for(i=1;i<list.size();i++){
//do something
//For Eg: move marker to a new position on the map
}
I want the above loop to complete all the iterations irrespective of the size of the list and also want the entire task to run for 1 minute. (60 seconds)
You can, for example, use System.nanoTime() to measure the duration of your loop, and then use TimeUnit.NANOSECONDS.sleep(...) to make it wait for the rest of time like this:
long start = System.nanoTime();
long desiredDuration = 60 * 1000 * 1000;
// your loop goes here
long duration = System.nanoTime() - start;
if (duration < desiredDuration)
TimeUnit.NANOSECONDS.sleep(desiredDuration - duration);
I don't really know if this is what you want but I hope this helps.
import java.util.concurrent.TimeUnit;
for(i=1;i<list.size();i++){
try {
TimeUnit.SECONDS.sleep(1);
} catch (InterruptedException e) {
e.printStackTrace();
}
// Execute thing you want to be executed every second
}
As explanation: you iterate through the for loop and the thread waits for one second before executing the code after the TimeUnit.SECONDS.sleep(1);.
If the list's size is 60 it would therefore take a minute for the loop to end.
Edit: It has occurred to me that it might be smarter to do a try-catch around the sleep function.
Check out this.
long start = System.currentTimeMillis();
long end = start + 60*1000; // 60 seconds * 1000 ms/sec
int i = 0;
while (System.currentTimeMillis() < end)
{
// do something, iterate your list
i++;
if (i == list.size()) { // check size of the list if iteration is completed
// if time has not yet expired, sleep for the rest of the time
Thread.sleep(end - System.currentTimeMillis());
}
}
Do not forget checking size of the list.
The best possible solution is to compute the desired time first and then run the loop to that extent.
long finish=System.currentTimeMillis() + 60000;
while(System.currentTimeMillis() != finish)
{
//statements;
//statements;
}
If you are trying to equip the CPU and keep it idle for this time the process is known as busy waiting but is not considered convenient in many cases so i recommend to use Thread.sleep(duration) for this purpose.
Would like to receive further queries from your side.
To spread N amount of invocations uniformly across a minute, you'll have to set the delay in between the invocations to the value 60/(N-1). The -1 is optional but causes the first and last invocations to be exactly 60 seconds apart. (just like how a ladder with N rungs has N-1 spaces)
Of course, using sleep() with the number calculated above is not only subject to round-off errors, but also drift, because you do stuff between the delays, and that stuff also takes time.
A more accurate solution is to subtract the time at which each invocation should occur (defined by startTime + 60*i/(N-1)) from the current time. Reorder and reformulate those formulas and you can subtract the 'time that should have elapsed for the next invocation' from the already elapsed time.
Of course 'elapsed time' should be calculated using System.nanoTime() and not System.currentTimeMillis() as the latter can jump when the clock changes or the computer resumes from stand-by.
For this example I changed 60 seconds to 6 seconds so you can more easily see what's going on when you run it.
public static void main(String... args) throws Exception {
int duration = 6; // seconds
List<Double> list = IntStream.range(0, 10).mapToDouble(i->ThreadLocalRandom.current().nextDouble()).boxed().collect(Collectors.toList());
long startTime = System.nanoTime();
long elapsed = 0;
for (int i = 0; i < list.size(); i++) { // Bug fixed: start at 0, not at 1.
if (i > 0) {
long nextInvocation = TimeUnit.NANOSECONDS.convert(duration, TimeUnit.SECONDS) * i / (list.size() - 1);
long sleepAmount = nextInvocation - elapsed;
TimeUnit.NANOSECONDS.sleep(sleepAmount);
}
elapsed = System.nanoTime() - startTime;
doSomething(elapsed, list.get(i));
}
}
private static void doSomething(long elapsedNanos, Double d) {
System.out.println(elapsedNanos / 1.0e9f + "\t" + d);
}
Of course when the task you preform per list element takes longer than 60/(N-1) seconds, you get contention and the 'elapsed time' deadlines are always exceeded. With this algorithm the total time just taking longer than a mnute. However if some earlier invocations exceed the deadline, and later invocations take much less time than 60/(N-1), this algorithm will show 'catch-up' behavior. This can be partially solved by sleeping at least a minimum amount even when sleepAmount is less.
