List<Throwable> exceptions = []
    Throwable exception = exceptions[0]
    List<TestFailure> failures
    long startTime = 0
    long endTime = startTime + 100
    long testCount = 1
    long successfulTestCount = 1
    long failedTestCount = 0
    long skippedTestCount = 0

    SimpleTestResult(long endTime = 100) {
        this.endTime = endTime
    }

        regOBD("_id");
        return this;
    }

    public void setEndTime_Equal(Long endTime) {
        setEndTime_Term(endTime, null);
    }

    public void setEndTime_Equal(Long endTime, ConditionOptionCall<TermQueryBuilder> opLambda) {
        setEndTime_Term(endTime, opLambda);
    }

    public void setEndTime_Term(Long endTime) {
        setEndTime_Term(endTime, null);
    }

    protected final long allRecordCount;
    protected final String allRecordCountRelation;
    protected final long queryTime;
    protected final boolean partialResults;
    protected final FacetResponse facetResponse;

    SearchResult(final List<Map<String, Object>> documentList, final long allRecordCount, final String allRecordCountRelation,

    }

    public static String getMemoryUsageLog() {
        final Runtime runtime = Runtime.getRuntime();
        final long freeBytes = runtime.freeMemory();
        final long maxBytes = runtime.maxMemory();
        final long totalBytes = runtime.totalMemory();
        final long usedBytes = totalBytes - freeBytes;

        return this;
    }

    public void setGidNumber_Equal(Long gidNumber) {
        setGidNumber_Term(gidNumber, null);
    }

    public void setGidNumber_Equal(Long gidNumber, ConditionOptionCall<TermQueryBuilder> opLambda) {
        setGidNumber_Term(gidNumber, opLambda);
    }

    public void setGidNumber_Term(Long gidNumber) {
        setGidNumber_Term(gidNumber, null);
    }

multiplication, along with a few n-digit additions and subtractions. The standard
n-by-n-digit multiplication algorithm requires O(n²) time, making the overall
algorithm require time T(n) where

	T(n) = 2T(n/2) + O(n) + O(n²)

which, by the Bentley-Haken-Saxe theorem, ends up reducing to T(n) = O(n²).
This is not an improvement over regular long division.

        given:
        def registry = new DefaultSerializerRegistry()
        registry.register(Long, longSerializer)
        registry.register(Integer, intSerializer)
        registry.useJavaSerialization(StringBuilder)

        expect:
        registry.canSerialize(Long)
        registry.canSerialize(Number)
        registry.canSerialize(Object)
        !registry.canSerialize(BigDecimal)

    }

    private static class Meminfo {
        private long total = -1;
        private long available = -1;
        private long free = -1;
        private long buffers = -1;
        private long cached = -1;
        private long reclaimable = -1;
        private long mapped = -1;

        public long getTotal() {
            return total;
        }

        /*
         * Linux 4.x: MemAvailable

      public long getLongLittleEndian(byte[] array, int offset) {
        long bigEndian = theUnsafe.getLong(array, (long) offset + BYTE_ARRAY_BASE_OFFSET);
        // The hardware is big-endian, so we need to reverse the order of the bytes.
        return Long.reverseBytes(bigEndian);
      }

      @Override
      public void putLongLittleEndian(byte[] array, int offset, long value) {

public class DocList extends ArrayList<Map<String, Object>> {

    private static final long serialVersionUID = 1L;

    private long contentSize = 0;

    private long processingTime = 0;

    @Override
    public void clear() {
        super.clear();
        contentSize = 0;
        processingTime = 0;
    }

    public long getContentSize() {
        return contentSize;
    }