Re: Fastest way to determine number is integer in plain Lua

["Javier Guerra" <javier@...>]

testing...

plain Lua:

time lua -e 'local a=1.5;local f=math.floor;for i=1,1e8 do local
t=(f(a)==a) end'

real    0m16.597s
user    0m16.161s
sys     0m0.008s

time lua -e 'local a=1.5;for i=1,1e8 do local t=(a%1==0) end'

real    0m10.110s
user    0m9.853s
sys     0m0.008s

time lua -e 'local a=1.5;for i=1,1e8 do local t=((a+2^52)-2^52==a) end'

real    0m9.944s
user    0m9.641s
sys     0m0.016s

LuaJIT 1.1.4

time luajit -e 'local a=1.5;local f=math.floor;for i=1,1e8 do local
t=(f(a)==a) end'

real    0m8.000s
user    0m7.680s
sys     0m0.000s

time luajit -e 'local a=1.5;for i=1,1e8 do local t=(a%1==0) end'

real    0m1.133s
user    0m1.116s
sys     0m0.008s

time luajit -e 'local a=1.5;for i=1,1e8 do local t=((a+2^52)-2^52==a) end'

real    0m1.874s
user    0m1.824s
sys     0m0.004s


conclusions:
- both a%1==0 and Mike's hack are much faster than math.floor()
- there's very little difference between them, Mike's is 3% faster
- math.floor() doesn't get much benefit from LuaJIT
- a%1==0 is the most improved by LuaJIT, (8.8 times!) becomes 40%
faster than Mike's hack

i'd use a%1==0, since it's the most readable

-- 
Javier