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• Subject: Re: Calculating estimate for size of a lua table
• From: Lars Müller <appgurulars@...>
• Date: Sat, 13 Aug 2022 21:34:47 +0200

Thanks for the correction.

I did some benchmarking using the following code:

`for i = 1, 10 do collectgarbage"collect" end; c = collectgarbage"count"; t = {}; for i = 1, 1e7 do t[i+.5] = i end; d = collectgarbage"count"; print((d - c) * 1e3 / 1e7, "bytes per entry")`

which consistently printed a size of about 39 bytes for each hash entry on both PUC Lua 5.4 and LuaJIT

For the list part I simply used "t[i] = i" in the above code. Surprisingly the list part only takes 13 bytes per entry on LuaJIT (!) vs 26 bytes per entry on Lua 5.1 and Lua 5.4.

On 13.08.22 19:32, Xmilia Hermit wrote:
Lars Müller:

I would assume the following:

For the list part: 8 bytes for each entry.
For the hash part: 8 bytes (key) + 8 bytes (value) for each entry.
It is more like 16 bytes for the list part as PUC Lua uses tagged values (8 byte value + 1 byte tag) which is rounded up to 16.
For the hast part it is likely 24. (2 * 8 byte value (for value + key) + 2 * 1 byte tag (for value + key) + 4 byte link to next element when multiple have the same hash rounded up to 24).
And the empty table has likley 64 byte of a constant cost.
So in total something like 64 + 16 * array_size + 24 * hash_size where the sizes are power of two's.

Regards,
Xmilia