PIC instruction set (continued)

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16.317
Microprocessor Systems Design I
Instructor: Dr. Michael Geiger
Fall 2013
Lecture 29:
PIC instruction set (continued)
Lecture outline

Announcements/reminders


HW 6 due 11/25
Today’s lecture: More PIC instructions
4/13/2015
Microprocessors I: Lecture 29
2
Review: PIC instructions

Logical operations:





andlw/andwf
iorlw/iorwf
xorlw/xorwf
Rotates: rrf/rlf
Jumps/calls/return



4/13/2015
goto
call
return/retlw/retfie
Microprocessors I: Lecture 29
3
Review: Conditional Execution


Conditional execution in PIC: skip next instruction if condition true
Two general forms


btfsc
btfss
decfsz
incfsz
Test bit and skip if bit clear/set
Increment/decrement register and skip if result is 0
f, b
f, b
f, F(W)
f, F(W)
Examples:

btfsc TEMP1, 0

btfss STATUS, C

decfsz TEMP1, F

incfsz TEMP1, W
4/13/2015
STATUS bits:
none
;Test bit b of register f, where b=0 to 7, skip if clear
;Test bit b of register f, where b=0 to 7, skip if set
;decrement f, putting result in F or W, skip if zero
;increment f, putting result in F or W, skip if zero
; Skip the next instruction if bit 0 of TEMP1 equals 0
; Skip the next instruction if C==1
; Decrement TEMP1, skip if TEMP1==0
; W <- TEMP1+1 , skip if W==0 (TEMP1==0xFF)
; Leave TEMP1 unchanged
Microprocessors I: Lecture 29
4
Example

Show the values of all changed registers
after each of the following sequences

(a)
What high-level operation does each perform?
movf
sublw
btfsc
goto
incf
goto
a, W
0xA
STATUS, Z
L1
b, W
L2
movf
subwf
btfss
goto
movf
goto
NUM2, W
NUM1, W
STATUS, C
BL
NUM1, W
Done
movf
NUM2, W
movwf
MIN
BL
L1
decf
b, W
movwf
a
L2
4/13/2015
(b)
Done
Microprocessors I: Lecture 29
5
Example solution (part a)
movf
sublw
btfsc
a, W
0xA
STATUS, Z
goto
L1
incf
goto
b, W
L2
decf
b, W
W=a
 W = 10 – a
 Skip goto if result
is non-zero
 Goto L1 if result == 0
 Reach this point if
result non-zero
W=b+1
L1
W=b-1
High-level operation:
if (a == 10)
a=b–1
else
a=b+1
L2
movwf a
4/13/2015
 a = W  value depends
on what’s executed before this
Microprocessors I: Lecture 29
6
Example solution (part b)
 W = NUM2
 W = NUM1 – W
= NUM1 – NUM2
 Carry indicates
“above”
 if set, NUM1 < NUM2
movf
subwf
NUM2, W
NUM1, W
btfss
STATUS, C
goto
movf
BL
NUM1, W
goto
Done
 Skip “below” section
movf
NUM2, W
 if (NUM1 >= NUM2)
W = NUM2
movwf
MIN
 if (NUM1 < NUM2)
W = NUM1
BL
High-level operation:
if (NUM1 < NUM2)
MIN = NUM1
else
MIN = NUM2
Done
4/13/2015
Microprocessors I: Lecture 29
7
Miscellaneous
clrwdt
sleep
nop
; clear watchdog timer
; go into standby mode
; no operation
STATUS bits:
clrwwdt, sleep:
NOT_TO, NOT_PD
nop: none
Examples:

clrwdt

sleep

nop
4/13/2015
; if watchdog timer is enabled, this instruction will reset
; it (before it resets the CPU)
; Stop clock; reduce power; wait for watchdog timer or
; external signal to begin program execution again
; Do nothing; wait one clock cycle
Microprocessors I: Lecture 29
8
Working with multiple registers



Can’t do simple data transfer or operation on
two registers
Usually must involve working register
Examples: x86  PIC (assume PIC registers
defined with same names as x86 registers)

MOV AL, BL
movf BL, W
movwf AL

ADD AL, BL
movf BL, W
addwf AL, F
4/13/2015
Microprocessors I: Lecture 29
9
Conditional jumps



Basic ones are combination of bit tests, skips
Remember that condition you’re testing is
opposite of jump condition
Examples: x86  PIC

JNC label
btfss
goto

JE label
btfsc
goto
4/13/2015
STATUS, C
label
STATUS, Z
label
Microprocessors I: Lecture 29
10
Conditional jumps (cont.)


To evaluate other conditions, may want to use
subtraction in place of compare
CMP X, Y turns into:
movf Y, W
subwf X, W

Possible results (unsigned comparison only):




X > Y  Z = 0, C = 1
X == Y  Z = 1, C = 1
X < Y  Z = 0, C = 0
More complex conditions



4/13/2015
X <= Y  Z == C
X != Y  Z = 0
X >= Y  C = 1
Microprocessors I: Lecture 29
11
Shift/rotate operations

May need to account for each of the following

Carry bit




Bit being shifted/rotated out



Always shifted in to register for rrf/rlf instructions
Basic shift: explicitly set carry to 0
Arithmetic shift right: set carry to sign bit
Basic rotate doesn’t rotate through carry
Can either pre-test or fix later
Multi-bit shift/rotate: loop where # iterations
matches shift amount
4/13/2015
Microprocessors I: Lecture 29
12
Shift/rotate operations (cont.)

Examples: x86  PIC

SHL
bcf
rlf


ROR
AL, 1
STATUS, C
AL, F
AL, 1
bcf
rrf
btfsc
STATUS, C
AL, F
STATUS, C
bsf
AL, 7
RCL
rlf
decfsz
goto
4/13/2015
; Clear carry bit
; Rotate AL one bit to right
; Skip next instruction if C clear
; C = bit shifted out of MSB
; Handle case where C = 1
; MSB of AL should be 1
AL, 3
movlw
movwf
Loop:
; Clear carry bit
; Rotate AL one bit to the left
3
; Initialize working register to 3 (# iterations)
COUNT ; Initialize count register
; Assumes you’ve declared variable COUNT
AL, F
; Rotate AL one bit to left
COUNT, F
; Decrement counter & test for 0
; Skip goto if result is zero
Loop
; Return to start to loop
Microprocessors I: Lecture 29
13
Examples








Translate these x86 operations to PIC code
Assume that there are registers defined for
each x86 register (e.g. AL, AH, BL, BH, etc.)
OR
SUB
JNZ
JL
SAR
ROL
4/13/2015
AL, BL
BL, AL
label
label
AL, 1
AL, 5
Microprocessors I: Lecture 29
14
Example solution

OR
AL, BL
movf BL, W
iorwf AL, F

SUB
BL, AL
movf AL, W
subwf BL, F

JNZ
; W = BL
; AL = AL OR W = AL OR BL
; W = AL
; BL = BL – W = BL – AL
label
btfss STATUS, Z ; Skip goto if Z == 1 (if
goto label
; previous result == 0)
4/13/2015
Microprocessors I: Lecture 29
15
Example solution (continued)

JL
label
btfsc
goto
btfss
goto
End:
STATUS, Z
End
STATUS, C
label
4/13/2015
; If Z == 0, check C
; Otherwise, no jump
; If C == 1, no jump
; Jump to label
; End of jump
Microprocessors I: Lecture 29
16
Example solution (continued)

SAR
AL, 1
bcf STATUS, C
btfsc AL, 7
bsf STATUS, C
rrf
4/13/2015
AL, F
;C=0
; Skip if MSB == 0
; C = 1 if MSB == 1
; C will hold copy of
; MSB (keeping sign
; intact)
; Rotate right by 1
Microprocessors I: Lecture 29
17
Example solution (continued)

ROL
L:
AL, 5
movlw 5
movwf COUNT
bcf
STATUS, C
btfsc AL, 7
bsf
STATUS, C
rlf
AL, F
decfsz COUNT
goto
4/13/2015
;W=5
; COUNT = W = 5
;C=0
; Skip if MSB == 0
; C = 1 if MSB == 1
; C will hold copy of
; MSB (bit rotated into
; LSB)
; Rotate left by 1
; If COUNT == 0, don’t
; restart loop
L
Microprocessors I: Lecture 29
18
Final notes

Next time:


Continue with complex operations—multi-byte
data
Reminders:

4/13/2015
HW 6 to be posted; due date TBD
Microprocessors I: Lecture 29
19
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