1. timing tests
Hi everybody,
Try the following test for a bit of an eye opener. Especially the last two divis
ion
results are quite startling! I keep a printout handy. Jiri
-- snip ------------------------------------------------------------------------
include machine.e -- tick_rate()
object x
sequence s
atom t,dt,dt0
integer j
tick_rate(100)
function abs(atom x)
if x<0 then x=-x end if
return x
end function
puts(1,"1,000,000 loop test\n")
t=time()
for i=1 to 1000000 do
-- empty for loop
end for
dt0=time()-t
printf(1,"empty for loops: %5.2f\n",dt0)
t=time()
j=0
while j<1000000 do
j=j+1
end while
dt=time()-t
printf(1,"empty while loops: %5.2f\n",dt)
t=time()
for i=1 to 1000000 do
x=3
end for
dt=time()-t
printf(1,"assignment: x=3 %5.2f\n",dt-dt0)
t=time()
for i=1 to 1000000 do
x=3.5
end for
dt=time()-t
printf(1,"assignment: x=3.5 %5.2f\n",dt-dt0)
s={1,2,3}
t=time()
for i=1 to 1000000 do
x=s[2]
end for
dt=time()-t
printf(1,"assignment: x=s[2] %5.2f\n",dt-dt0)
s={{1,2,3},{1,2,3},{1,2,3}}
t=time()
for i=1 to 1000000 do
x=s[2][2]
end for
dt=time()-t
printf(1,"assignment: x=s[2][2] %5.2f\n",dt-dt0)
s={1,2,3}
t=time()
for i=1 to 1000000 do
s[2]=3
end for
dt=time()-t
printf(1,"assignment: s[2]=3 %5.2f\n",dt-dt0)
s={{1,2,3},{1,2,3},{1,2,3}}
t=time()
for i=1 to 1000000 do
s[2][2]=3
end for
dt=time()-t
printf(1,"assignment: s[2][2]=3 %5.2f\n",dt-dt0)
t=time()
for i=1 to 1000000 do
x=2+2
end for
dt=time()-t
printf(1,"addition: x=2+2 %5.2f\n",dt-dt0)
t=time()
for i=1 to 1000000 do
x=1.5+1.5
end for
dt=time()-t
printf(1,"addition: x=1.5+1.5 %5.2f\n",dt-dt0)
t=time()
for i=1 to 1000000 do
x=5*2
end for
dt=time()-t
printf(1,"multiplication: x=5*2 %5.2f\n",dt-dt0)
t=time()
for i=1 to 1000000 do
x=5*3
end for
dt=time()-t
printf(1,"multiplication: x=5*3 %5.2f\n",dt-dt0)
t=time()
for i=1 to 1000000 do
x=2.5*1.5
end for
dt=time()-t
printf(1,"multiplication: x=2.5*1.5 %5.2f\n",dt-dt0)
t=time()
for i=1 to 1000000 do
x=256/2
end for
dt=time()-t
printf(1,"division: x=256/2 %5.2f\n",dt-dt0)
t=time()
for i=1 to 1000000 do
x=256/4
end for
dt=time()-t
printf(1,"division: x=256/4 %5.2f\n",dt-dt0)
for i=1 to 1000000 do
x=5/2
end for
dt=time()-t
printf(1,"division: x=5/2 %5.2f\n",dt-dt0)
t=time()
for i=1 to 1000000 do
x=5/3
end for
dt=time()-t
printf(1,"division: x=5/3 %5.2f\n",dt-dt0)
t=time()
for i=1 to 1000000 do
x=2.5/1.5
end for
dt=time()-t
printf(1,"division: x=2.5/1.5 %5.2f\n",dt-dt0)
t=time()
for i=1 to 1000000 do
x=floor(3.5)
end for
dt=time()-t
printf(1,"floor: x=floor(3.5) %5.2f\n",dt-dt0)
t=time()
for i=1 to 1000000 do
x=remainder(5,3)
end for
dt=time()-t
printf(1,"remainder: x=remainder(5,3) %5.2f\n",dt-dt0)
t=time()
for i=1 to 1000000 do
x=remainder(5.5,3)
end for
dt=time()-t
printf(1,"remainder: x=remainder(5.5,3) %5.2f\n",dt-dt0)
t=time()
for i=1 to 1000000 do
x=abs(-5)
end for
dt=time()-t
printf(1,"abs: x=abs(-5) %5.2f\n",dt-dt0)
t=time()
for i=1 to 1000000 do
x=abs(-5.5)
end for
dt=time()-t
printf(1,"abs: x=abs(-5.5) %5.2f\n",dt-dt0)
2. Re: timing tests
> Hi everybody,
> Try the following test for a bit of an eye opener. Especially the last two div
is
> ion
> results are quite startling! I keep a printout handy. Jiri
Yes, some things were very supprising..
I added a few tests.. they also show some very weird results..
For example, calling a function, that's a huge slowdown.. so only
have functions for large pieces of code, that you have to call at a
lot of different places in you code.
And this:
puts(1,"Euphoria rules!!")
Is a LOT slower than:
sequence s
s = "Euphoria rules!!"
puts(1,s)
Also the type of object you want to assign a value to determines
speed. Assigning to an object is faster than to a atom if the value
is integer.... (?? That's weird..)
The function sabs included... is the same as abs.. but handles all
kinds of objects... integers, atoms, sequences until any depth.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
~~~~>> Ralf Nieuwenhuijsen <<~~~~
~~~>> nieuwen at xs4all.nl <<~~~~
3. Re: timing tests
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Oops, like always i forgot to attach the file.. here it is..
I am really sorry, i don't want to clutter up the list-serv.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
~~~~>> Ralf Nieuwenhuijsen <<~~~~
~~~>> nieuwen at xs4all.nl <<~~~~
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4. Re: timing tests
- Posted by Jacques Deschenes <desja at GLOBETROTTER.QC.CA>
Aug 07, 1997
-
Last edited Aug 08, 1997
At 14:54 07-08-97 +1200, Jiri wrote:
>Hi everybody,
>Try the following test for a bit of an eye opener. Especially the last two
divis
>ion
>results are quite startling! I keep a printout handy. Jiri
>
Talking about timing test try this one.
------------------- code begin here ---------------------------------------
atom t,dt0, dt1
constant format = "%03x %03x %03x "
t = time()
for i = 1 to 10000 do
printf(1,format,{23,46,25})
end for
dt0 = time() - t
clear_screen()
t = time()
for i = 1 to 10000 do
puts(1,sprintf(format,{23,46,25}))
end for
dt1 = time() - t
printf(1,"using printf() %5.2f\n",{dt0})
printf(1,"using puts(1,sprintf()) %5.2f\n",{dt1})
-------------- code end ------------------------------------------------
On my computer it's more than 2.4 times faster to use puts(1,sprintf())
Jacques Deschenes
Baie-Comeau, Quebec
Canada
desja at globetrotter.qc.ca
