Re: Calculating estimate for size of a lua table

[Lars Müller <appgurulars@...>]

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
>