Instruction Set Summary (Arithmetic Operations)
ADDC A, (src-byte)
Function: Add with Carry
Description: ADDC simultaneously adds the byte variable
indicated, the carry flag and the Accumulator contents, leaving
the result in the Accumulator. The carry and auxiliary carry flags
are set respectively, if there is a carry-out from bit 7 or bit 3,
and cleared
otherwise. When adding unsigned integers, the carry flag indicates
an overflow occurred.
OV is set if there is a carry-out of bit 6 but not out of bit 7, or
a carry-out of bit 7 but not out of bit 6; otherwise OV is cleared.
When adding signed integers, OV indicates a negative number
produced as the sum of two positive operands or a positive sum from
two negative operands.
Four source operand addressing modes are allowed: register, direct,
register-indirect, or
immediate.
Example: The Accumulator holds 0C3H (11000011B) and register
0 holds 0AAH (10101010B) with the carry flag set. The following
instruction,
ADDC A, R0
leaves 6EH (01101110B) in the Accumulator with AC cleared and both
the Carry flag and OV set to 1.
ADDC A,Rn
Bytes: 1
Cycles: 1
Operation: ADDC
(A) ← (A) + (C) + (Rn)
ADDC A,direct
Bytes: 2
Cycles: 1
Operation: ADDC
(A) ← (A) + (C) + (direct)
ADDC A,@Ri
Bytes: 1
Cycles: 1
Operation: ADDC
(A) ← (A) + (C) + ((Ri))
ADDC A,#data
Bytes: 2
Cycles: 1
Operation: ADDC
(A) ← (A) + (C) + #data
SUBB A,(src-byte)
Function: Subtract with borrow
Description: SUBB subtracts the indicated variable and the
carry flag together from the Accumulator, leaving the result in the
Accumulator. SUBB sets the carry (borrow) flag if a borrow is
needed for bit 7 and clears C otherwise. (If C was set before
executing a SUBB instruction, this indicates that a borrow was
needed for the previous step in a multiple- precision subtraction,
so the carry is subtracted from the Accumulator along with the
source operand.)
AC is set if a borrow is needed for bit 3 and cleared otherwise. OV
is set if a borrow is needed into bit 6, but not into bit 7, or
into bit 7, but not bit 6.
When subtracting signed integers, OV indicates a negative number
produced when a negative value is subtracted from a positive
value, or a positive result when a positive number is subtracted
from a negative number.
The source operand allows four addressing modes: register, direct,
register-indirect, or immediate.
Example: The Accumulator holds 0C9H (11001001B), register 2
holds 54H (01010100B), and the carry flag is set. The instruction,
SUBB A, R2
will leave the value 74H (01110100B) in the accumulator, with the
carry flag and AC cleared but OV set.
Notice that 0C9H minus 54H is 75H. The difference between this and
the above result is due to the carry (borrow) flag being set before
the operation. If the state of the carry is not known before
starting a single or multiple-precision subtraction, it should be
explicitly cleared by CLR C instruction.
SUBB A,Rn
Bytes: 1
Cycles: 1
Operation: SUBB
(A) ← (A) - (C) – (Rn)
SUBB A,direct
Bytes: 2
Cycles: 1
Operation: SUBB
(A) ← (A) - (C) – (direct)
SUBB A,@Ri
Bytes: 1
Cycles: 1
Operation: SUBB
(A) ← (A) - (C) – ((Ri))
SUBB A,#data
Bytes: 2
Cycles: 1
Operation: SUBB
(A) ← (A) - (C) - #data
INC (byte)
Function: Increment
Description: INC increments the indicated variable by 1. An
original value of 0FFH overflows to 00H. No flags are affected.
Three addressing modes are allowed: register, direct, or
register-indirect.
Note: When this instruction is used to modify an output port, the
value used as the original port data will be read from the output
data latch, not the input pins.
Example: Register 0 contains 7EH (011111110B). Internal RAM
locations 7EH and 7FH contain 0FFH and 40H, respectively. The
following instruction sequence,
INC @R0
INC R0
INC @R0
leaves register 0 set to 7FH and internal RAM locations 7EH and 7FH
holding 00H and 41H, respectively.
INC A
Bytes: 1
Cycles: 1
Operation: INC
(A) ← (A) + 1
INC Rn
Bytes: 1
Cycles: 1
Operation: INC
(Rn) ← (Rn) + 1
INC direct
Bytes: 2
Cycles: 1
Operation: INC
(direct) ← (direct) + 1
INC @Ri
Bytes: 1
Cycles: 1
Operation: INC
((Ri)) ← ((Ri)) + 1
INC DPTR
Function: Increment Data Pointer
Description: INC DPTR increments the 16-bit data pointer by
1. A 16-bit increment (modulo 2 16 ) is performed, and an overflow
of the low-order byte of the data pointer (DPL) from 0FFH to 00H
increments the high-order byte (DPH). No flags are affected.
This is the only 16-bit register which can be incremented.
Example: Registers DPH and DPL contain 12H and 0FEH,
respectively. The following instruction sequence,
INC DPTR
INC DPTR
INC DPTR
changes DPH and DPL to 13H and 01H.
Bytes: 1
Cycles: 2
Operation: INC
(DPTR) ← (DPTR) + 1
DEC byte
Function: Decrement
Description: DEC byte decrements the variable indicated by 1.
An original value of 00H underflows to 0FFH. No flags are
affected. Four operand addressing modes are allowed: accumulator,
register, direct, or register-indirect.
Note: When this instruction is used to modify an output port, the
value used as the original port data will be read from the output
data latch, not the input pins.
Example: Register 0 contains 7FH (01111111B). Internal RAM
locations 7EH and 7FH contain 00H and 40H, respectively. The
following instruction sequence,
DEC @R0
DEC R0
DEC @R0
leaves register 0 set to 7EH and internal RAM locations 7EH and 7FH
set to 0FFH and 3FH.
DEC A
Bytes: 1
Cycles: 1
Operation: DEC
(A) ← (A) – 1
DEC Rn
Bytes: 1
Cycles: 1
Operation: DEC
(Rn) ← (Rn) – 1
DEC direct
Bytes: 2
Cycles: 1
Operation: DEC
(direct) ← (direct) – 1
DEC @Ri
Bytes: 1
Cycles: 1
Operation: DEC
((Ri)) ← ((Ri)) - 1
MUL AB
Function: Multiply
Description: MUL AB multiplies the unsigned 8-bit integers in
the Accumulator and register B. The low-order byte of the 16-bit
product is left in the Accumulator, and the high-order byte in B.
If the product is greater than 255 (0FFH), the overflow flag is set;
otherwise it is cleared. The carry flag is always cleared.
Example: Originally the Accumulator holds the value 80 (50H).
Register B holds the value 160 (0A0H). The instruction,
MUL AB
will give the product 12,800 (3200H), so B is changed to 32H
(00110010B) and the Accumulator is cleared. The overflow flag is
set, carry is cleared.
Bytes: 1
Cycles: 4
Operation: MUL
(A)7-0 ← (A) X (B)
(B)15-8
DIV AB
Function: Divide
Description: DIV AB divides the unsigned eight-bit integer in
the Accumulator by the unsigned eight- bit integer in register B.
The Accumulator receives the integer part of the quotient;
register B receives the integer remainder. The carry and OV flags
are cleared.
Exception: if B had originally contained 00H, the values returned
in the Accumulator and B-register are undefined and the overflow
flag are set. The carry flag is cleared in any case.
Example: The Accumulator contains 251 (0FBH or 11111011B) and
B contains 18 (12H or 00010010B). The following instruction,
DIV AB
leaves 13 in the Accumulator (0DH or 00001101B) and the value 17
(11H or 00010001B) in B, since 251 = (13 x 18) + 17. Carry and OV
are both cleared.
Bytes: 1
Cycles: 4
Operation: DIV
(A)15-8 ← (A)/(B)
(B)7-0
DA A
Function: Decimal-adjust Accumulator for Addition
Description: DA A adjusts the eight-bit value in the
Accumulator resulting from the earlier addition of two variables
(each in packed-BCD format), producing two four-bit digits. Any ADD
or ADDC instruction may have been used to perform the addition.
If Accumulator bits 3 through 0 are greater than nine
(xxxx1010-xxxx1111), or if the AC flag is one, six is added to the
Accumulator producing the proper BCD digit in the low- order
nibble. This internal addition sets the carry flag if a carry-out of
the low-order four-bit field propagates through all high-order
bits, but it does not clear the carry
flag otherwise.
If the carry flag is now set, or if the four high-order bits now
exceed nine (1010xxxx- 1111xxxx), these high-order bits are
incremented by six, producing the proper BCD digit in the
high-order nibble. Again, this sets the carry flag if there is a
carry-out of the high- order bits, but does not clear the carry.
The carry flag thus indicates if the sum of the original two BCD
variables is greater than 100, allowing multiple precision decimal
addition. OV is not affected.
All of this occurs during the one instruction cycle. Essentially,
this instruction performs the decimal conversion by adding 00H,
06H, 60H, or 66H to the Accumulator, depending on initial
Accumulator and PSW conditions.
Note: DA A cannot simply convert a hexadecimal number in the
Accumulator to BCD notation, nor does DAA apply to decimal
subtraction.
Example: The Accumulator holds the value 56H (01010110B),
representing the packed BCD digits of the decimal number 56.
Register 3 contains the value 67H (01100111B), representing the
packed BCD digits of the decimal number 67. The carry flag is set.
The following instruction sequence
ADDC A,R3
DA A
first performs a standard two’s-complement binary addition,
resulting in the value 0BEH (10111110) in the Accumulator. The
carry and auxiliary carry flags are cleared.
The Decimal Adjust instruction then alters the Accumulator to the
value 24H (00100100B), indicating the packed BCD digits of the
decimal number 24, the low-order two digits of the decimal sum of
56, 67, and the carry-in. The carry flag is set by the Decimal
Adjust instruction, indicating that a decimal overflow occurred. The
true sum of
56, 67, and 1 is 124.
BCD variables can be incremented or decremented by adding 01H or
99H. If the Accumulator initially holds 30H (representing the
digits of 30 decimal), then the following instruction sequence,
ADD A, # 99H
DA A
leaves the carry set and 29H in the Accumulator, since 30 + 99 =
129. The low-order byte of the sum can be interpreted to mean 30 -
1 = 29.
Bytes: 1
Cycles: 1
Operation: DA
-contents of Accumulator are BCD
IF [[(A 3-0 ) > 9] [(AC) = 1]]
THEN (A 3-0 ) ← (A 3-0 ) + 6
AND
IF [[(A 7-4 ) > 9] [(C) = 1]]
THEN (A 7-4 ) ← (A 7-4 ) + 6
What will be the problem if I use the Mnemonic ADD A,@Ri instead of ADD A,direct? and When will I use these two?
ReplyDeleteHi Karen! One thing that I can think of the advantage of using ADD A, @Ri instead of ADD A, direct is that it only uses 1 byte of instruction. ADD A, direct consumes 2 bytes so it is one way of using your Internal RAM wisely. Also, direct addressing (ADD A, direct) can only be use on the first 128 bytes of the memory, meaning from 00H-07FH only. But, if you want to access the remaining higher bytes, you use Indirect Addressing (ADD A, @Ri) Hope this helps! :)
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