/* * Copyright (c) 2013, Oracle and/or its affiliates. All rights reserved. * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER. * * This code is free software; you can redistribute it and/or modify it * under the terms of the GNU General Public License version 2 only, as * published by the Free Software Foundation. * * This code is distributed in the hope that it will be useful, but WITHOUT * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or * FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License * version 2 for more details (a copy is included in the LICENSE file that * accompanied this code). * * You should have received a copy of the GNU General Public License version * 2 along with this work; if not, write to the Free Software Foundation, * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA. * * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA * or visit www.oracle.com if you need additional information or have any * questions. */ import java.util.*; import java.util.function.*; import java.util.stream.*; /* * @test * @bug 8006572 * @summary Test for use of non-naive summation in stream-related sum and average operations. */ public class TestDoubleSumAverage { public static void main(String... args) { int failures = 0; failures += testForCompenstation(); failures += testZeroAverageOfNonEmptyStream(); if (failures > 0) { throw new RuntimeException("Found " + failures + " numerical failure(s)."); } } /** * Compute the sum and average of a sequence of double values in * various ways and report an error if naive summation is used. */ private static int testForCompenstation() { int failures = 0; /* * The exact sum of the test stream is 1 + 1e6*ulp(1.0) but a * naive summation algorithm will return 1.0 since (1.0 + * ulp(1.0)/2) will round to 1.0 again. */ double base = 1.0; double increment = Math.ulp(base)/2.0; int count = 1_000_001; double expectedSum = base + (increment * (count - 1)); double expectedAvg = expectedSum / count; // Factory for double a stream of [base, increment, ..., increment] limited to a size of count Supplier ds = () -> DoubleStream.iterate(base, e -> increment).limit(count); DoubleSummaryStatistics stats = ds.get().collect(DoubleSummaryStatistics::new, DoubleSummaryStatistics::accept, DoubleSummaryStatistics::combine); failures += compareUlpDifference(expectedSum, stats.getSum(), 3); failures += compareUlpDifference(expectedAvg, stats.getAverage(), 3); failures += compareUlpDifference(expectedSum, ds.get().sum(), 3); failures += compareUlpDifference(expectedAvg, ds.get().average().getAsDouble(), 3); failures += compareUlpDifference(expectedSum, ds.get().boxed().collect(Collectors.summingDouble(d -> d)), 3); failures += compareUlpDifference(expectedAvg, ds.get().boxed().collect(Collectors.averagingDouble(d -> d)),3); return failures; } /** * Test to verify that a non-empty stream with a zero average is non-empty. */ private static int testZeroAverageOfNonEmptyStream() { Supplier ds = () -> DoubleStream.iterate(0.0, e -> 0.0).limit(10); return compareUlpDifference(0.0, ds.get().average().getAsDouble(), 0); } /** * Compute the ulp difference of two double values and compare against an error threshold. */ private static int compareUlpDifference(double expected, double computed, double threshold) { double ulpDifference = Math.abs(expected - computed) / Math.ulp(expected); if (ulpDifference > threshold) { System.err.printf("Numerical summation error too large, %g ulps rather than %g.%n", ulpDifference, threshold); return 1; } else return 0; } }