Search results
Results from the WOW.Com Content Network
The post-increment and post-decrement operators increase (or decrease) the value of their operand by 1, but the value of the expression is the operand's value prior to the increment (or decrement) operation. In languages where increment/decrement is not an expression (e.g., Go), only one version is needed (in the case of Go, post operators only).
The advantage of skew binary is that each increment operation can be done with at most one carry operation. This exploits the fact that (+) + = +.Incrementing a skew binary number is done by setting the only two to a zero and incrementing the next digit from zero to one or one to two.
The most significant bit of the first number is 1 and that of the second number is also 1 so the most significant bit of the result is 1; in the second most significant bit, the bit of second number is zero, so we have the result as 0. [2]
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:
0101 (decimal 5) AND 0011 (decimal 3) = 0001 (decimal 1) The operation may be used to determine whether a particular bit is set (1) or cleared (0). For example, given a bit pattern 0011 (decimal 3), to determine whether the second bit is set we use a bitwise AND with a bit pattern containing 1 only in the second bit:
The 1620 was a decimal-digit machine which used discrete transistors, yet it had hardware (that used lookup tables) to perform integer arithmetic on digit strings of a length that could be from two to whatever memory was available. For floating-point arithmetic, the mantissa was restricted to a hundred digits or fewer, and the exponent was ...
Integer overflow can be demonstrated through an odometer overflowing, a mechanical version of the phenomenon. All digits are set to the maximum 9 and the next increment of the white digit causes a cascade of carry-over additions setting all digits to 0, but there is no higher digit (1,000,000s digit) to change to a 1, so the counter resets to zero.
The increment ensures that a value of 5, incremented and left-shifted, becomes 16 (10000), thus correctly "carrying" into the next BCD digit. Essentially, the algorithm operates by doubling the BCD value on the left each iteration and adding either one or zero according to the original bit pattern.