Pausing with sleep for 1/100 of a second

So I wanted have a script pause for 1/100 of a second using sleep, but it seemed to have a bit of overhead.

A little bash script to test it says that I should sleep for about 0.0084 seconds to get an effective 0.01 second delay.

It's a little tricky to get those bash one-liners right with ifs and having to put semicolons in just the right places to parse correctly.


This is with Ubuntu 17.10.


Also, why don't we measure how much overhead there actually is for a "sleep 0" command.

A lua5.3 one-liner is:

sleeptime = 0 iterations = 1000 print("sleeptime = "..sleeptime.." iterations = "..iterations) function popenread(executestr) local p=io.popen(executestr,"r") local retval=p:read() p:close() return retval end time0secs=popenread("date +%s.%N") for i = 1,iterations do os.execute("sleep "..sleeptime) end time1secs=popenread("date +%s.%N") print(time0secs) print(time1secs) timeperiteration = ((time1secs-time0secs)/iterations) print("timeperiteration = "..timeperiteration,"iterations per second = "..1/timeperiteration,"time0seconds = "..time0secs,"time1seconds = "..time1secs)

sleeptime = 0 iterations = 1000
1520255731.296324963
1520255733.145632126
timeperiteration = 0.0018493070602417 iterations per second = 540.743082368 time0seconds = 1520255731.296324963 time1seconds = 1520255733.145632126
 

so we can see about .0018 of overhead, so if we want .01 delay, we'd sleep for approx.01-.0018 = .0082 seconds.  .0078 seems to be about the right value.


sleeptime = 0.0078 iterations = 1000 print("sleeptime = "..sleeptime.." iterations = "..iterations) function popenread(executestr) local p=io.popen(executestr,"r") local retval=p:read() p:close() return retval end time0secs=popenread("date +%s.%N") for i = 1,iterations do os.execute("sleep "..sleeptime) end time1secs=popenread("date +%s.%N") print(time0secs) print(time1secs) timeperiteration = ((time1secs-time0secs)/iterations) print("timeperiteration = "..timeperiteration,"iterations per second = "..1/timeperiteration,"time0seconds = "..time0secs,"time1seconds = "..time1secs)

sleeptime = 0.0078 iterations = 1000
1520256433.576907853
1520256443.544411099
timeperiteration = 0.0099675033092499 iterations per second = 100.32602638536 time0seconds = 1520256433.576907853 time1seconds = 1520256443.544411099

and why don't we try it with bash with a sleep 0:

a=`date +%s.%N`;for i in `seq 1 1000`; do sleep 0; done; b=`date +%s.%N`;echo $a; echo $b;echo scale = 9 \; \( $b - $a \) / 1000  | bc; echo scale = 9 \; 1 / \( \( $b - $a \) / 1000 \) | bc
1520255846.353004034
1520255847.453091183
.001100087
909.019013950

so a sleep 0 will be about 909 cycles per second

a=`date +%s.%N`;for i in `seq 1 1000`; do sleep 0.0084; done; b=`date +%s.%N`;echo $a; echo $b;echo scale = 9 \; \( $b - $a \) / 1000  | bc; echo scale = 9 \; 1 / \( \( $b - $a \) / 1000 \) | bc
1520256098.974860366
1520256108.944800873
.009969940
100.301506328

and a sleep 0.0084 gives us about 100 cycles per second