Transcript Chap. 6 Programming the Basic Computer

Chap. 6 Programming the Basic Computer

6-1
6-1 Introduction
 Translate user-oriented symbolic program(alphanumeric character set) into
binary programs recognized by the hardware
 25 Instruction Set of the basic computer




Memory Reference Instruction
Register Reference Instruction
Input-output Instruction
6-2 Machine Language
 Program Categories
 1) Binary Code(Machine Language)
» Program Memory의 내용 : Tab. 6-2

2) Octal or Hexadecimal Code
» Binary Code와 동일 : Tab. 6-3

3) Symbolic Code : Tab. 6-4
» Assembly Language : Tab. 6-5

4) High Level Language
» C, Fortran,.. : Tab 6-6
Computer System Architecture
* Tab. 6-4와 차이점
Pseudoinstruction,
Label 사용 가능
Symbol
AND
ADD
LDA
STA
BUN
BSA
ISZ
CLA
CLE
CMA
CME
CIR
CIL
INC
SPA
SNA
SZA
SZE
HLT
INP
OUT
SKI
SKO
ION
IOF
Hex Code
0xxx 8xxx
1xxx 9xxx
2xxx Axxx
3xxx Bxxx
4xxx Cxxx
5xxx Dxxx
6xxx Exxx
7800
7400
7200
7100
7080
7040
7020
7010
7008
7004
7002
7001
F800
F400
F200
F100
F080
F040
Chap. 6 Programming the Basic Computer
Description
And memory word to AC
Add memory word to AC
Load memory word to AC
Store content of AC in memory
Branch unconditionally
Branch and Save return address
Increment and skip if zero
Clear AC
Clear E
Complement AC
Complement E
Circulate right AC and E
Circulate left AC and E
Increment AC
Skip next instruction if AC positive
Skip next instruction if AC negative
Skip next instruction if AC zero
Skip next instruction if E is 0
Halt computer
Input character to AC
Output character from AC
Skip on input flag
Skip on output flag
Interrupt On
Interrupt Off
© Korea Univ. of Tech. & Edu.
Dept. of Info. & Comm.
6-2

6-3 Assembly Language
 The rules for writing assembly language program
 Documented and published in manuals(from the computer manufacturer)
 Rules of the Assembly Language
 Each line of an assembly language program is arranged in three columns
» 1) Label field : empty or symbolic address
» 2) Instruction field : machine instruction or pseudoinstruction
» 3) Comment field : empty or comment

Field
Label Instruction Comment
Symbolic Address(Label field)
» One, two, or three, but not more than three alphanumeric characters
» The first character must be a letter; the next two may be letters or numerals
» A symbolic address is terminated by a comma(recognized as a label by the assembler)

Instruction Field
» 1) A memory-reference instruction(MRI)

Ex) ADD OPR(direct address MRI), ADD PTR I(indirect address MRI)
» 2) A register-reference or input-output instruction(non-MRI)

Ex) CLA(register-reference), INP(input-output)
» 3) A pseudoinstruction with(ORG N) or without(END) an operand : Tab. 6-7

Computer System Architecture
Pseudoinstruction 은 Machine Instruction이 아니고 Assembler에게 필요한 정보만 제공
Chap. 6 Programming the Basic Computer
© Korea Univ. of Tech. & Edu.
Dept. of Info. & Comm.
6-3

Comment field
» Comment filed must be preceded by a slash(recognized by assembler as comment)
 An Example Program : Tab. 6-8
 83 - ( - 23 ) = 83 + ( 2’s Complement of -23)
= 83 + 23
교과서에서는
같은 의미로 사용됨
 Translation to Binary : Tab. 6-9
 Assembler = the translation of the symbolic(= assembly) program into binary
 Address Symbol Table = Hexadecimal address of symbolic address
» MIN = 106, SUB = 107, DIF = 108
 Two Pass Assembler : in next Sec. 6-4
 1) 1st scan pass : generate user defined address symbol table
 2) 2nd scan pass : binary translation

Ex) LDA SUB
1) SUB = 107
2) 2107
6-4 The Assembler
 Source Program
asm
a96
a51
c(cpp)
for
pas
Computer System Architecture
Object Code
Assembler
(Compiler)
obj
Binary Code
Linker
bin
exe
com
hex
Chap. 6 Programming the Basic Computer
Library 또는 외부 함수를
사용하여 Relocation
© Korea Univ. of Tech. & Edu.
Dept. of Info. & Comm.
6-4
 Representation of Symbolic Program in Memory : Tab. 6-11
 Line of Code :
PL3, LDA SUB I
(Carriage return)
» The assembler recognizes a CR code as the end of a line of code
First pass
 Two Pass Assembler
 1) 1st pass : Generate userdefined address symbol table
LC
0
Scan next line of code
Set LC
» Flowchart of first pass :
Fig. 6-1
yes
» Address Symbol Table for
Program in Tab. 6-8 :
no
Label
no
ORG
yes
yes
Tab. 6-12
yes
END
no
Store symbol
in addresssymbol table
together with
value of LC
Fig. 6-1 Flowchart for first pass of assembler
Computer System Architecture
Go to
second
pass
Increment LC
Chap. 6 Programming the Basic Computer
© Korea Univ. of Tech. & Edu.
Dept. of Info. & Comm.
6-5
Second pass
LC

2) 2nd pass : Binary translation
Tab. 6-9 에서의
Contents를 결정
이 경우는
Hand Assemble
Fig. 6-2
Set LC
Yes
Pseudo- Yes
instruction
Yes
3
4
I Opcode
5
DEC or HEX
Valid
No
non-MRI
instruction
Convert
operand
to binary
and store
in location
given by LC
Yes
Check for possible errors in the
symbolic program
Ex) Invalid Machine Code Error
2
No
MRI
Search addresssymbol table for
binary equivalent
of symbolic address
and set bits 5-16
» Error Diagnostics
1
Yes
END
No
Get operation code
and set bits 2-4
Tab. 6-9 의 Content
Pseudoinstruction,
MRI, non-MRI 에
속하지 않음
No
ORG
No
» Binary Code translation 예제 :

Done
Scan next line of code
» 다음의 4 개의 Table을 참고하여
Instruction Format에 의한 Binary
Code 형성 (Pseudoinstruction Table,
MRI Table, Non-MRI Table, Address
Symbol Table)
» Flowchart of second pass :

0
6 ….
…
Yes
No
I
15 16
Set first
bit to 1
Store binary
equivalent of
instruction
in location
given by LC
Error in
line of
code
Error
Set first
bit to 0
Address
Assembley all parts of
binary instruction and
store in location given by LC
Increment LC
Fig. 6-2 Flowchart for second pass of assembler
Computer System Architecture
Chap. 6 Programming the Basic Computer
© Korea Univ. of Tech. & Edu.
Dept. of Info. & Comm.
6-6

6-5 Program Loops
Tab. 6-13 Symbolic Program to Add 100 numbers
 Program Loops
 A sequence of instructions that
are executed many times
 Example of program loop
 Sum of 100 integer numbers
» Fortran
DIMENSION A(100)
INTEGER SUM, A
SUM = 0
DO 3 J = 1, 100
3 SUM = SUM + A(J)

Symbolic Program : Tab 6-13
» Address 150 부터 100 개의
Data를 더하기
Computer System Architecture
Line
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
,
,
118
119
LOP,
ADS,
PTR,
NBR,
CTR,
SUM,
ORG
LDA
STA
LDA
STA
CLA
ADD
ISZ
ISZ
BUN
STA
HLT
HEX
HEX
DEC
HEX
HEX
ORG
DEC
,
,
DEC
END
100
ADS
PTR
NBR
CTR
PTR I
PTR
CTR
LOP
SUM
150
0
-100
0
0
150
75
,
.
23
Chap. 6 Programming the Basic Computer
/
/
/
/
/
/
/
/
/
/
A = 150
PTR = 150
A = -100
CTR = -100
A=0
A + 75 (150 번지의 내용)
150 + 1 = 151
-100 + 1 = -99
Loop until CTR = 0
Store A to SUM
/ 150
/ -100
/ Result of Sum
Data
© Korea Univ. of Tech. & Edu.
Dept. of Info. & Comm.
6-7

6-6 Programming Arithmetic & Logic Operations
 Hardware implementation

Operations are implemented in a computer with one machine instruction

Ex) ADD, SUB : 그러나 자리수가 늘어나면 Software subroutine 처리
 Software implementation

Operations are implemented by a set of instruction(Subroutine)

Ex) MUL, DIV : 그러나 이와 같은 명령어를 갖는 CPU도 있음
 Multiplication Program

Positive Number Multiplication
» X = multiplicand
Y = multiplier
P = Partial Product Sum
» Y 를 AC 에 저장한 후 E 로
Circular Right
Algorithm
Fig. 6-3
Computer System Architecture

E = 1 : P에 1111을 더함

E = 0 : 더하지 않음
X=1111
Y=1011
1111
1111
0000
1111
10100101
Chap. 6 Programming the Basic Computer
P=00000000
+00001111
P=00001111
+00011110
P=00101101
+00000000
P=00101101
+01111000
P=10100101
© Korea Univ. of Tech. & Edu.
Dept. of Info. & Comm.
6-8
Fig. 6-3 flowchart for Multiplication Program
CTR
P
-8
0
E
0
AC
Y
cir EAC
Y
AC
=0
=1
E
P
P+X
E
AC
0
X
cil EAC
X
CTR
AC
CTR + 1
≠0
=0
CTR
Computer System Architecture
Stop
Tab. 6-14 Program to Multiply Two Positive numbers
Line
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
LOP,
ONE,
ZRO,
CTR,
X,
Y,
P,
ORG
CLE
LDA
CIR
STA
SZE
BUN
BUN
LDA
ADD
STA
CLE
LDA
CIL
STA
ISZ
BUN
HLT
DEC
HEX
HEX
HEX
END
100
Y
Y
ONE
ZRO
X
P
P
X
X
CTR
LOP
/A=0
/ A = Y (000B)
/ Circular Right to E
/ Store shifted Y
/ Check if E = 0
/E=1
/E=0
A = X (000F)
/X=X+P
/ Store A to P
/ Clear E
/A=X
/ A = 00011110 (00001111)
/ Store A to X
/ CTR = - 7 = -8 + 1
/ Repeat until CTR = 0
-8
000F
000B
0
Chap. 6 Programming the Basic Computer
© Korea Univ. of Tech. & Edu.
Dept. of Info. & Comm.
6-9
 Double Precision Addition : 32 bits
 하위 AL + BL 먼저 수행하여
Line
E 를 상위에 반영(AH + BH + E) 1
31
+
16 15
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0
AH
AL
BH
BL
CH
CL
AL,
AH,
BL,
BH,
CL,
CH,
LDA
ADD
STA
CLA
CIL
ADD
ADD
STA
HLT
DEC
DEC
DEC
DEC
HEX
HEX
AL
BL
CL
?
?
?
?
0
0
/ Operand
LDA
CMA
STA
LDA
CMA
AND
CMA
A
/
/
/
/
/
/
/
 Logic Operations
 Logic Operation 중에서 OR 명령이 없다(Tab. 6-1)
» 추가 하려면 더 길은 Instruction Format 필요

해결 방법 : DeMorgan’s theorem
»
A B  A B  A  B
Computer System Architecture
AH
BH
CH
TMP
B
TMP
Chap. 6 Programming the Basic Computer
/
/
/
/
/
/
/
/
A = AL
A = AL + BL
Store A to CL
A=0
0000 0000 0000 000?(?=E )
A = 00(E=0) or 01(E=1)
A = A + AH + BH
Store A to CH
Load A
Complement A
Store to TMP
Load B
Complement B
AND
Complement
© Korea Univ. of Tech. & Edu.
Dept. of Info. & Comm.
6-10
 Shift Operations
 Logical Shift : Zero must added to the extreme position
» Shift Right
» Shift Left

0
CLE
CIL
Arithmetic Shift Right
» Positive ( + = 0 )
» Negative ( - = 1 )

E
0
CLE
CIR
0
E
0
E
1
0
1
E
0
CLE
SPA
CME
CIR
/
/
/
/
E= 0
Skip if A= +, E= 0
Toggle E(=1) if A= Circulate A with E
6-7 Subroutines
 Subroutine
 A set of common instruction that can be used in a program many times
 In basic computer, the link between the main program and a subroutine is the
BSA instruction(Branch and Save return Address)
 Subroutine example : Tab. 6-16
Computer System Architecture
Chap. 6 Programming the Basic Computer
© Korea Univ. of Tech. & Edu.
Dept. of Info. & Comm.
6-11
Location
100
101
102
103
104
105
106
107
108
109
10A
10B
10C
10D
10E
10F
110
110
X,
Y,
SH4,
MSK,
ORG
LDA
BSA
STA
LDA
BSA
STA
HLT
HEX
HEX
100
X
SH4
X
Y
SH4
Y
/
/
/
/
/
/
/
Main Program
Load X
Call SH4 with X
Store result X
Load Y
Call SH4 with Y
Store result Y
1234
4321
HEX
CIL
CIL
CIL
CIL
AND
BUN
HEX
END
0
/
/
/
/
Result = 2340
Result = 3210
Subroutine
Save Return Address
MSK
SH4 I
FFF0
/ Mask lower 4 bit
/ Indirect Return to main
/ Mask pattern
Subroutine
CALL hear
Ex) 1234
- CIL 4 회 = 2340
- Mask : AND FFF0
- 결과 = 2340
X = 102
Y = 105
Tab. 6-16 Program to Demonstrate the use of Subroutines
Computer System Architecture
Chap. 6 Programming the Basic Computer
© Korea Univ. of Tech. & Edu.
Dept. of Info. & Comm.
6-12
 Subroutine Parameters & Data Linkage
 Parameter(or Argument) Passing
» When a subroutine is called, the main
program must transfer the data

2 가지 Parameter Passing 방법
» 1) Data transfer through the Accumulator

Used for only single input and single
output parameter
» 2) Data transfer through the Memory

Call 후에 Return
Address를 이용


여러 개의 Operand 전달 가능
Operand are often placed in memory
locations following the CALL
2 개의 Parameter Passing 예 : Tab. 6-17
» First Operand and Result : Accumulator
» Second Operand : Inserted in location
following the BSA

Tab. 6-17 Program to Demonstrate Parameter Linkage
Location
200
201
202
203
204
205
206
207
208
209
20A
20B
20C
20D
20E
20F
210
X,
Y,
OR,
TMP,
ORG
LDA
BSA
HEX
STA
HLT
HEX
HEX
HEX
CMA
STA
LDA
CMA
AND
CMA
ISZ
BUN
HEX
END
200
X
OR
3AF6
Y
7B95
0
0
TMP
OR I
TMP
OR
OR I
0
/
/
/
/
Load first operand X
Call OR with X
Second operand
Subroutine return here(
Y=result)
/
/
/
/
/
/
/
/
/
/
/
First operand
Result store here
Return address =202
Complement X
TMP = X
A = 3AF6 (202 번지의 내용)
Complement Second operand
AND
전체 Complement
Return Address = 202 + 1 =
203
Return to main
BSA후에 2개 Operand 예 : Tab. 6-18
» BSA 후에 2개의 Operand 사용
» Block 전송 Source 와 Destination
Address로 사용
Computer System Architecture
* OR Subroutine
First Operand : X = 7B95
Second Operand : BSA후 = 3AF6
Chap. 6 Programming the Basic Computer
© Korea Univ. of Tech. & Edu.
Dept. of Info. & Comm.
6-13
Tab. 6-18 Subroutine to Move a Block of Data
100
101
102
103
104
105
106
107
108
109
10A
10B
10C
10D
10E
10F
110
111
112
113
114
115
116
117
118
MVE,
LOP,
PT1,
PT2,
CTR,
Computer System Architecture
ORG
BSA
HEX
HEX
DEC
HLT
HEX
LDA
STA
ISZ
LDA
STA
ISZ
LDA
STA
ISZ
LDA
STA
ISZ
ISZ
ISZ
BUN
BUN
HEX
HEX
DEC
100
MVE
200
300
-16
/
/
/
/
0
MVE
PT1
MVE
MVE
PT2
MVE
MVE
CTR
MVE
PT1
PT2
PT1
PT2
CTR
LOP
MVE
?
?
?
/
I /
/
/
I /
/
/
I /
/
/
I /
I /
/
/
/
/
I /
/
/
/
Subroutine Call
Source Address
Destination Address
Number of data to move
2 개의
Operand
Return address=101
A= 200
PT1= 200
Return address=102
A= 300
PT2= 300
Return address=103
A= -16
CTR= -16
Return address=104
A= Address 200의 내용
Address 300에 저장
PT1= 201
PT2= 301
CTR= -15 if 0 skip
Loop until CTR= 0
104 로 Return = HLT
Source
Destination
Counter
Chap. 6 Programming the Basic Computer
* Block Move Subroutine
메모리 200 번지 부터
16개의 데이터를 메모리
300 번지로 이동
© Korea Univ. of Tech. & Edu.
Dept. of Info. & Comm.
6-14

6-8 Input-Output Programming
 One-character I/O
 Programmed I/O 방식
 Two-character I/O
 Two character Packing
Tab. 6-19 Program to input and output One character
Tab. 6-20 Subroutine to input and pack Two character
(a) Input a character
CIF,
SKI
BUN
CIF
INP
OUT
STA
CHR
HLT
CHR,
?
IN2,
FST,
(b) Output a character
LDA
CHR
COF,
SKO
BUN
COF
OUT
HLT
CHR,
HEX
0057
/
/
/
/
/
Check FGI = 1 ?
Go to CIF if FGI= 0
Input character (FGI = 1)
Echo Back
Store character
/ Store character here
/
/
/
/
Load output character
Check FGO = 1 ?
Go to COF if FGO= 0
Output character (FGO = 1)
SCD,
HEX
SKI
BUN
INP
OUT
BSA
BSA
SKI
BUN
INP
OUT
BUN
?
/
/
/
/
/
/
/
FST
SH4
SH4
Save return address
Check if FGI= 1 ?
Loop (FGI = 0)
Input first character
Echo back
Shift left 4 bit
Again(total 8 bit shift)
SCD
/ Input second character
/ Echo back
/ Return
IN2 I
15
8 7
0
1st Char
/ Output character = "W"
1st Char
1st Char
Computer System Architecture
Chap. 6 Programming the Basic Computer
2nd Char
© Korea Univ. of Tech. & Edu.
Dept. of Info. & Comm.
6-15
 Compare Two Word
 Store Input Character in Buffer
Tab. 6-21 Program to store input character in buffer
LOP,
ADS,
PTR,
LDA
STA
BSA
STA
ISZ
BUN
HLT
HEX
HEX
ADS
PTR
IN2
PTR I
PTR
LOP
/
/
/
/
/
/
500
0
/ Buffer address
/ Pointer
Computer System Architecture
Tab. 6-22 Program to compare Two word
Load buffer address A= 500
PTR= 500
Get a character Tab.
(
6-20 )
500 번지에 character 저장
PTR= 501
Endless Loop
WD1,
WD2,
LDA
CMA
INC
ADD
SZA
BUN
BUN
HEX
HEX
WD1
/ Load first word A= WD1
/ Make 2's complement
WD2
/
/
/
/
/
/
UEQ
EQL
?
?
Chap. 6 Programming the Basic Computer
WD2 - WD1
Skip if A=0 E
( qual)
Unequal
Equal
first word
second word
© Korea Univ. of Tech. & Edu.
Dept. of Info. & Comm.
6-16
 Interrupt Program
 Interrupt Condition
Location
0
1
» Interrupt F/F R = 1
when IEN = 1 and [FGI or FGO = 1]
» Save return address at 0000
Interrupt
» Jump to 0001 (Interrupt Start)
Here

ZRO,
100
101
102
103
104
ORG
HEX
BUN
0
?
SRV
/ Save Interrupt Return Address
/ Jump to ISR
ORG
CLA
ION
LDA
ADD
STA
100
/ Main program
ORG
STA
CIR
STA
SKI
BUN
INP
OUT
STA
ISZ
SKO
BUN
LDA
OUT
ISZ
LDA
CIL
LDA
ION
BUN
HEX
HEX
HEX
HEX
200
SAC
/ Turn on Interrupt(IEN= 1)
X
Y
Z
/ Interrupt occurs here
/ Return Address(104 )
Interrupt Service Routine(ISR)
»
»
»
»
»
»
1) Save Register (AC, E)
2) Check Input or Output Flag
3) Input or Output Service Routine
4) Restore Register (AC, E)
5) Interrupt Enable (ION)
6) Return to the running program
Modified Fetch Cycle 과
Reset 시에 IEN = 0 이 된다
(p. 158, Fig. 5-15)
200
201
202
203
204
205
206
207
208
SRV,
NXT,
EXT,
SAC,
SE,
PT1,
PT2,
Computer System Architecture
Chap. 6 Programming the Basic Computer
SE
NXT
PT1
PT1
EXT
PT2
I
I
PT2
SE
SAC
ZRO
?
?
300
400
I
/
/
/
/
/
/
/
/
/
/
/
/
/
/
Save A to SAC
Move A into A(15)
Save E to SE
Check if FGI= 1?
No, FGI= 0, Check FGO
Yes, FGI= 1, Character Input
Echo back
Store in input buffer(PT1)
PT1 + 1
Check if FGO= 1?
No, FGO= 0, Exit
Yes, FGO= 1, Get output character
Character output
PT2 + 1
/
/
/
/
Restore E
Restore A
Interrupt ON
Return to running program(104 )
/ Input Buffer Address
/ Output Buffer Address
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