1) If you XOR a bit with its own negated value, you will always get 1. Therefore thesolution to a ^ (~a) will be the sequence of 1s.
consider a is of data-type byte.
example:
a = 00000001
~a = 11111110
Now a ^ (~a) = 11111111
2) An operation like x & (~0 << n) clears the n rightmost bits of x. The value ~o is the sequences of 1s, so by shifting it by n, we have a bunch of ones followed by n zeros. By doing AND we clear the the rightmost n digits
of x.
3) Bit Facts and Tricks:
x ^ 0′s = x x & 0′s = 0 x | 0′s = x
x ^ 1′s = ~x x & 1′s = x x | 1′s = 1′s
x ^ x = 0 x & x = x x | x = x
4) Get Bit :
If I want to get the bit at i-th position, this method shifts1 over by i bits , creating a value and performing an
AND operation with the numbernum. If the resultant value is zero, that bit is unset else, that bit is set.
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boolean int num,inti) {
return((num & (1<< i)) != 0);
} |
5) Set Bit:
If I want to set the bit at i-th position, this method shifts1 over by i bits , creating a value and performing an
OR operation with the numbernum.
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public setBit(intnum,int i) {
returnnum | (1<< i);
} |
6) Clear Bit:
This method operates in reverse way as the setBit method works. To clear the
i-th bit we first negate the result obtained from 1 << i and perform AND operation with the numbernum.
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public clearBit(intnum,int i) {
intmask = ~(1<< i);
returnnum & mask;
} |
7) Clear all the MSB through i (inclusive) 高位到低位清零（包括低位）
This method is meant for clearing all the MSB’s (Most Significant Bits) through i (inclusive) we do:
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public clearBitsMSBThrough(intnum,int i) {
intmask = (1 << (i + 1) ) - 1;
returnnum & mask;
} |
8) Clear all the bits from i through o(inclusive) 低位到0位清零 （包括低位）
This method is meant for clearing all the bits from i through 0 from the number num:
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public clearBitsIthrough0(intnum,int i) {
intmask = ~(1<< (i + 1)) -1;
returnnum & mask;
} |