Question

1- Assume that the state of the 8086 registers and memory just prior to the execution of each instruction is given as below:
(AX) = 0010 H
(BX) = 0020 H

(CX) = 0030 H (DX) = 0040 H (SI) = 0100 H
(DI) = 0200 H (CF) = 1 (DS:100H) = 10 H (DS:101H) = 00 H (DS:120H) = FF H (DS:121H) = FF H (DS:130H) = 08 H (DS:131H) = 00 H (DS:150H) = 02 H (DS:151H) = 00 H (DS:200H) = 30 H (DS:201H) = 00 H (DS:210H) = 40 H (DS:211H) = 00 H (DS:220H) = 30 H (DS:221H) = 00 H

Note: (DS:100H) = 10 H ; this means that the data written in the address that is obtained by combining DS register (as base address) and 100H(offset) , is 10H.

Which result is produced in the destination operand in each of the following cases (instructions are independent): (32 points)

1. a) ADDAX,00FFH

2. b) ADC SI, AX

3. c) INCBYTEPTR[0100H]

4. d) SUB DL, BL

5. e) SBB DL, [0200H]

6. f) MUL DX

7. g) IMULBYTEPTR[SI]

8. h) IDIV BX

Answer 1

a) ADD AX,00FFH

AX= 0010H = 0000 0000 0001 0000 +

0000 0000 1111 1111

They are added together and we will get 0000 0001 0000 1111 = 010F H

b) ADC SI, AX

ADC will work with the syntax operand1 = operand1 + operand2 + CF so here it wil be

SI = SI +AX+CF

= 0100H+0010H+1

= 0111 H

c) INCBYTEPTR[0100H] hope you mean INC BYTE PTR[0100H]

INC BYTE PTR [0100H]; This indicates the size of memory data addressed by thememory pointer

Here the value at location [0100] is given as 10H but as we increment it by using INC instruction it will become 11H

INC BYTE PTR[0100H] = 11H

d) SUB DL, BL - This instruction performs subtraction of lower 8-bits in both DX and BX registers . The lowe part of these registers are DL and BL respectively. Now thier values become.

DX = 0040H So DL = 40H

BX = 0020H So BL = 20 H

Now SUB DL, BL 40H - 20H and result will store in DL register

DL = 40H = 0100 0000

BL = 20H = 0010 0000

So  SUB DL, BL = 0010 0000 = 20H

  

e) SBB DL, [0200H]

Solution:Value in DL = DL – (byte at DS:0200) – CF = 40H – 30H – 1 = 0FH

To understand better you can try to convert these Hexadecimalnumber to binary and solve it

that is 40H = 64 , 30H = 48 and 1 remains same. So 64-48-1 = 64 - 49 = 15 = 0FH

f) MUL DX - The MUL instruction will multiply the value in AX with DX and result will be stored in DXAX

So here Multiply AX= 10H with DX = 0040H and result will stored in DXAX = 10H * 40H = 400H

DX = upper 16 bits of result = 0000H; AX = lower 16 bits of result = 0400H

g) IMUL BYTE PTR[SI]

Anser Will be : AX = AL * byte at address DS:SI (signed multiplication of 8-bit values)

Here DS:SI = 0010H Al = 0010H

AX = 10H * 10H = 100H

h)  IDIV BX

As this this a division with 16 bit register and uses IDIV it will perform signed divion of BX register with values in DXAX together as 32 bit register

DIV BX = DX AX/BX = 0040H 0010H / 0020H = 20000H

Since it is a word IDIV divides double-word value in DX:AX bysource, returning quotient in AX and the remainder in DX.

In Decimal = 4194320 / 32 = 131072

Here in the cases of 'g' and 'h' we are not using any signed numbers so all the operations will work similar to the unsigned operations