if (field == null) {
            return null;
        }
        if (clazz.equals(Integer.class)) {
            final Number value = (Number) field;
            return (T) Integer.valueOf(value.intValue());
        }
        if (clazz.equals(Long.class)) {
            final Number value = (Number) field;
            return (T) Long.valueOf(value.longValue());
        }
        return (T) field;
    }

    /**

     */
    public Smb2WriteResponse(final Configuration config) {
        super(config);
    }

    /**
     * Returns the number of bytes written.
     *
     * @return the count
     */
    public final int getCount() {
        return this.count;
    }

    /**
     * Returns the number of bytes remaining to be written.
     *
     * @return the remaining
     */
    public final int getRemaining() {

   * provides more control over the underlying data structure.
   *
   * @param expectedKeys the expected number of distinct keys
   * @param expectedValuesPerKey the expected average number of values per key
   * @throws IllegalArgumentException if {@code expectedKeys} or {@code expectedValuesPerKey} is
   *     negative
   */

			}
		})
	}
}

func getHexSequences(start int, number int, paddinglen int) (seq []string) {
	for i := start; i <= number; i++ {
		if paddinglen == 0 {
			seq = append(seq, fmt.Sprintf("%x", i))
		} else {
			seq = append(seq, fmt.Sprintf(fmt.Sprintf("%%0%dx", paddinglen), i))
		}
	}
	return seq
}

func getSequences(start int, number int, paddinglen int) (seq []string) {

      }
      return array;
    }

    /** Number of bits */
    long bitSize() {
      return (long) data.length() * Long.SIZE;
    }

    /**
     * Number of set bits (1s).
     *
     * <p>Note that because of concurrent set calls and uses of atomics, this bitCount is a (very)
     * close *estimate* of the actual number of bits set. It's not possible to do better than an

with 4 drives we get a total of 256 possible sets. So algorithm automatically chooses 64 sets, which is *16* 64 = 1024* drives in total.

- *If total number of nodes are of odd number then GCD algorithm provides affinity towards odd number erasure sets to provide for uniform distribution across nodes*. This is to ensure that same number of drives are pariticipating in any erasure set. For example if you have 2 nodes with 180 drives then GCD is 15 but this would lead to uneven distribution,...

		}
	}
}

// Tests maximum allowed part number.
func TestMaxPartID(t *testing.T) {
	sizes := []struct {
		isMax bool
		partN int
	}{
		// Test - 1 part number within max part number.
		{
			false,
			globalMaxPartID - 1,
		},
		// Test - 2 part number bigger than max part number.
		{
			true,
			globalMaxPartID + 1,
		},
	}

	for i, s := range sizes {

                    entity.put(fessConfig.getIndexFieldId(), newId);
                    entity.remove(fessConfig.getIndexFieldVersion());
                    final Number seqNo = (Number) entity.remove(fessConfig.getIndexFieldSeqNo());
                    final Number primaryTerm = (Number) entity.remove(fessConfig.getIndexFieldPrimaryTerm());
                    if (seqNo != null && primaryTerm != null && oldId != null) {

                                    <div class="form-inline col-sm-9">
                                        <la:errors property="depth"/>
                                        <input type="number" name="depth" id="depth"
                                               value="${f:h(depth)}" class="form-control"
                                               min="0" max="10000">
                                    </div>

    /**
     * Channel binding is required
     */
    public static final int CHANNEL_BINDING_REQUIRED = 2;

    /**
     * Default maximum number of channels per session
     */
    public static final int DEFAULT_MAX_CHANNELS = 4;

    /**
     * Absolute maximum number of channels supported
     */
    public static final int ABSOLUTE_MAX_CHANNELS = 32;

    /**
     * Network interface capability flag for RSS support