Donc, pendant ce temps, l'ordinateur pourra effectuer d'autres tâches, pendant que "fichier-lent" 📝 se termine.

    }

    private static SpnegoToken getToken(final byte[] token, final int off, final int len) throws SpnegoException {
        byte[] b = new byte[len];
        if (off == 0 && token.length == len) {
            b = token;
        } else {
            System.arraycopy(token, off, b, 0, len);
        }
        return getToken(b);
    }

      this.elemCount = 0;
      successor(pred(), succ());
      if (left == null) {
        return right;
      } else if (right == null) {
        return left;
      } else if (left.height >= right.height) {
        AvlNode<E> newTop = pred();
        // newTop is the maximum node in my left subtree
        newTop.left = left.removeMax(newTop);
        newTop.right = right;

          return classify(Integer.compare(left, right));
        }

        @Override
        public ComparisonChain compare(long left, long right) {
          return classify(Long.compare(left, right));
        }

        @Override
        public ComparisonChain compare(float left, float right) {
          return classify(Float.compare(left, right));
        }

      this.elemCount = 0;
      successor(pred(), succ());
      if (left == null) {
        return right;
      } else if (right == null) {
        return left;
      } else if (left.height >= right.height) {
        AvlNode<E> newTop = pred();
        // newTop is the maximum node in my left subtree
        newTop.left = left.removeMax(newTop);
        newTop.right = right;

    for (int trial = 0; trial < 100; trial++) {
      byte[] left = new byte[1 + rnd.nextInt(32)];
      rnd.nextBytes(left);
      byte[] right = left.clone();
      assertThat(comparator1.compare(left, right)).isEqualTo(0);
      assertThat(comparator2.compare(left, right)).isEqualTo(0);
      int i = rnd.nextInt(left.length);
      left[i] ^= (byte) (1 + rnd.nextInt(255));

    INSTANCE;

    @Override
    public int compare(long[] left, long[] right) {
      int minLength = Math.min(left.length, right.length);
      for (int i = 0; i < minLength; i++) {
        int result = Long.compare(left[i], right[i]);
        if (result != 0) {
          return result;
        }
      }
      return left.length - right.length;
    }

    @Override
    public String toString() {

	if len(users1) != 3 {
		t.Errorf("should find two users using left join, but got %v", len(users1))
	}

	var users2 []User
	DB.Joins("inner join pets on pets.user_id = users.id AND pets.name = ?", user.Pets[0].Name).Where("users.name = ?", user.Name).First(&users2)
	if len(users2) != 1 {
		t.Errorf("should find one users using left join with conditions, but got %v", len(users2))
	}

	var users3 []User

     * entities such as devices or pipes, in which case we must fall back on comparing the bytes
     * directly.
     */
    long len1 = file1.length();
    long len2 = file2.length();
    if (len1 != 0 && len2 != 0 && len1 != len2) {
      return false;
    }
    return asByteSource(file1).contentEquals(asByteSource(file2));
  }

  /**

Então, durante esse tempo, o computador pode ir e fazer outro trabalho, enquanto o "arquivo lento" 📝 termina.