timing a function / tasks in lua - timer

Question

I'm trying to determine how long certain statements take to run in my lua code.
My code looks like this:
function test(self)
local timer1
local timer2
local timer3
timer1 = os.time()
print('timer1 start time is:'.. timer1)
--do some stuff.
print( 'Timer1 end time is:' , os.difftime(os.time(), timer1) )
timer2 = os.time()
print('timer2 start time is:'.. timer2)
-- do a lot of stuff
print( 'Timer2 end time is:' , os.difftime(os.time(), timer2) )
timer3=os.time()
print('timer3 start time is:'.. timer3)
-- a lot of processing...
print( 'Timer3 end time is:' , os.difftime(os.time(), timer3) )
end
This is what the output looks like:
timer1 start time is:1401798084
Timer1 end time is: = 0
timer2 start time is:1401798084
Timer2 end time is: = 0
timer3 start time is:1401798084
Timer3 end time is: = 2
Other things I've tried:
Lua - Current time in milliseconds
In the above post, I found this snippet of code:
local x = os.clock()
local s = 0
for i=1,100000 do s = s + i end
os.execute("sleep "..1)
print(string.format("elapsed time: %.2f\n", os.clock() - x))
I added the sleep time... but when it runs, I get the output:
elapsed time: 0.00
I'm clearly doing something wrong. If you have suggestions on how I can fix / improve this, I'm all ears. In the interim, I'm going to revisit the lua site to read up on os.difftime() in case I'm using it incorrectly.
EDIT 1
I changed the test code to look like this:
local x = os.clock()
local s = 0
for i=1,100000 do
s = s + i
os.execute("sleep "..1)
end
print(string.format("elapsed time: %.2f\n", os.clock() - x))
and now I'm getting some values that make sense!

Answer 1

os.clock measures CPU time, not wall time. CPU time does not include time spent in sleep. So the script below still prints zero elapsed time:
local x = os.clock()
os.execute("sleep 60")
print(string.format("elapsed time: %.2f\n", os.clock() - x))
When you move os.execute into the loop, what you're probably measuring is the time to fork a shell. The script below print nonzero elapsed time, even if it is a short loop:
local x = os.clock()
for i=1,1000 do os.execute("true") end
print(string.format("elapsed time: %.2f\n", os.clock() - x))
Finally, you got zero elapsed time in the first loop because Lua is fast. Try changing the limit to 1000000:
local x = os.clock()
local s = 0
for i=1,1000000 do s = s + i end
print(string.format("elapsed time: %.2f\n", os.clock() - x))

Answer 2

This snippet does many rounds of addition and then executes one sleep call for one second.
for i=1,100000 do s = s + i end
os.execute("sleep "..1)
This snippet does the same amount of addition but sleeps for one second each time through the loop.
for i=1,100000 do
s = s + i
os.execute("sleep "..1)
end
That is a big difference.