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SP Mid-Term Exam
Chang-Yu Wu
2006/12/03
Q1. Consider the memory contents shown in the following figure.
What would be loaded to register A with the instruction 022030
(hexidecimal)? (10%)
.
.
.
.
.
.
3030
003600
.
.
.
.
.
.
3600
103000
.
.
.
.
.
.
6390
00C303
.
.
.
.
.
.
C303
003030
.
.
.
.
.
.
(B)=006000
(PC)=003000
(X)=000090
Answer
Machine instruction
Hex
Binary
op
022030
n i x b p e
000000 1 0 0 0 1 0
disp/address
0000 0011 0000
.
.
.
.
.
.
3030
003600
.
.
.
.
.
.
3600
103000
.
.
.
.
.
.
(B)=006000
(PC)=003000
(X)=000090
Target
address
3030
Value loaded
into register A
103000
Q2. Briefly answer the following questions (40%)
1. Why does Beck's "System Software"
textbook design a hypothetical computer SIC
to present the concepts in his book?
Ans:
illustrate the most commonly encountered
hardware features and concepts, while
avoiding most of the idiosyncrasies. (p.4)

2. What are the maximum memory size of
SIC and SIC/XE, respectively?
Ans:
215/220 bytes(p.5/p.7)
 3.How many addressing modes do SIC and
SIC/XE support, respectively?
Ans:
2/18 (p.6/p.499)

4.How many instructions are there in the
SIC/XE instruction set?
Ans:
59 (p.496~498)
 5.How does SIC determine whether the result
of a TD instruction is successful?
Ans:
CC (condition code) = LT means the device is
ready (p.7)

6. What are the two general purpose
registers in SIC/XE?
Ans:
S,T (p.7)
 7. What is the size of F register?
Ans:
48bits (p.7)

8. Why does the assembler need to handle
the source files in two passes?
Ans:
Pass1: assign addresses to all symbol
Pass2: generate object code in order to solve
the problem of forward reference

Q3. Translate (by hand) the following assembly program to SIC
object code. (The output format will look like Figure 2.3, which
contains H record, T record, and E record.) (10%)
Source statement
STRCPY
START
1000
FIRST
LDX
ZERO
MOVECH
LDCH
STR1,X
STCH
STR2,X
TIX
ELEVEN
JLT
MOVECH
STR1
BYTE
C’TEST STRING’
STR2
RESB
11
ZERO
WORD
0
ELEVEN
WORD
11
END
FIRST
Loc
Source statement
Object code
1000
STRCPY
START
1000
1000
FIRST
LDX
ZERO
041025
1003
MOVECH
LDCH
STR1,X
50900F
1006
STCH
STR2,X
54901A
1009
TIX
ELEVEN
2C1028
100C
JLT
MOVECH
381003
544553…
100F
STR1
BYTE
C’TEST STRING’
101A
STR2
RESB
11
1025
ZERO
WORD
0
000000
1028
ELEVEN
WORD
11
00000B
END
FIRST
102B
Answer
HSTRCPY00100000002B
^
^
^
T0010000F04102550900F54901A2C1028381003
^
^ ^
^
^
^
^
T00100F0B5445535420535453494E47
^
^ ^
T0010250600000000000B
^
^ ^
^
E001000
^
Q4. Translate (by hand) the following assembly program to
SIC/XE object code.
Source statement
STRCP2
START
1000
FIRST
LDT
#11
LDX
#0
LDCH
STR1,X
STCH
STR2,X
TIXR
T
JLT
MOVECH
STR1
BYTE
C'TEST STRING'
STR2
RESB
11
END
FIRST
MOVECH
Loc
Source statement
Object code
1000
STRCP2
START
1000
1000
FIRST
LDT
#11
75000B
LDX
#0
050000
LDCH
STR1,X
53A008
1009
STCH
STR2,X
57A010
100C
TIXR
T
B850
100E
JLT
MOVECH
3B2FF5
544553…
1003
1006
MOVECH
1011
STR1
BYTE
C'TEST STRING'
101C
STR2
RESB
11
END
FIRST
1027
Answer
HSTRCP2001000000027
^
^
^
T0010001175000B05000053A00857A010B8503B2FF5
^ ^
^
^
^
^
^
^
T0010110B5445535420535452494E47
^
^ ^
E001000
^
Q5. Disassemble (convert object code back into assembly language)
the following SIC/XE program. (10%)
HQUIZ 000000000019
T000000130100031320100500006F10001690010F900013
E000000
Answer:
0000
0003
0006
0009
000D
000F
0013
0016
QUIZ
FIRST
RESULT
THREE
START 0
LDA
#3
STX
THREE
LDX
#0
+LDS
THREE
ADDR
A,X
+STA
RESULT,X
RESW
1
RESW
1
END
FIRST
010003
132010
050000
6F100016
9001
0F900013
Q6. Simulate (by hand) the execution of the above program and let
the breakpoint by 000013. What would be the value of register X
when (pc) = 000013? (10%)
LOC
0000
0003
0006
0009
000D
000F
0013
A
F
3
3
3
3
3
3
X
F
F
F
0
0
3
3
REGISTER
A=3
X=3
Q7. Simulate (by hand) the execution of the object code below. Let
the breakpoint be 001011. What is the value of register A when (PC)
=001011? (10%)
HGAUSS 001000000017
T0010000C01000169200B1B4003232008
T00100C0B6F40009C40000002000064
E001000
Answer
Loc
Source statement
Object code
1000
GAUSS
START
#1000
1000
FIRST
LDA
#1
010001
LDB
#TWO
69200B
BASE
#TWO
1006
ADD
#K100
1B4003
1009
MUL
#K100
232008
100C
LDS
#TWO
6F4000
100F
DIVR
#S,A
9C40
1003
1011
TWO
WORD
#2
000002
1014
K100
WORD
#100
000064
END
#FIRST
1017
Machine instruction
Hex
Binary
Code
op
n i x b p e
disp/address
69200B
0110 10
0 1 0 0 1 0
0000 0000 1011
LDB
#TWO
1B4003
0001 10
1 1 0 1 0 0
0000 0000 0011
ADD
#K100
232008
0010 00
1 1 0 0 1 0
0000 0000 1000
MUL
#K100
6F4000
0110 11
1 1 0 1 0 0
0000 0000 0000
LDS
#TWO
Loc
1000
1003
1006
1009
100C
100F
1011
A
F
1
1
101
10100
10100
5050
B
F
F
2
2
2
2
2
S
F
F
F
F
F
2
2
Register
[(1+100)*100]/2=5050
A=13BA