1. System Programming Mid-Term 2
Date: December 30th, 2005
Time: 09:10-12:00
2006/1/6

2. Q1(10%)
Translate (by hand) the following assembly program to SIC/XE object code.
Loc
Source statement
Object code
SUM2
#START #0
FIRST
#LDX
#LDA
+LDB
#TABLE2
#BASE
LOOP
#ADD
#TABLE,X
#TABLE2,X
#TIX
#COUNT
#JLT
#LOOP
+STA
#TOTAL
#RSUB
COUNT
#RESW #1
TABLE
#2000
TABLE2
TOTAL
#END
#FIRST

3. Q1(10%)
Translate (by hand) the following assembly program to SIC/XE object code.
Loc
Source statement
Object code
SUM2
#START #0
0000
FIRST
#LDX
050000
0003
#LDA
010000
0006
+LDB
#TABLE2
#BASE
000A
LOOP
#ADD
#TABLE,X
1BA013
000D
#TABLE2,X
1BC000
0010
#TIX
#COUNT
2F200A
0013
#JLT
#LOOP
3B2FF4
0016
+STA
#TOTAL
0F102F00
001A
#RSUB
4F0000
001D
COUNT
#RESW #1
0020
TABLE
#2000
1790
TABLE2
2F00
TOTAL
2F03
#END
#FIRST

4. Q1(10%)
HSUM F03
T BA0131BC000
T D2F200A3B2FF40F102F004F0000
E000000 ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^

5. Q2(10%)
Disassemble (convert object code back into assembly language) the following SIC/XE program.
　HSUM E
　T E D000B3B2FF8
　E001000

6. Q2(10%) Loc Source statement Object code 1000 SUM START #1000 FIRST
LDA #0
010000
1003
LDX #1
050001
1006
LOOP
ADDR
#X,A
9010
1008
TIX
#11
2D000B
100B
JLT
#LOOP
3B2FF8
100E
END
#FIRST

7. Q3(20%)
Disassemble the following SIC program. (You may choose meaningful labels for those addresses as you wish.)
HEXAM A T B E C E C004B3C0006E000 T ED E1C F1C004E4C T A000010F E000000

8. Q3(20%) Loc Source statement Object code EXAM2 START 0000 FIRST LDA
0000
FIRST
LDA
ZERO
00004B
0003
LDX
K128
040048
0006
LOOP
JSUB
GETC
48001E
0009
MUL
K16
200054
000C
STA
ZERO2
0C0058
000F
0012
ADD
180058
0015
STCH
FIRST,X
548000
0018
TIX
2C004B
001B J
3C0006
001E TD
DEV
E00057
0021
JEQ
30001E
0024 RD
D80057
0027
COMP K4
280042
002A 80
300080

9. Q3(20%) 002D COMP K48 280045 0030 JLT GETC 38001E 0033 SUB 1C0045 0036
280051
0039
RETURN
38003F
003C K7
1C004E
003F
RSUB
4C0000
0042 K4
WORD 4
000004
0045 48
000030
0048
K128
128
000080
004B
ZERO
000000
004E 7
000007
0051 10
00000A
0054
K16 16
000010
0057
DEV
BYTE
X’F1’ F1
0058
ZERO2
005A
END
FIRST

10. Q4(20%) Explain the function of the above program.
Bootstrap loader (p.128)
If we modify the first three bytes from 00004B to , will it make any difference? Why?
No, because of both contents loaded are zero.
If we modify Bytes[18:1A] from 2C004B to 2C0000, will it make any difference? Why?
No, because of following instruction is “J”.
What is the ID of the device accessed by this program? F1

11. Q5(20%)
Consider the three control sections in Figure 3.8. Suppose they are linked in the sequence of PROGC+PROGB+PROGA, and the beginning load address is Print the load map (generated after Pass 1) that shows the external symbols and their addresses.

12. Q5(20%) Control section Symbol name Address Length PROGC 1000 0051
LISTC
1030
ENDC
1042
PROGB
1051
007F
LISTB
10B1
ENDB
10C1
PROGA
10D0
0063
LISTA
1110
ENDA
1124

13. Q6(20%) REF5 in PROGA ENDC-LISTC-10 001042-001030+FFFFF6=000008
REF5 in PROGC
000008
REF8 in PROGA
LISTB-LISTA-PROGA
0010B1+FFFFC0-0010D0=FFFFA1
REF8 in PROGB
LISTB+PROGB-LISTA
=FFFFA1
REF8 in PROGC
LISTB-LISTA
0010B =FFFFA1