}
                }
            }
        }

        if (dr == null && path != null) {
            /* We did not match a domain based root. Now try to match the
             * longest path in the list of stand-alone referrals.
             */
            if (referrals != null && now > referrals.expiration) {
                referrals = null;
            }
            if (referrals == null) {

    // waited and assert based on that time. Empirically, the race where the thread ends up waiting
    // for 5.5 seconds happens about 2% of the time.
    boolean longWait = TimeUnit.NANOSECONDS.toSeconds(thread.timeSpentBlocked) >= 5;
    // Count how long it actually took to return; we'll accept any number between the expected delay

=== "Python 3.10 and above"

    ```Python hl_lines="18-19"
    {!> ../../../docs_src/path_operation_configuration/tutorial003_py310.py!}
    ```

## Descrição do docstring

import com.google.common.base.Objects;
import com.google.common.base.Predicate;
import com.google.common.hash.BloomFilterStrategies.LockFreeBitArray;
import com.google.common.math.DoubleMath;
import com.google.common.math.LongMath;
import com.google.common.primitives.SignedBytes;
import com.google.common.primitives.UnsignedBytes;
import com.google.errorprone.annotations.CanIgnoreReturnValue;
import java.io.DataInputStream;

    // Start by looking for exact matches. We start at the leftmost label. For example, foo.bar.com
    // will look like: [foo, bar, com], [bar, com], [com]. The longest matching rule wins.
    var exactMatch: String? = null
    for (i in domainLabelsUtf8Bytes.indices) {
      val rule = publicSuffixListBytes.binarySearch(domainLabelsUtf8Bytes, i)
      if (rule != null) {

As chaves do `dict` identificam cada exemplo, e cada valor é outro `dict`.

Cada `dict` de exemplo específico em `examples` pode conter:

* `summary`: Pequena descrição do exemplo.
* `description`: Uma descrição longa que pode conter texto em Markdown.
* `value`: O próprio exemplo mostrado, ex: um `dict`.

      // non-negative int, we can do long-arithmetic on index * (dataset.length - 1) / scale to get
      // a rounded ratio and a remainder which can be expressed as ints, without risk of overflow:
      int quotient = (int) LongMath.divide(numerator, scale, RoundingMode.DOWN);
      int remainder = (int) (numerator - (long) quotient * scale);
      selectInPlace(quotient, dataset, 0, dataset.length - 1);
      if (remainder == 0) {

/**
 * A bounded {@linkplain BlockingQueue blocking queue} backed by an array. This queue orders
 * elements FIFO (first-in-first-out). The head of the queue is that element that has been
 * on the queue the longest time. The tail of the queue is that element that has been on
 * the queue the shortest time. New elements are inserted at the tail of the queue, and the queue
 * retrieval operations obtain elements at the head of the queue.

            log.trace("No match for domain based root, checking standalone " + domain);
        }
        /*
         * We did not match a domain based root. Now try to match the
         * longest path in the list of stand-alone referrals.
         */

        CacheEntry<DfsReferralDataInternal> refs;
        synchronized ( this.referralsLock ) {
            refs = this.referrals;

/**
 * A bounded {@linkplain BlockingQueue blocking queue} backed by an array. This queue orders
 * elements FIFO (first-in-first-out). The head of the queue is that element that has been
 * on the queue the longest time. The tail of the queue is that element that has been on
 * the queue the shortest time. New elements are inserted at the tail of the queue, and the queue
 * retrieval operations obtain elements at the head of the queue.