sp mid-term exam
DESCRIPTION
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%). Answer. Q2. Briefly answer the following questions (40%). - PowerPoint PPT PresentationTRANSCRIPT
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%).
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(B)=006000(PC)=003000
(X)=000090
3030 003600
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3600 103000
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6390 00C303
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C303 003030
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Answer
Machine instruction Value loaded into register AHex Binary Target
addressop n i x b p e disp/address
022030 000000 1 0 0 0 1 0 0000 0011 0000 3030 103000
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(B)=006000(PC)=003000
(X)=000090
3030 003600
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3600 103000
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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 S
IC/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
100F STR1 BYTE C’TEST STRING’ 544553…
101A STR2 RESB 11
1025 ZERO WORD 0 000000
1028 ELEVEN WORD 11 00000B
102B END FIRST
Answer
HSTRCPY00100000002BT0010000F04102550900F54901A2C1028381003
T00100F0B5445535420535453494E47T0010250600000000000BE001000
^
^
^
^
^
^ ^
^
^
^
^ ^ ^ ^
^
^
^ ^
Q4. Translate (by hand) the following assembly program to SIC/XE object code.
Source statement
STRCP2 START 1000
FIRST LDT #11
LDX #0
MOVECH LDCH STR1,X
STCH STR2,X
TIXR T
JLT MOVECH
STR1 BYTE C'TEST STRING'
STR2 RESB 11
END FIRST
Loc Source statement Object code
1000 STRCP2 START 1000
1000 FIRST LDT #11 75000B
1003 LDX #0 050000
1006 MOVECH LDCH STR1,X 53A008
1009 STCH STR2,X 57A010
100C TIXR T B850
100E JLT MOVECH 3B2FF5
1011 STR1 BYTE C'TEST STRING' 544553…
101C STR2 RESB 11
1027 END FIRST
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:
QUIZ START 0
FIRST LDA #3
STX THREE
LDX #0
+LDS THREE
ADDR A,X
+STA RESULT,X
RESULT RESW 1
THREE RESW 1
END FIRST
0000000300060009000D000F00130016
0100031320100500006F10001690010F900013
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
1003 LDB #TWO 69200B
BASE #TWO
1006 ADD #K100 1B4003
1009 MUL #K100 232008
100C LDS #TWO 6F4000
100F DIVR #S,A 9C40
1011 TWO WORD #2 000002
1014 K100 WORD #100 000064
1017 END #FIRST
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
LocRegister
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
[(1+100)*100]/2=5050A=13BA