Re: Calculating estimate for size of a lua table

[Xmilia Hermit <xmilia.hermit@...>]

Lars Müller:

>       
>       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")
>     
    You should look out since as you noted the length of the array and
    hash part are power of two's, so in the worst case this can give
    double the cost when using random lengths. So using power of two's
    instead of 1e7 would be better.

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.

    I would argue that this is not surprising as LuaJIT uses NaN boxing
    which allows to store all values in 8 bytes instead of the 16 used
    by PUC Lua which is reflected in your benchmarking.
    
    Regards,
    Xmilia