Binary system, primitives in Java and bit manipulation.
22^3 = 82^4 = 162^7 = 1282^8 = 2562^10 = 1 0242^20 = 1 048 5768-bit signed two’s complement integer, minimum -128, maximum 12716-bit signed two’s complement integer, minimum-32,768, maximum 32,76732-bit signed two’s complement integer, minimum -2^31, maximum 2^31-1; from Java 8 can be used to represent an unsigned 32-bit integer, minimum 0, maximum value of 2^32-1 (extra method on Integer class)64-bit two’s complement integer, minimum -2^63, maximum 2^63-1; from Java 8 like with int32-bit IEEE 754 floating point64-bit IEEE 754 floating point1 bit16-bit Unicode character, minimum '\u0000' (or 0), maximum '\uffff' (or 65 535)0100 equals 4 - the 2^0 is on the right.
Floating point number is represented in 32 bits. The most significant bit is the sign bit, then come 8 bits of the exponent, and 23 bits of the fraction, so e.g.: 001100100.01011010110010101110001.
1 1
0 1 1 0 = 6
+ 0 1 1 1 = 7
-------
1 1 0 1 = 13
1 / 1 / / 1
1 1 1 1 1 1
1 0 0 0 = 8 / 0 0 0 / 0 0 0 / 0 0 0
- 0 0 1 1 = 3 - 0 0 1 1 - 0 0 1 1 - 0 0 1 1
------- ------- ------- -------
? 0 1 0 1 = 5
We move elerything to the left and fill the empty space with 0. It is equlivalent to *2:
40 « 1 = 80
We move everything to the right and fill the empty space with 0. It is equlivalent to /2:
40 » 1 = 20
40 » 5 = 1 - we drop the remainder, this is integer division
Is like right shift but does not preserve the sign. So, 32th bit (2^31, the one on the left) is the sign bit. 0 means +, 1 means -. In normal shift it would be overwritten after the shift, in this shift it stays, so:
-1 »> 1 = 2147483647
-1 » 1 = -1
00001000 - 1 = 000001110000 = x1111 = ~x00000000 = 00001111 = x0000 = x1111 = 1111return ((num & (1 << i)) != 0)return num | (1 << i)return num & ~(1 << i)return num & ((1 << i) - 1)return num & ~((1 << i) - 1)return num & ~(1 << i) | (v << i)Comments: