public void flip(int fromIndex, int toIndex) { checkRange(fromIndex, toIndex); if (fromIndex == toIndex) return; int startWordIndex = wordIndex(fromIndex); int endWordIndex = wordIndex(toIndex - 1); expandTo(endWordIndex); long firstWordMask = WORD_MASK << fromIndex; long lastWordMask = WORD_MASK >>> -toIndex; if (startWordIndex == endWordIndex) { // Case 1: One word words[startWordIndex] ^= (firstWordMask & lastWordMask); } else { // Case 2: Multiple words // Handle first word words[startWordIndex] ^= firstWordMask; // Handle intermediate words, if any for (int i = startWordIndex+1; i < endWordIndex; i++) words[i] ^= WORD_MASK; // Handle last word words[endWordIndex] ^= lastWordMask; } recalculateWordsInUse(); checkInvariants(); } 6. 设置某一指定位 （or 操作）
/**
* Sets the bit at the specified index to {@code true}.
*
* @param bitIndex a bit index
* @throws IndexOutOfBoundsException if the specified index is negative
* @since JDK1.0
*/
public void set(int bitIndex) {
if (bitIndex < 0)
throw new IndexOutOfBoundsException("bitIndex < 0: " + bitIndex);
int wordIndex = wordIndex(bitIndex);
expandTo(wordIndex);
words[wordIndex] |= (1L << bitIndex); // Restores invariants
checkInvariants();
}
思路与flip是一样的，只是执行的是与1的or操作。
同时jdk中提供了，具体设置成0或1的操作，以及设置某一区间的操作。
public void set(int bitIndex, boolean value) {
if (value)
set(bitIndex);
else
clear(bitIndex);
}
7. 获取某一位置的状态
/**
* Returns the value of the bit with the specified index. The value
* is {@code true} if the bit with the index {@code bitIndex}
* is currently set in this {@code BitSet}; otherwise, the result
* is {@code false}.
*
* @param bitIndex the bit index
* @return the value of the bit with the specified index
* @throws IndexOutOfBoundsException if the specified index is negative
*/
public boolean get(int bitIndex) {
if (bitIndex < 0)
throw new IndexOutOfBoundsException("bitIndex < 0: " + bitIndex);
checkInvariants();
int wordIndex = wordIndex(bitIndex);
return (wordIndex < wordsInUse)
&& ((words[wordIndex] & (1L << bitIndex)) != 0);
}
同样的两步走，这里的位操作时&。可以看到，如果指定的bit不存在的话，返回的是false，即没有设置。
jdk同时提供了一个获取指定区间的bitset的方法。当然这里的返回值会是一个bitset，是一个仅仅包含需要查询位的bitset。注意这里的大小也仅仅是刚刚能够容纳必须的位（当然，规整到long的整数倍）。代码如下：
public BitSet get(int fromIndex, int toIndex) {
checkRange(fromIndex, toIndex);
checkInvariants();
int len = length();
// If no set bits in range return empty bitset
if (len <= fromIndex || fromIndex == toIndex)
return new BitSet(0);
// An optimization
if (toIndex > len)
toIndex = len;
BitSet result = new BitSet(toIndex - fromIndex);
int targetWords = wordIndex(toIndex - fromIndex - 1) + 1;
int sourceIndex = wordIndex(fromIndex);
boolean wordAligned = ((fromIndex & BIT_INDEX_MASK) == 0);
// Process all words but the last word
for (int i = 0; i < targetWords - 1; i++, sourceIndex++)
result.words[i] = wordAligned ? words[sourceIndex] :
(words[sourceIndex] >>> fromIndex) |
(words[sourceIndex+1] << -fromIndex);
// Process the last word
long lastWordMask = WORD_MASK >>> -toIndex;
result.words[targetWords - 1] =
((toIndex-1) & BIT_INDEX_MASK) < (fromIndex & BIT_INDEX_MASK)
? /* straddles source words */
((words[sourceIndex] >>> fromIndex) |
(words[sourceIndex+1] & lastWordMask) << -fromIndex)
:
((words[sourceIndex] & lastWordMask) >>> fromIndex);
// Set wordsInUse correctly
result.wordsInUse = targetWords;
result.recalculateWordsInUse();
result.checkInvariants();
return result;
}
这里有一个tricky的操作，即fromIndex的那个bit会存在返回bitset的第0个位置，以此类推。如果fromIndex不是word对齐的话，那么返回的bitset的第一个word将会包含fromIndex所在word的从fromIndex开始的到fromIndex+1开始的的那几位（总共加起来是一个word的大小）。
其中>>>是无符号位想右边移位的操作符。
8. 获取当前bitset总bit的大小
/**
* Returns the "logical size" of this {@code BitSet}: the index of
* the highest set bit in the {@code BitSet} plus one. Returns zero
* if the {@code BitSet} contains no set bits.
*
* @return the logical size of this {@code BitSet}
* @since 1.2
*/
public int length() {
if (wordsInUse == 0)
return 0;
return BITS_PER_WORD * (wordsInUse - 1) +
(BITS_PER_WORD - Long.numberOfLeadingZeros(words[wordsInUse - 1]));
}
9. hashcode
hashcode是一个非常重要的属性，可以用来表明一个数据结构的特征。bitset的hashcode是用下面的方式实现的：
/**
* Returns the hash code value for this bit set. The hash code depends
* Note that the hash code changes if the set of bits is altered.
*
* @return the hash code value for this bit set
*/
public int hashCode() {
long h = 1234;
for (int i = wordsInUse; --i >= 0; )
h ^= words[i] * (i + 1);
return (int)((h >> 32) ^ h);
}
这个hashcode同时考虑了没给word以及word的位置。当有bit的状态发生变化时，hashcode会随之改变。