CSc 343
Computer Organization and Programming
Fall Quarter 1997
Mid-Term Exam (24 October 1997, Friday) With Solutions
1. Convert this UNSIGNED BINARY number to DECIMAL: 10011001
A. 88 B. 102 C. 143 D. 200 E. None of the previous
ANSWER: E
2. Convert this SIGNED MAGNITUDE BINARY number to DECIMAL: 10011001
A. -10 B. -25 C. -50 D. -100 E. None of the previous
ANSWER: B
3. Convert this SIGNED ONE'S COMPLEMENT BINARY number to DECIMAL: 10011001
A. -29 B. -88 C. -102 D. -200 E. None of the previous
ANSWER: C
4. Convert this SIGNED TWO'S COMPLEMENT BINARY number to DECIMAL: 10011001
A. -29 B. -88 C. -102 D. -200 E. None of the previous
ANSWER: E
5. What range of DECIMAL NUMBERS fit in a 12 bit UNSIGNED binary field.
A. 0...2047 B. 0...2048 C. 0...4095 D. 0...4096 E. None of those
ANSWER: C
6. What range of DECIMAL NUMBERS fit in a 12 bit binary SIGNED TWO'S
COMPLEMENT field.
A. -2047...+2048 B. -2048...+2047 C. -4095...+4096 D.-4096...+4095
E. None of the above
ANSWER: B
7. Convert the number -117.375 to IEEE single-precision floating point format.
Give your answer in HEX.
ANSWER: C2EAC000
8. What does the following IEEE single-precision floating point
number represent?
1 11111111 00000000000000000000000
ANSWER: - infinity
For questions 9-12, perform the specified 8 bit binary operation.
Generate your answers and the values of the Condition Code (SF, ZF, OF,
CF) the way they would be set if these operations were performed by the
Intel 8086.
9. What is the result of performing the following binary operation.
10111011
+ 10111011
--------
A. 10111011 B. 11110110 C. 01110110 D. 01111000 E. None of those
ANSWER: C
10. Based on the operation performed by the previous question, how will
the Condition Code bits be set.
A. SF=0 ZF=0 OF=0 CF=1 B. SF=0 ZF=0 OF=1 CF=1
C. SF=0 ZF=1 OF=0 CF=1 D. SF=1 ZF=0 OF=1 CF=1 E. None of the above
ANSWER: B
11. What is the result of performing the following binary operation.
00010001
- 01000100
--------
A. 10000001 B. 11001111 C. 11000001 D. 11001101 E. None of those
ANSWER: D
12. Based on the operation performed by the previous question, how will
the Condition Code bits be set.
A. SF=1 ZF=0 OF=0 CF=1 B. SF=1 ZF=0 OF=1 CF=1
C. SF=0 ZF=0 OF=0 CF=1 D. SF=1 ZF=0 OF=1 CF=0 E. None of the above
ANSWER: E
13. Convert the address EE11:EE22 to a 20-bit address, where EE11 is the
contents of the segment register and EE22 is the offset within the
segment. Express the answer in HEX notation.
ANSWER: FCF32
14. Convert the 20-bit address 00100H to segment:offset notation.
Express the segment and offset as 4-digit HEX numbers.
ANSWER: 0000:0100
15. What is the SOURCE ADDRESSING MODE for the instruction: ADD AX,[1000]
A. Register Direct B. Immediate C. Memory Direct
D. Register Indirect E. It can not be determined
ANSWER: C
16. What is the DESTINATION ADDRESSING MODE for the instruction: ADD AX,[1000]
A. Register Direct B. Immediate C. Memory Direct
D. Register Indirect E. It can not be determined
ANSWER: A
17. Given these data declaration and instructions and initial hex
value of AX=1234 then what is in AX after the last instruction
completes execution.
LIST DB 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
MOV SI,OFFSET LIST
ADD SI,5
MOV AL,[SI]
A. 1203 B. 1204 C. 1205 D. 1206 E. None of the above
ANSWER: C
18. Given these data declaration and instructions and initial hex
value of AX=1234 then what is in AX after the last instruction
completes execution.
LIST DB 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
MOV SI,OFFSET LIST
ADD SI,10
MOV AX,[SI]
A. 120A B. 120B C. 0A0B D. 0B0A E. None of the above
ANSWER: D
19. Given the following data declarations and instructions, how
many characters will be written to the standard output stream
(display or file).
MSG0 DB `HELLO'
MSG1 DB `VAL='
VALUE DB 36D
FIN DB 13,10,`$'
MSG2 DB `DONE',`$'
MOV DX,OFFSET MSG0
MOV AH,9
INT 21H
A. 0 (decimal)
B. 9 (decimal)
C. 11 (decimal)
D. 12 (decimal)
E. The number of characters can not be determined
ANSWER: B
Questions 20-23 refer to the following code:
MOV AX,0
MOV CX,10
REPEAT: INC AX
SUB CX,1
XXX REPEAT
DONE:
20. If the conditional jump (XXX) is replaced by JNE then what
hex value will be in AX when the loop terminates (we reach the
instruction DONE:)
A. 0000 B. 0001 C. 000A D. 000B E. None of the above
ANSWER: C
21. If the conditional jump (XXX) is replaced by LOOP then what
hex value will be in AX when the loop terminates (we reach the
instruction DONE:)
A. 0004 B. 0005 C. 000A D. 000B E. None of the above
ANSWER: B
22. If the conditional jump (XXX) is replaced by JLE then what hex
value will be in AX when the loop terminates (we reach the
instruction DONE:)
A. 0000 B. 0001 C. 000A D. 000B E. None of the above
ANSWER: B
23. If the conditional jump (XXX) is replaced by JGE then what
hex value will be in AX when the loop terminates (we reach the
instruction DONE:)
A. 0000 B. 0001 C. 000A D. 000B E. None of the above
ANSWER: D
24. How does the instruction ADD AX,100D convert into hex machine code.
A. 00 01 00
B. 01 64 00
C. 03 00 64
D. 04 01 00
E. 05 64 00
ANSWER: E
25. What symbolic instruction is converted into the hex machine code 51
A. MOV AL,1
B. ADD AX,BX
C. PUSH CX
D. CMP DX,1
E. INC BX
ANSWER: C
26. The following instructions use indirect addressing to point
to data that is moved to the AX register. Which of these
instructions refers to data from the Stack Segment that
is moved to AX.
A. MOV AX,[SI]
B. MOV AX,[DI]
C. MOV AX,[BX]
D. MOV AX,[BP]
E. None of the above
ANSWER: D
27. Write a single instruction that clears (sets to 0) the AX register.
Do not use the SUB or the MOV instruction.
XOR AX,AX
28. Name the instruction that could be used to test the bit that was
``shifted off'' during a shift operation.
JC
29. Suppose that AL contains 93H and BL contains 44H. What is the outcome of
the following instructions?
CMP AL,BL
JE A10 Jumps_______ Does not jump___X___
CMP AL,BL
JNZ A10 Jumps___X___ Does not jump_______
CMP AL,BL
JAE A10 Jumps___X___ Does not jump_______
CMP AL,BL
JL A10 Jumps___X___ Does not jump_______
CMP AL,BL
JB A10 Jumps_______ Does not jump___X___
30. List the bytes that would be generated by each of the
following directives. Show byte values in HEX; Write one
byte per line. (You may not need all the lines provided).
Show bytes in the order they will appear in memory.
DB 'Foo','bar' _46_ _6F_ _6F_ _62_ _61_ _72_ ____ ____ ____ ____
DB -128 _80_ ____ ____ ____ ____ ____ ____ ____ ____ ____
DB 110B,110D _06_ _6E_ ____ ____ ____ ____ ____ ____ ____ ____
DB '110' _31_ _31_ _30_ ____ ____ ____ ____ ____ ____ ____
DW 110 _6E_ _00_ ____ ____ ____ ____ ____ ____ ____ ____
DW 0FFFFH,-1 _FF_ _FF_ _FF_ _FF_ ____ ____ ____ ____ ____ ____
DW 4 DUP (123H) _23_ _01_ _23_ _01_ _23_ _01_ _23_ _01_ ____ ____
31. The following assembler instructions have been partially
translated to machine code. Finish the translation by
filling in the missing words.
Offset Machine Code Source Code
______ _____________ __________________________
0000 .model small
0000 .stack 32
0000 .data
0000 .code
0000 proc1 proc near
0000 E8 ____ (ANSWER: 0004) call proc2
0003 05 0001 add ax,1
0006 C3 ret
0007 proc1 endp
0007 proc2 proc near
0007 03 C0 add ax,ax
0009 C3 ret
000A proc2 endp
000A main proc far
000A B8 0001 mov ax,1
000D E8 ____ (ANSWER: FFF0) call proc1
0010 E8 ____ (ANSWER: FFF4) call proc2
0013 B4 4C mov ah,4cH
0015 CD 21 int 21h
0017 main endp
end main
32. Show the top two words on the stack after PROC2 is called from
PROC1 but before PROC2 has executed the RET instruction. Give
your answers in HEX words.
Top Word: _0300__
Next Word: _1000__
33. Show the contents (in HEX) of the AX register immediately after
the call to PROC2 is completed in the main program.
AX __0006___
Raj Sunderraman
Tue Oct 28 12:25:52 PST 1997