Matakuliah
Tahun
Versi
: T0324 / Arsitektur dan Organisasi Komputer
: 2005
:1
Pertemuan 5
Bahasa Rakitan: I
1
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Mendemonstrasikan penggunaan bahasa
rakitan dalam instruksi mesin ( C3 ) ( No
TIK : 3 )
2
Chapter 2.
Assembly Language: I
3
LOOP
100
104
Move
Move
N,R1
#NUM1,R2
108
112
116
Clear
Add
Add
R0
(R2),R0
#4,R2
120
124
128
132
Decrement
Branch>0
Move
R1
LOOP
R0,SUM
SUM
N
200
204
NUM1
NUM2
208
212
NUM n
604
100
Figure 2.17. Memory arrangement for the program in Figure 2.12.
4
Memory
address
label
Assembler directives
SUM
N
NUM1
Statemen
ts that
generate
machine
instructions
Assemblerdirectives
START
LOOP
Operation
EQU
ORIGIN
DATAWORD
RESERVE
ORIGIN
MOVE
MOVE
CLR
ADD
ADD
DEC
BGTZ
MOVE
RETURN
END
Addressing
or data
information
200
204
100
400
100
N,R1
#NUM1,R2
R0
(R2),R0
#4,R2
R1
LOOP
R0,SUM
START
Figure 2.18. Assembly language representation for the program in Figure 2.17.
5
Figure 2.19 Bus connection for processor
, keyboard, and display
.
6
Move
#LOC,R0
READ
TestBit
Branch=0
MoveByte
#3,INSTATUS
READ
DATAIN,(R0)
ECHO
TestBit
Branch=0
MoveByte
#3,OUTSTATUS
ECHO
(R0),DATAOUT
Compare
#CR,(R0)+
Branch0
READ
Initialize pointer registerR0 to point to the
addressof the first locationin memory
wherethe charactersare to be stored.
Wait for a characterto be entered
in the keyboard buffer DATAIN.
Transferthe characterfrom DATAIN into
the memory(this clears SIN to 0).
Wait for the display to becomeready.
Move thecharacterjust read to the display
buffer register(this clearsSOUT to 0).
Check if the characterjust read is CR
(carriagereturn). If it is not CR, then
branch back and read anothercharacter.
Also, increment the pointer to store the
next character.
Figure 2.20. A program that reads a line of characters and displays it.
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0
Stack
pointer
register
SP
•
•
•
- 28
Current
top element
17
739
Stack
BOTTOM
2k - 1
•
•
•
43
Bottom
element
•
•
•
Figure 2.21. A stack of words in the memory.
8
SP
19
- 28
17
739
- 28
17
739
SP
Stack
NEWITEM
•
•
•
•
•
•
43
43
19
(a) After push from NEWITEM
ITEM
- 28
(b) After pop into ITEM
Figure 2.22. Effect of stack operations on the stack in Figure 2.21.
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SAFEPOP
Compare
Branch> 0
#2000,SP
EMPTYERROR
Move
(SP)+,ITEM
Check to seeif the stack pointer contains
an addressvaluegreaterthan 2000. If it
does,the stack is empty . Branch to the
routine EMPTYERROR for appropriate
action.
Otherwise,pop the top of the stack into
memory location ITEM.
(a) Routine for a safe pop operation
SAFEPUSH
Compare
Branch 0
#1500,SP
FULLERROR
Move
NEWITEM, – (SP)
Check to seeif the stack pointer
contains an addressvalueequal
to or less than 1500. If it does, the
stack is full. Branch to the routine
FULLERROR for appropriateaction.
Otherwise,push the element in memory
location NEWITEM onto the stack.
(b) Routine for a safe push operation
Figure 2.23. Checking for empty and full errors in pop and push operations.
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Memory
location
Memory
location
Calling program
200
204
Call
SUB
next instruction
Subroutine SUB
1000
first instruction
Return
1000
PC
204
Link
204
Call
Return
Figure 2.24. Subroutine linkage using a link register.
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