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/*
* Copyright (C) 2015 Google Inc.
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
*
* http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*/
package benchmarks;
import java.math.BigInteger;
import java.util.Random;
/**
* This pretends to measure performance of operations on small BigIntegers.
* Given our current implementation, this is really a way to measure performance of
* finalization and JNI.
* We manually determine the number of iterations so that it should cause total memory
* allocation on the order of a few hundred megabytes. Due to BigInteger's reliance on
* finalization, these may unfortunately all be kept around at once.
*/
public class SmallBigIntegerBenchmark {
// We allocate about 2 1/3 BigIntegers per iteration.
// Assuming 100 bytes/BigInteger, this gives us around 500MB total.
static final int NITERS = 2 * 1000 * 1000;
static final BigInteger BIG_THREE = BigInteger.valueOf(3);
static final BigInteger BIG_FOUR = BigInteger.valueOf(4);
public static void main(String args[]) {
final Random r = new Random();
BigInteger x = new BigInteger(20, r);
final long startNanos = System.nanoTime();
long intermediateNanos = 0;
for (int i = 0; i < NITERS; ++i) {
if (i == NITERS / 100) {
intermediateNanos = System.nanoTime();
}
// We know this converges, but the compiler doesn't.
if (x.and(BigInteger.ONE).equals(BigInteger.ONE)) {
x = x.multiply(BIG_THREE).add(BigInteger.ONE);
} else {
x = x.shiftRight(1);
}
}
if (x.signum() < 0 || x.compareTo(BIG_FOUR) > 0) {
throw new AssertionError("Something went horribly wrong.");
}
final long finalNanos = System.nanoTime();
double firstFewTime = ((double) intermediateNanos - (double) startNanos) / (NITERS / 100);
double restTime = ((double) finalNanos - (double) intermediateNanos) / (99 * NITERS / 100);
System.out.println("First Few: " + firstFewTime
+ " nanoseconds per iteration (2.33 BigInteger ops/iter)");
System.out.println("Remainder: " + restTime + " nanoseconds per iteration");
}
}