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

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^ ^

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