Microprocess Assignment
1.Name all eight 32-bit general-purpose registers. Name all six segment
.registers
Name at least four CPU status flags
Answer
-Name all eight 32-bit general-purpose registers
EAX, EBX, ECX, EDX, ESI, EDI, ESP, EBP
-Name all six segment registers
CS, DS, ES, FS, GS, SS
-Name at least four CPU status flags
Most common: Carry, Sign, Zero, Overflow. Less common: Auxiliary
.Carry, Parity
2.What is the purpose of the IP/EIP register? What special purpose does
the ECX register serv?
Answer
What special purpose does the ECX register?
Loop counter
3.Besides the stack pointer (ESP), which other register points to variables
on the stack
Answer
EBP
1
4.Which flag is set when the result of a signed arithmetic operation is too
wide to fit into the destination.
Answer
Overflow
Which flag is set when an operation generates a negative result?
Answer… Sign
5.What is the range of addressable memory in protected mode?
Answer
GByte 0 to FFFFFFFFh
What is the range of addressable memory in Real-address mode?
Answer
MByte 0 to FFFFFh
6.In Real-address mode, convert the following segment-offset address to
.a linear address: 0CD1:02E0
Answer.
0CFF0h
7.In Protected mode, which register references the descriptor for the stack
segment?
Answer
SS Register
2
In Protected mode, which table contains pointers to the various segments
used by a single program.
Answer.
Local descriptor table
8.In the real mode, show the starting and ending addresses of each
segment located by the following segment register values: (a) 1000H (b)
1234H (c) 2300H (d) E000H.
Answer
Staring Address
Ending Address
1000H
0000H
1000 H
1234H
1001 H
1001 H
2300H
2000 H
2000 H
E000H
E000 H
E000 H
9.Find the memory address of the next instruction executed by the
,microprocessor
:when operated in the real mode, for the following CS: IP combinations
(a) CS =1000H and IP = 2000H
(b) CS = 2000H and IP = 1000H
(c) CS = 2300H and IP = 1A00H
(d) CS =1A00H and IP =B000H
3
Answer
(a) CSx10+IP= 10000H + 2000H=12000H
(b) CSx10+IP= 20000H + 1000H=21000H
(c) CSx10+IP= 23000H + 1A00H=24A00H
(d) CSx10+IP= 1A000H + B000H=25000H
10. Which 32-bit register or registers are used as an offset address for .
data segment data in the Pentium 4 microprocessor
Answer
EAX, EBX, ECX, EDX, ESI, EDI, EBP AND ESP
11.Determine the memory location addressed by the following real mode
80286 register combinations :
(a) DS = 1000H and DI = 2000H
DS = 2000H and SI = 1002H
SS = 2300H and BP = 3200H
DS = A000H and BX = 1000H
Answer
(a) DSx10+DI =10000H +2000H=12000H
(b) DS x10+SI= 20000H +1002H=21002H
(c) SS x10+BP= 23000H +3200H=26200H
(d) DSx10+BX=A0000H +1000H=A1000H
4
12.Determine the memory location addressed by the following real
mode Pentium 4 register combinations:
(a) DS = 2000H and EAX = 00003000H
(b)DS = 1A00H and ECX = 00002000H
)c)DS = C000H and ESI = 0000A000H
(d) SS = 8000H and ESP = 00009000H
Answer
DSx10+EAX=20000H+00003000H=23000H
DSx10+ECX=1A000H+00002000H=1C000H
DSx10+ESI=C0000H+0000A000H=0CA000H
DSx10+ESP=80000H+00009000H=89000H
13.Protected mode memory addressing allows access to
which area of the memory in the 80286 microprocessors and to
which area of the memory in the Pentium II microprocessor?
Answer
Protect mode tending to (80286or more) permits admittance to
information located over the main 1m byte just as inside the
primary 1m byte.
14.What is the purpose of the segment register in
protected mode memory addressing.
Answer
In protect mode, segment register simply point to data structure
called segment descriptors that contain the information needed
to access physical memory location.
5
15.How is the stack top address calculated.
Answer
The SP store the offset of the stack top from the base of stack segment.
For example
Ss=0*7000 and SP=0*000 ss left shifted by 4bits =0*70000
SP=0*0000 the sum =0*70000 is the physical address of stack top
Assembly
[1*] Which of the following instructions cannot be coded in the
8086/88 assembly language and why?
(a) MOV AX,27
(d) MOV CX, 397
(g) MOV DS, BX
(b) MOV AL,978
(e) MOV SI,9516
(h) MOV BX, CS
(c) MOV DS, 9BF2
(f) MOV CS, 3490
(I) MOV AX, DL
(j) MOV BL, CX (k) MOV CS, SS (l) MOV [BX], [DI]
(m) MOV [BX],10H
Answer
6
(b) MOV AL,978
(c) MOV DS, 9BF2
7
(f) MOV CS, 3490
(k) MOV CS, SS
8
MOV [BX], [DI]
9
[2*] Suppose that DS = 0200H, BX = 0200H and DI = 0300H.
Determine the data memory address accessed by each of the
following instructions
(a) MOV AL, [2000H]
AL
(b) MOV AL, [BX] (c) MOV [DI],
Answer
(a)DSx10+2000H=2000H+2000H=4000H
(b)DSx10+200H=2000H+200H=2200H
(c)DSx10+300H=2000H+300H=2300H
10
[3*] Assume memory locations of the following
contents:DS:6826=48 and DS: 6827=22.Show the content of
BX after the execution of the instruction c MOV BX,
[6826].
Answer
[4*] If AX=3245H, BX=5632H, CX=672BH, CS=2000H,
SI=62F4H, find the result of each line of the
following program :( the memory map is shown
below)
11
(a) MOV DS, AX (b) MOV SI,100H (c) MOV BX,04H
(d) ADD AX, [SI] Memory location data:
Memory location
Data
32557H
2DH
32556H
24H
32555H
3CH
32554H
8BH
32553H
79H
32552H
6FH
32551H
6AH
32550H
34H
Answer
12
13
5] Identify the addressing modes, the source, and the
destination operands in the following instructions:
(a) MOV DL, BL (b) MOV AX, 00FFH (c) MOV [DI], AX
(d) MOV DI, [SI] (e) MOV [BX+100H], CX
(f) MOV 100H[DI], AH (g) MOV [BX][DI]+100H, AL
Answer
Source
destination
(a) MOV DL, BL
Register
Register
(b) MOV AX, 00FFH
Immediate
Register
(c) MOV [DI], AX
Register
Register
indirect
(d) MOV DI, [SI]
Register
indirect
Register
(e) MOV [BX+100H], CX
Register
Based
(f) MOV 100H[DI], AH
Register
Indexed
(g) MOV [BX][DI]+100H, AL
Register
Based with
16bit
displacement
14
[6*] Given that: DS = 1000H, SS = 2000H, BP = 1000H,
DI = 0100H, BX = 0200H, LIST = 0250H, SI = 0500H and
LABEL
= 1234H. Determine the data memory address and
the addressing modes for each of the following
instructions:
(a) MOV AL, [BP][DI]+LABEL (b) MOV CX, [DI]
(b) MOV DX, [BP] (d) MOV LIST[SI], DX
(e) MOV BL, [BX][SI]+LIST (f) MOV BH, [BX]+LIST
Answer
(a)DSx10+[1000H] [0500H] +1234H= 502BE2h
(b)DSx10+0100H= 2774h
(c)Ds 10+1000H= 2AF8h
(d)DSx10+ [0250] [0500] = 20F58h
(e)DSx10+[0200H] [0500H] +0250H=1AEAAh
(f) DSx10+0200+0250=28D2h
[7*] Write a program to add a data byte located at offset
0500H in 2000H segment to another data byte available at
0600H in the same segment and store the result at 0700H in
the same segment.
Answer
15
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