How to time interval between event in lua - lua

Question

I am writing a simple script in lua. If the LED is on, it will increment the counter by 1. If the led is off for more than 1 seconds, it reset the counter.
So how exactly do we time event like that in lua ?
This is what i have and been testing multiple ways so far.
function ReadADC1()
local adc_voltage_value = 0
adc_voltage_value = tonumber(adc.readadc()) * 2 -- 0.10 --get dec number out of this -- need to know where package adc come from
--convert to voltage
adc_voltage_value = adc_voltage_value *0.000537109375 --get V
adc_voltage_value = math.floor(adc_voltage_value *1000 +0.5) --since number is base off resolution
--print (adc_voltage_value)
return adc_voltage_value
end
-- end of readADC1() TESTED
function interval()
local counter1 =0
ledValue = ReadADC1()
if (ledValue >= OnThreshHold) then
ledStatus = 1
t1=os.time()
else
ledStatus = 0
t2 = os.time()
end
--counter1 =0
for i=1,20 do
if (ledStatus == 1) then -- if led is off for more than 1 second, reset counter = 0
counter1 = counter1 + 1
elseif ((ledStatus ==0 and ((os.difftime(os.time(),t2)/100000) > 1000))) then -- increment counter when led is on
counter1 = 0
end
end
print (counter1)
end
I know for sure the logic for interval is wrong since os.time return a huge number (i assume it in psecond instead of second).
Any suggest or solution is welcome. I tried vmstarttimer and vmstoptimmer before this but not sure how it work.
EDIT:
function counter()
local ledValue = ReadADC1()
local t1 = os.clock()
while true do
local t2 = os.clock()
local dt = t2 - t1
t1 = t2
if ((ledValue < OnThreshHold) and (dt < 1)) then -- if led is off for less than 1 second
ledCounter = ledCounter + 1
elseif ((ledValue < OnThreshHold) and (dt > 1)) then-- if led is off for more than 1 second
ledCounter = 0;
else
ledCounter = ledCounter
end
print (ledCounter)
end
end
Ultimately, it will be return ledCounter instead of print counter since I will plug the value of counter to another function that will print a message correspond to the number of counter

Answer 1

You could use os.clock which returns your programs runtime since start in seconds.
Returns an approximation of the amount in seconds of CPU time used by the program.
Source
This function could be used in this way.
local t1 = os.clock() -- begin
local t2 = os.clock() -- end
local dt = t2 - t1 -- calulate delta time
-- or looped
local t1 = os.clock() -- begin
while true do
local t2 = os.clock() -- end
local dt = t2 - t1 -- calulate delta time
t1 = t2 -- reset t1
-- use dt ...
end
-- or wait for time elapsed
-- runs until 1 second passed
local t1 = os.clock()
while (os.clock() - t1) < 1 do
-- do stuff while dt is smaller than 1
-- could even reset timer (t1) to current to
-- repeat waiting
-- t1 = os.clock() | ...
end
-- logic for your example
function counter()
local time = os.clock()
local lastLedOn = false
local counter = 0
while true do
if os.clock() - time > 1.0 then
break
end
if getLedValue() == on then -- replace
time = os.clock()
if not lastLedOn then
lastLedOn = true
counter = counter + 1
-- print(counter) | or here if you want to print repeatedly
end
end
end
print(counter)
end -- was unable to test it