Logical OR
The OR function of two bits is given by the following truth table:
Since any bit ORed with a one produces a one, and any bit ORed with a zero
remains unchanged, the OR function is often used to set groups of bits to one.
The following subroutine produces the OR, bit by bit,of the two 4 bit quantities held in index registers 0 and 1. The result is placed in register 0 while register 1 is set to 0. Index registers 2 and 3 are also used.
For example, if register 0 is set to 0100 and register 1 to 0011, register 0 will be replaced with 0111.
0 1 0 0
AND 0 0 1 1
-------
0 1 1 1
The subroutine produces the OR of two bits by placing the bits in the leftmost position of the accumuiator and register 2, respectively, and zeroing the rightmost three bits of the accumulator and register 2. Register 2 is then added to the accumulator. If the resulting carry = 1, the OR of the two bits = 1. If the resulting carry = 0, the OR of the two bits is equal to the leftmost bit of the accumulator.
OR, FIM 1P 11 / Register 2 = 0, Register 3 = 11
L1, LDM 0 / Get bit of Register 0, Set accumulator = 0
XCH 0 / Set Register 0 to accumulator; Register 0 = 0
RAL / Move first 'OR' bit to carry
XCH 0 / Save shifted data in Register 0, set accumulator = 0
INC 3 / Done if Register 3 = 0
XCH 3 / Register 3 to accumulator
JCN 4 L2 / Return if accumulator = 0
XCH 3 / Otherwise, restore accumulator and Register 3
RAR / Bit of Register 0 is alone in the accumulator
XCH 2 / Save first 'OR' bit in Register 2
LDM 0 / Get bit in Register 1, set accumulator = 0
XCH 1 / Get bit of Register 1
RAL / Move leftmost bit to carry
XCH 1 / Save shifted data in Register 1
RAR / Move second 'OR' bit to carry
ADD 2 / 'ADD' gives the 'OR' of the bits in carry
JCN 2 L1 / Jump if carry = 1 because 'OR' = 1
RAL / Otherwise, 'OR' leftmost bit of the accumulator,
/ transmit to carry by RAL
JUN L1
L2, BBL 0 / Return to main program
