Consumed Time of System.out.println(); Java statement?

Question

I hear that the System.out.println(); Java statement is costly (it consumes a lot of time)
So I try to evaluate its cost:
When I evaluate 5 statements... The cost = 1.0
So I expect the cost of 1 statement = 0.2
But actually I found The cost = 0.0 !!

double t1 = 0;
double t2 = 0;

t1 = System.currentTimeMillis();
System.out.println("aa");
System.out.println("aa");
System.out.println("aa");
System.out.println("aa");
System.out.println("aa");
t2 = System.currentTimeMillis();
System.out.println("The cost = " + (t2-t1) ); 
// The cost = 1.0 


t1 = System.currentTimeMillis();
System.out.println("aa");
t2 = System.currentTimeMillis();
System.out.println("The cost = " + (t2-t1) ); 
// The cost = 0.0 
// Expected : 1.0/5 = 0.2 -- But Actual : 0.0

Why that?

You should increase the count of calls in your benchmark to something like 5000 in a loop. **Then** you probably will get a better output! — bobbel, Dec 19 '13 at 13:48
You do know that even one millisecond is damn costly as a normal cpu can handle a few million instructions in that time? — Matthias, Dec 19 '13 at 13:53

Maroun · Accepted Answer · 2013-12-19T13:56:49.870

11

System#currentTimeMillis returns long and not double. Thus you're loosing .2.

Testing 5 statements is not a good idea, specially when you almost don't feel the time it takes to perform it. I advise you to have more than 5 statements to test and then reduce the amount to something more than 1 as well.

You want to do a precise measurements time, it's better to use System#nanoTime, since it gives time in nano seconds:

long startTime = System.nanoTime();
// ... the code being measured ...
long estimatedTime = System.nanoTime() - startTime;

Look for "nanoTime vs currentTimeMillis" and you'll get hundreds of articles.

edited Dec 19 '13 at 13:56

answered Dec 19 '13 at 13:48

Maroun

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3

Would you also educate on the difference between `currentTimeMillis` and `nanoTime`? Then I can skip an own attempt at answering yet another invalid benchmarking question :) – Matthias Dec 19 '13 at 13:51

kai · Answer 2 · 2013-12-19T13:55:56.647

When i tested your code, the output is always 0.0. Pls try your benchmark with System.nanoTime(); Or do it with more System.out.println()

    double t1 = 0;
    double t2 = 0;

    t1 = System.currentTimeMillis();
    for(int i = 0; i<1000; i++)
        System.out.println("aa");
    t2 = System.currentTimeMillis();
    System.out.println("The cost = " + (t2-t1) ); 


    t1 = System.currentTimeMillis();
    for(int i = 0; i<100; i++)
        System.out.println("aa");
    t2 = System.currentTimeMillis();
    System.out.println("The cost = " + (t2-t1) );

Output: 62.0 and 5.0

And this difference of 1.2 ms is because the loop gets faster while runtime.

score 3 · Answer 3 · answered Dec 19 '13 at 13:51

If the execution time is less than one millisecond, you will see nothing, because the clock hasn't 'ticked' yet. For this type of micro benchmarking, you should use something like:

long t1 = System.nanoTime()

for your time measurements. (this measures time in nanosecond units, although not always with that much granularity)

score 1 · Answer 4 · answered Dec 19 '13 at 15:59

1

Also, the time to complete System.out.println will be different on a more or less powerful computer, and how busy that computer is performing other tasks at that particular moment.

answered Dec 19 '13 at 15:59

user2810910

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Consumed Time of System.out.println(); Java statement?

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