PIC18F87J11_12 Datasheet, PDF (383/466 Pages) Microchip Technology – 64/80-Pin, High-Performance, 1-Mbit Flash Microcontrollers

PIC18F87J11 FAMILY
SUBLW
Subtract W from Literal
Syntax:
Operands:
Operation:
Status Affected:
Encoding:
Description:
Words:
Cycles:
Q Cycle Activity:
Q1
Decode
SUBLW k
0 k 255
k – (W) W
N, OV, C, DC, Z
0000 1000 kkkk kkkk
W is subtracted from the eight-bit
literal ‘k’. The result is placed in W.
1
1
Q2
Read
literal ‘k’
Q3
Process
Data
Q4
Write to
W
Example 1:
SUBLW 02h
Before Instruction
W
= 01h
C
=?
After Instruction
W
= 01h
C
= 1 ; result is positive
Z
=0
N
=0
Example 2:
SUBLW 02h
Before Instruction
W
= 02h
C
=?
After Instruction
W
= 00h
C
= 1 ; result is zero
Z
=1
N
=0
Example 3:
SUBLW 02h
Before Instruction
W
= 03h
C
=?
After Instruction
W
= FFh ; (2’s complement)
C
= 0 ; result is negative
Z
=0
N
=1
 2007-2012 Microchip Technology Inc.
SUBWF
Subtract W from f
Syntax:
Operands:
Operation:
Status Affected:
Encoding:
Description:
Words:
Cycles:
Q Cycle Activity:
Q1
Decode
SUBWF f {,d {,a}}
0 f 255
d  [0,1]
a  [0,1]
(f) – (W) dest
N, OV, C, DC, Z
0101 11da ffff ffff
Subtract W from register ‘f’ (2’s
complement method). If ‘d’ is ‘0’, the
result is stored in W. If ‘d’ is ‘1’, the result
is stored back in register ‘f’.
If ‘a’ is ‘0’, the Access Bank is selected.
If ‘a’ is ‘1’, the BSR is used to select the
GPR bank.
If ‘a’ is ‘0’ and the extended instruction
set is enabled, this instruction operates
in Indexed Literal Offset Addressing
mode whenever f 95 (5Fh). See
Section 26.2.3 “Byte-Oriented and
Bit-Oriented Instructions in Indexed
Literal Offset Mode” for details.
1
1
Q2
Read
register ‘f’
Q3
Process
Data
Q4
Write to
destination
Example 1:
SUBWF REG, 1, 0
Before Instruction
REG
W
C
=3
=2
=?
After Instruction
REG
W
C
Z
N
=1
=2
=1
=0
=0
; result is positive
Example 2:
SUBWF REG, 0, 0
Before Instruction
REG
W
C
=2
=2
=?
After Instruction
REG
W
C
Z
N
=2
=0
=1
=1
=0
; result is zero
Example 3:
SUBWF REG, 1, 0
Before Instruction
REG =
W
=
C
=
After Instruction
REG =
W
=
C
=
Z
=
N
=
1
2
?
FFh ;(2’s complement)
2
0 ; result is negative
0
1
DS39778E-page 383