Use the following data declarations:```.data byteVal sbyte 1, 2, 3, FCh wordVal word 1000h, 2000h, 3000h, 4000h dwordVal dword 34567890h, 90785634h, 0Ah, 33445566h```Show the value of the **final destination operand** after each of the following code fragments has executed: *(If any instruction is invalid, indicate "INV" as the answer and briefly explain why)*a. ```assemblymov bh, byteVal+2```**Answer: bh =**b.```assemblymov edx, 1add dx, [wordVal+4]```**Answer: edx =**c.```assemblymov ecx, 5xchg ecx, [dwordVal+12]```**Answer: ecx =**d.```assemblymov ah, byteVal+3sub ah, [byteVal+0]sub ah, [byteVal+2]```**Answer: ah =**e.```assemblymov eax, dword ptr dwordVal+7```**Answer: eax =**f.```assemblymovsx cx, byteVal+3```**Answer (show your answer in binary): cx =**

Let's go through each code fragment step by step:a.The value at byteVal+2 is 3. So, bh will be…

Use the following data declarations: .data sbyte 1, 2, 3, FCh word 1000h, 2000h, 3000h, 4000h dwordVal dword 34567890h, 90785634h, 0Ah, 33445566h byte Val word Val Show the value of the final destination operand after each of the following code fragments has executed: (If any instruction/s is invalid, indicate "INV" as the answer and briefly explain why) a. mov bh,byteVal+2 C. b. mov edx,1 add dx,[wordVal+4] mov ecx,5 xchg ecx,[dwordVal+12] d. mov ah, byteVal+3 sub ah, [ byte Val+0] sub ah,[ byte Val +2] e. mov eax, dword ptr dwordVal+7 f. movsx cx,byteVal+3 answer bh= answer edx= answer ecx= answer ah= answer eax= answer (show your answer in binary) CX=

Use the following data declarations: .data sbyte 1, 2, 3, FCh word 1000h, 2000h, 3000h, 4000h dwordVal dword 34567890h, 90785634h, 0Ah, 33445566h byte Val word Val Show the value of the final destination operand after each of the following code fragments has executed: (If any instruction/s is invalid, indicate "INV" as the answer and briefly explain why) a. mov bh,byteVal+2 C. b. mov edx,1 add dx,[wordVal+4] mov ecx,5 xchg ecx,[dwordVal+12] d. mov ah, byteVal+3 sub ah, [ byte Val+0] sub ah,[ byte Val +2] e. mov eax, dword ptr dwordVal+7 f. movsx cx,byteVal+3 answer bh= answer edx= answer ecx= answer ah= answer eax= answer (show your answer in binary) CX=

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Concept explainers

Computer Fundamentals

The computer is termed computation. For calculating or computing something the device that has been used is known as the computer. Or we can say that for performing a fast arithmetic operation the device that has been used is a computer. Storing, processing, and retrieving are the main features of a computer. 

Question

Use the following data declarations:

```
.data
byteVal sbyte 1, 2, 3, FCh
wordVal word 1000h, 2000h, 3000h, 4000h
dwordVal dword 34567890h, 90785634h, 0Ah, 33445566h
```

Show the value of the **final destination operand** after each of the following code fragments has executed: *(If any instruction is invalid, indicate "INV" as the answer and briefly explain why)*

a.
```assembly
mov bh, byteVal+2
```
**Answer: bh =**

b.
```assembly
mov edx, 1
add dx, [wordVal+4]
```
**Answer: edx =**

c.
```assembly
mov ecx, 5
xchg ecx, [dwordVal+12]
```
**Answer: ecx =**

d.
```assembly
mov ah, byteVal+3
sub ah, [byteVal+0]
sub ah, [byteVal+2]
```
**Answer: ah =**

e.
```assembly
mov eax, dword ptr dwordVal+7
```
**Answer: eax =**

f.
```assembly
movsx cx, byteVal+3
```
**Answer (show your answer in binary): cx =**

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