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