Search results
Results from the WOW.Com Content Network
Bit manipulation is the act of algorithmically manipulating bits or other pieces of data shorter than a word. ... an operand with exactly one '1' bit is a power of 2 ...
The result in each position is 1 if only one of the bits is 1, but will be 0 if both are 0 or both are 1. In this we perform the comparison of two bits, being 1 if the two bits are different, and 0 if they are the same. For example: 0101 (decimal 5) XOR 0011 (decimal 3) = 0110 (decimal 6)
For example, when shifting a 32 bit unsigned integer, a shift amount of 32 or higher would be undefined. Example: If the variable ch contains the bit pattern 11100101, then ch >> 1 will produce the result 01110010, and ch >> 2 will produce 00111001. Here blank spaces are generated simultaneously on the left when the bits are shifted to the right.
It is frequently stated that arithmetic right shifts are equivalent to division by a (positive, integral) power of the radix (e.g., a division by a power of 2 for binary numbers), and hence that division by a power of the radix can be optimized by implementing it as an arithmetic right shift. (A shifter is much simpler than a divider.
The most significant digit is an exception to this: for an n-bit Gray code, the most significant digit follows the pattern 2 n-1 on, 2 n-1 off, which is the same (cyclic) sequence of values as for the second-most significant digit, but shifted forwards 2 n-2 places. The four-bit version of this is shown below:
XOR returns 1 if and only if an odd number of bits are 1. Therefore, if two corresponding bits are 1, the result will be a 0, but if only one of them is 1, the result will be 1. Therefore inversion of the values of bits is done by XORing them with a 1. If the original bit was 1, it returns 1 XOR 1 = 0. If the original bit was 0 it returns 0 XOR ...
The function 2 ⌈log 2 (x)⌉ (round up to the nearest power of two) using shifts and bitwise ORs [40] is not efficient to compute as in this 32-bit example and even more inefficient if we have a 64-bit or 128-bit operand:
So the carry-less product of a and b would be c = 101100011101100 2. For every bit set in the number a, the number b is shifted to the left as many bits as indicated by the position of the bit in a. All these shifted versions are then combined using an exclusive or, instead of the regular addition which would be used for regular long ...