Endless Series Java Stream-API Performance

Question

I am observing some peculiar behavior with Java8 and the new Stream-API .

I expect that the performance of the next two statements will be the same, but it is not.

 LongStream.iterate(1, n -> n + 1).limit(5000) .anyMatch(n -> isPerfectCube((n*n*n)+((n*n)*p))); 

vs

 LongStream.iterate(1, n -> n + 1) .anyMatch(n -> isPerfectCube((n*n*n)+((n*n)*p))); 

Both operators should return true , and I would not expect any performance difference, given that they can both short circuits in the first match found. The only difference in the statements is that one is limited to the upper limit of the range of numbers for iteration, and the other is not.

Can someone explain to me why you need to work faster and use less memory than another?

Answer 1

There are some values ​​of p , where the condition is true for large values ​​of n . For example, with p = 3 condition becomes true for n = 50_331_648 . In this case, the limit of 5000 will certainly win in terms of performance, but two calculations will not return the same result.

I accidentally chose p (3002), which returns true for n less than 5000, and the results are very close (although the version with limit bit slower, probably due to the additional condition n < 5000 ).

Test results (in microseconds per call to anyMatch ):

 Benchmark Mode Samples Mean Mean error Units capSO24003674.limit avgt 5 130.165 2.663 us/op capSO24003674.noLimit avgt 5 126.876 2.440 us/op 

Benchmark code (using jmh):

 @BenchmarkMode(Mode.AverageTime) @OutputTimeUnit(TimeUnit.MICROSECONDS) @State(Scope.Thread) @Warmup(iterations = 5, time = 500, timeUnit = TimeUnit.MILLISECONDS) @Measurement(iterations = 5, time = 1000, timeUnit = TimeUnit.MILLISECONDS) @Fork(1) public class SO24003674 { private int p = 3002; @GenerateMicroBenchmark public boolean limit() { return LongStream.iterate(1, n -> n + 1).limit(5000) .anyMatch(n -> isPerfectCube((n * n * n) + ((n * n) * p))); } @GenerateMicroBenchmark public boolean noLimit() { return LongStream.iterate(1, n -> n + 1) .anyMatch(n -> isPerfectCube((n * n * n) + ((n * n) * p))); } private static boolean isPerfectCube(long n) { long tst = (long) (Math.cbrt(n) + 0.5); return tst * tst * tst == n; } }