/* * 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"); } }