Nelua slower than luajit in this little benchmark?

[fenix272]

`require 'os'
local function x()
local j = 0
for i=2.0, 1000000000.0 do
j = i
end
return j
end

x()

local t1 = os.clock()
local y = x();
local t2 = os.clock()
print(y)

--assert(y == 1000000000.0)
print("time taken ", t2-t1);`

Results:
Nelua 4.18
Luajit 0.61
If i change the numbers to integer in the x function Nelua improves getting 1.65 whereas luajit is still about 0.61
Why is nelua slower?

[jrfondren]

With those times you're probably comparing nelua bench.lua vs. luajit bench.lua. The first step includes a lengthy compilation process in the time. Try nelua -M -o bench bench.nelua and then comparing the times of ./bench and luajit bench.lua. With floats luajit's still 3 times faster. With integers Nelua's 880 times faster. I imagine that the speed advantage with floats is that luajit (due to the JIT, and due to Lua 5.1's weak integer support making this a very important optimization) notices at runtime that it doesn't need to use floating point, and uses integers instead.

[fenix272]

No. First i compiled the program and then i lauched the exe file. But either way the result is the same, 4.2

[jrfondren]

Right, you're comparing os.clock() in the running program, so you're excluding startup time. That's quite significant though:

$ hyperfine './bench' 'luajit bench.nelua'
Summary
  luajit bench.nelua ran
    3.00 ± 0.04 times faster than ./bench

$ hyperfine 'nelua bench.nelua' 'luajit bench.nelua'
Summary
  luajit bench.nelua ran
   12.78 ± 0.13 times faster than nelua bench.nelua

[0komo]

Try with highest optimization C flags.