一.源码

1.构造方法

  1. private long[] words;
  3. public BitSet() {
  4.     // 64
  5.     initWords(BITS_PER_WORD);
  6. }
  8. public BitSet(int nbits) {
  9.     initWords(nbits);
  10. }
  12. private void initWords(int nbits) {
  13.     words = new long[wordIndex(nbits-1) + 1];
  14. }
  16. private static int wordIndex(int bitIndex) {
  17.     // bitIndex >> 6 == bitIndex / 64
  18.     return bitIndex >> ADDRESS_BITS_PER_WORD;
  19. }

2.set

  1. public void set(int bitIndex, boolean value) {
  2.     if (value)
  3.         set(bitIndex);
  4.     else
  5.         clear(bitIndex);
  6. }
  8. public void set(int bitIndex) {
  9.     // 数组下标
  10.     int wordIndex = wordIndex(bitIndex);
  11.     expandTo(wordIndex);
  12.     // 1L << 4 == 1L << (4 + 64) == 1L << (4 + 64 + 64)
  13.     words[wordIndex] |= (1L << bitIndex);
  14. }
  16. private void expandTo(int wordIndex) {
  17.     // 数组长度
  18.     int wordsRequired = wordIndex+1;
  19.     if (wordsInUse < wordsRequired) {
  20.         // 扩容
  21.         ensureCapacity(wordsRequired);
  22.         // 使用的数组元素个数
  23.         wordsInUse = wordsRequired;
  24.     }
  25. }
  27. private void ensureCapacity(int wordsRequired) {
  28.     if (words.length < wordsRequired) {
  29.         int request = Math.max(2 * words.length, wordsRequired);
  30.         words = Arrays.copyOf(words, request);
  31.     }
  32. }
  34. public void clear(int bitIndex) {
  35.     // 数组下标
  36.     int wordIndex = wordIndex(bitIndex);
  37.     // wordIndex下标的值为0，不需要操作
  38.     if (wordIndex >= wordsInUse)
  39.         return;
  40.     words[wordIndex] &= ~(1L << bitIndex);
  41.     recalculateWordsInUse();
  42. }
  44. private void recalculateWordsInUse() {
  45.     int i;
  46.     for (i = wordsInUse-1; i >= 0; i--)
  47.         if (words[i] != 0)
  48.             break;
  49.     wordsInUse = i+1;
  50. }

3.get

  1. public boolean get(int bitIndex) {
  2.     //数组下标
  3.     int wordIndex = wordIndex(bitIndex);
  4.     // wordIndex >= wordsInUse 直接返回false，因为wordIndex下标的值一定是0
  5.     return (wordIndex < wordsInUse)
  6.         && ((words[wordIndex] & (1L << bitIndex)) != 0);
  7. }