R – Lua garbage collection of Tables, nested Tables

lualua-tablememory-management

[I've read the Lua manual, but it did not provide solid answers.]

Let's say I have a Lua Table, acting as an indexed array:

local myArray = {};
myArray[1] = "Foo";
myArray[2] = "Bar";

How do I best dispose of this Table? Do I just set myArray to nil? Or do I have to iterate through array and set each indexed element to nil?

Similarly, let's say I have I have a Lua Table, acting as a dictionary:

local myDictionary = {};
myDictionary["key1"] = "Foo";
myDictionary["key2"] = "Bar";

Can I just set 'myDictionary' to nil, or do I have to iterate through?

Lastly, what do I do, memory-management wise, where I have nested Tables? e.g.

local myNestedCollection = {};
myNestedCollection[1] = {1, 2, 3};
myNestedCollection[2] = {4, 5, 6};

Do I need to iterate through each of these sub-tables, setting them to nil? Thanks for any help.

Best Answer

In most GC an object will be collected when there are no references to it. Setting the top of your reference chain to nil removes a reference to the children. If that was the only reference then the children will be collected.

Related Solutions

Why do Lua arrays(tables) start at 1 instead of 0

Lua is descended from Sol, a language designed for petroleum engineers with no formal training in computer programming. People not trained in computing think it is damned weird to start counting at zero. By adopting 1-based array and string indexing, the Lua designers avoided confounding the expectations of their first clients and sponsors.

Although I too found them weird at the beginning, I have learned to love 0-based arrays. But I get by OK with Lua's 1-based arrays, especially by using Lua's generic for loop and the ipairs operator—I can usually avoid worrying about just how arrays are indexed.

Reverse string.find() or string.gmatch in Lua

You could use

local pattern = ".*abc '([^']+)'"

The .* is greedy so it chews up as much as it can before it matches (in this case, it chews up all the earlier matches and gives you the last).

Or if you really wanted, you could reverse your string and (sort of) your pattern too, but I think it's better to rely on the greedy .* :P

pattern = "'([^']+)' cba"
print(text:reverse():gmatch(pattern)())           -- rab
print(text:reverse():gmatch(pattern)():reverse()) -- bar

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