That is, they have at least one set bit in the same index: BitSet first = new BitSet() ĪssertThat(first.intersects(second)).isTrue() The intersects(BitSet)method takes another BitSet and returns true when two BitSets have something in common. It's also worth mentioning that this method will return zero if all bits are clear.įinally, the isEmpty()method returns false when there is at least one set bit in the BitSet. When we set the index 100 to true, then the length() method returns 101. After setting the [10, 30) range to true, then the cardinality() method call returns 20.Īlso, the length()method returns the one index after the index of the last set bit: assertThat(bitSet.length()).isEqualTo(30) ĪssertThat(bitSet.length()).isEqualTo(101) Īt first, the last set index is 29, so this method returns 30. Moreover, the cardinality()method represents the number of set bits in a BitSet: assertThat(bitSet.cardinality()).isEqualTo(0) ĪssertThat(bitSet.cardinality()).isEqualTo(30 - 10) Īt first, this method returns zero as all bits are false. Of course, this will change if we pass the number of bits explicitly: BitSet bitSet = new BitSet(1024) ĪssertThat(bitSet.size()).isEqualTo(1024) With one 64-bit number, we can only represent 64 bits. For instance, since the no-arg constructor allocates a long array with one element, then the size() will return 64 for it: BitSet defaultBitSet = new BitSet() ĪssertThat(defaultBitSet.size()).isEqualTo(64) The size()method returns the number of bits the internal array can represent. There are three length-like methods for a BitSet. To solve this problem, we can use a combination of numeric data types (such as long) and bit-wise operations. The addressability issues and word tearing are the main reasons why booleans are more than just one single bit. That's 700% more memory than what we expected. If we ignore the overhead of object header, the array elements are consuming 1024 bytes, instead of the expected 1024 bits. Ideally, we expect a 1024-bit memory footprint from this array. I made this article quite a while ago to make it easier to work with binary strings, you may find it of some (ClassLayout.parseInstance(bits).toPrintable()) If you want to work with binary values directly, unfortunately there is no binary literal in C#. By ANDing that with the original integer, we can check if the bit is setĬonsole.WriteLine( " Base Value = ", otherValue & ( byte)(Bits.Bit0 | Bits.Bit1)) So, shifting 1 left by N will give us a binary pattern where only the Nth bit is set. Those numbers are in binary format by the way. So 1 << 5 would result in 100000, 1 << 2 would result in 100, etc. It will perform a left binary shift on the left operand. To get this, we just use the << operator. Now we just have to get a binary pattern to use against the AND. So, we can apply this further, and say that if q AND n is equal to n, then bit n was set. Now, unfortunately, you don't get a true or false value. You use the AND against a bit pattern to see if that bit is set. For example, 39 AND 4 would be worked out like so:Īs you can see, you end up with a binary sequence, which is represented by 4 in decimal (base-10). To get a bit, you have to AND it, which gets all the bits which are set in both of the numbers given to it. For example, 39 would be represented by 100111. Basically, you use the formula:īitNSet = (originalInteger & (1 << N) = 1 << N)Įffectively, every integer is represented by a binary sequence. It's a matter of using ANDs, and other boolean stuff (if you want to read multiple bits).
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