Download System Programming Mid-Term

System Programming Mid-Term
Date: October 28th, 2005
Time: 09:10-12:00
Q1(10%)

Consider the memory contents
shown in the following figure.
What would be loaded to register A
with the instruction 022030
(hexidecimal)?
.
.
.
.
.
.
3030
003600
.
.
.
.
.
.
3600
103000
.
.
.
.
.
.
6390
00C303
.
.
.
.
.
.
C303
003030
.
.
.
.
.
.
(B)=006000
(PC)=003000
(X)=000090
2
Q1(10%)
Machine instruction
Hex
Binary
022030
op
n i x b p e
disp/address
000000
1 0 0 0 1 0
0000 0011 0000
.
.
.
.
.
.
3030
003600
.
.
.
.
.
.
3600
103000
.
.
.
.
.
.
Value loaded into
register A
Target
address
3030
103000
(B)=006000
(PC)=003000
(X)=000090
3
Q2(40%)





Why does Beck's "System Software" textbook
design a hypothetical computer SIC to present the
concepts in his book?
illustrate the most commonly encountered hardware
features and concepts, while avoiding most of the
idiosyncrasies (p.4)
easy to learn
What are the maximum memory size of SIC and
SIC/XE, respectively?
215/220 (p.5/p.7)
4
Q2(40%)






How many addressing modes do SIC and SIC/XE
support, respectively?
2/18 (p.6/p.499)
How many instructions are there in the SIC/XE
instruction set?
59 (p.496~498)
How does SIC determine whether the result of a TD
instruction is successful?
CC (condition code) = LT means the device is ready (p.7)
5
Q2(40%)








What are the two general purpose registers in
SIC/XE?
S,T (p.7)
What is the size of F register?
48bits (p.7)
Why does the assembler need to handle the source
files in two passes?
Pass1 assign addresses to all symbol
Pass2 generate object code
in order to solve the problem of forward reference
6
Q3(10%)

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.)
Loc
STRCPY
FIRST
MOVECH
STR1
STR2
ZERO
ELEVEN
Source statement
START
1000
LDX
ZERO
LDCH
STR1,X
STCH
STR2,X
TIX
ELEVEN
JLT
MOVECH
BYTE
C’TEST STRING’
RESB
11
WORD
0
WORD
11
END
FIRST
Object code
7
Q3(10%)

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.)
Loc
1000
1000
1003
1006
1009
100C
100F
101A
1025
1028
102B
STRCPY
FIRST
MOVECH
STR1
STR2
ZERO
ELEVEN
Source statement
START
1000
LDX
ZERO
LDCH
STR1,X
STCH
STR2,X
TIX
ELEVEN
JLT
MOVECH
BYTE
C’TEST STRING’
RESB
11
WORD
0
WORD
11
END
FIRST
Object code
041025
50900F
54901A
2C1028
381003
544553…
000000
00000B
8
Q3(10%)
Machine instruction
Code
Binary
op
x
disp/address
Hex
LDCH
STR1,X
01010000
1
001 0000 0000 1111
50900F
STCH
STR2,X
01010100
1
001 0000 0001 1010
54901A
9
Q3(10%)
HSTRCPY00100000002B
^
^
^
T0010000F04102550900F54901A2C1028381003
^
^ ^
^
^
^
^
T00100F0B5445535420535453494E47
^
^ ^
T0010250600000000000B
^
^ ^
^
E001000
^
10
Q4(10%)

Translate (by hand) the following assembly program
to SIC/XE object code.
Loc
Source statement
STRCP2
START
1000
FIRST
LDT
LDX
#11
#0
MOVECH
LDCH
STR1,X
STCH
STR2,X
TIXR
T
JLT
MOVECH
STR1
BYTE
C'TEST STRING'
STR2
RESB
11
END
FIRST
Object code
11
Q4(10%)

Translate (by hand) the following assembly program
to SIC/XE object code.
Loc
Source statement
Object code
1000
STRCP2
START
1000
1000
FIRST
LDT
LDX
#11
#0
75000B
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
050000
12
Q4(10%)
Machine instruction
Code
Binary
op
n i x bpe
Hex
disp/address
LDCH
STR1,X
0101 00 1 1 1 0 1 0 0000 0000 1000
53A008
STCH
STR2,X
0101 01 1 1 1 0 1 0 0000 0001 0000
57A010
JLT
MOVECH
0011 10 1 1 0 0 1 0 1111 1111 0101
3B2FF5
13
Q4(10%)
HSTRCP2001000000027
^
^
^
T0010001175000B05000053A00857A010B8503B2FF5
^
^ ^
^
^
^
^
^
T0010110B544553542053545249
^
^ ^
E001000
^
14
Q5(10%)

Disassemble (convert object code back into
assembly language) the following SIC program.
HSUM 00100000000E
T0010000E01000005000190102D000B3B2FF8
E001000
15
Q5(10%)
Loc
Source statement
Object code
1000
SUM
START
#1000
1000
FIRST
LDA
#0
010000
LDX
#1
050001
ADDR
#X,A
9010
1008
TIX
#11
2D000B
100B
JLT
#LOOP
3B2FF8
100E
END
#FIRST
1003
1006
LOOP
16
Q5(10%)
Machine instruction
Hex
Binary
op
3B2FF8
n i x bpe
0011 10 1 1 0 0 1 0
Code
disp/address
1111 1111 1000 JLT
LOOP
17
Q6(10%)



Simulate (by hand) the execution of the above program and let
the breakpoint be 00100E. What would be the value of register
A and register X when (PC) = 00100E?
Iteration
Register
1
2
3
4
5
6
7
8
9
10
11
A
0
1
3
6
10
15
21
27
36
45
55
X
1
2
3
4
5
6
7
8
9
10
11
1+2+3+……+9+10=55
A=37, X=B
18
Q7(10%)

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?
HGAUSS 001000000017
T0010000C01000169200B1B4003232008
T00100C0B6F40009C40000002000064
E001000
19
Q7(10%)
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
20
Q7(10%)
Machine instruction
Hex
Binary
op
n i x bpe
Code
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
21
Q7(10%)
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
22
Homework





The code can work well  greater than 7
If I need to correct it  less then 5
Second chance  23:59 on Nov. 4th (tonight)
Subject of email: [SP HW1v2] ########
Email to me: [email protected]
23