Logical XOR
The XOR function of two bits is given by the following truth table:
Since the exclusive OR of two equal bits produces a zero and the exclusive
OR of two unequal bits produces a one, the exclusive OR function can be used to
test two quantities for equality.
If the quantities differ in any bit position, a one will be produced in the result.
The following subroutine produces the exclusive - OR 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 = 0011 and register 1 = 0010, register 0 will be replaced with 0001.
0 0 1 1
AND 0 0 1 0
-------
0 0 0 1
The subroutine produces the XOR of two bits by placing the bits in the leftmost position of the accumulator and register 2, respectively, and zeroing the rightmost three bIts of the accumulator and register 2. Register 2 is then added to the accumulator. The XOR of the two bits is then equal to the leftmost bit of the accumulator.
XOR, 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 'XOR' 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 'XOR' bit in Register 2
LDM 0 / Get bit in Register 1, set accumulator = 0
XCH 1
RAL / Move leftmost bit to carry
XCH 1 / Save shifted data in Register 1
RAR / Move second 'XOR' bit to carry
ADD 2 / 'ADD' gives the 'XOR' of the bits in carry
RAL / Otherwise, 'XOR' leftmost bit of the accumulator,
/ transmit to carry by RAL
JUN L1
L2, BBL 0 / Return to main program
