayersInQueue,plr.Name)
end
end)

In a serverscript.

_G.PlayersInQueue = {}
1
Something to note is that there is a separate _G table for *every* Client, including the Server. adark 5487 — 9y

Locked by User#24403

This question has been locked to preserve its current state and prevent spam and unwanted comments and answers.

Why was this question closed?

1 answer

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12
Answered by 9 years ago
Edited 6 years ago

You should really be more specific about your question. However, from the code you've provided, I see the common (but false) assumption you're making about table.remove. The remove function is meant for arrays {1, 2, 3, ...} rather than dictionaries {key = value, ...}. This function will not traverse a table looking for a given value, instead, it will search for a given index. Here's a quick example:

table.remove(table, index)

table.remove({"a", "b", "c"}, 2)

The snippet above passes an array with three elements (a, b, and c) to table.remove, and requests to remove the second position in the table. This is what the 2 represents as the second argument -- the position in the array to be removed and replaced by the next element (if there is one). If you seek a function that removes an entry from an array depending on values, then you must traverse the array and compare all of them...

local function removeEntryFromArray(array, entry)
    for i = 1, #array do
        if array[i] == entry then
            table.remove(array, i)
        end
    end
end

removeEntryFromArray({"a", "b", "c"}, "b")

This is called a linear search, opposed to other searching algorithms such as a binary search. This, however, is beyond the scope of what we're talking about, so I'll leave it to you to further investigate that.

The code above accomplishes the same thing as previous example, except this time, we're choosing the entry to be removed by it's value. Notice how we're still using table.remove in the function. This is because whenever you're removing a value from an array, you want the table to remain an array. Let me explain: If we just set a random index equal to nil in an array, the numeric indices of the table are no longer consecutive (such as 1, 2, 3, and so on). Therefore, it's no longer an array.

Setting an index to nil

local t = {"a", "b", "c"}
t[2] = nil

If we used the code above in place of our removeEntryFromArray function, we would run into some problems. As I mentioned, table t is now no longer an array. Now it looks like this: t = {"a", nil, "b"}. Not good. Now you lose the ability to accurately get the length of the table with the # operator, or the table.getn function. This means you'll run into problems traversing it with a numeric for loop, since you will only iterate over the "array" part of the table.

How does table.remove behave differently, then...?

Good question. Internally, table.remove function does set the index to nil. However, it also updates all the indices in the array after the index it sent to nil and pushes them down the list to fill the gap it created. Thus, the array remains an array since all indices are in sequential order.

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