Question

In: Computer Science

9- Show how the word ABCD H is stored in the memory starting at 0A002 H?...

9- Show how the word ABCD H is stored in the memory starting at 0A002 H?

11- How the decimal number 500 is stored in the memory?

13- Which two values are needed for memory address calculations?

15- What are the steps of an instruction execution? Briefly explain each step.

please answer all of them I will upvote!

19- Calculate the value of memory address in each case: a) 2000:33BB b) 0200:AA00 c) 0220:00BC

20- A data segment is to be located from A0000 to AFFFF H, which value is stored in DS?

Expert Solution

Q 9

The character values are stored as their corresponding ASCII values

The ASCII values of A - 065 binary form 01000001

B - 066 binary form 01000010

C - 067 binary form 01000011

D - 068 binary form 01000100

A letter is stored in a single byte that is 8 bits.

So A B C D would get stored as

01000001 01000010 01000011 01000100

Q 11

The number 500 after converting into binary becomes

111110100 here we get a 9 bit number.

Assuming it as a 32 bit register

The remaining 23 bits are filled with 0.

Hence the value stored in memory is

00000000 00000000 00000001 11110100

Starting address would have 00000000

And then moving up to store the subsequent bytes.

Q 13

The segment and offset values are needed for calculating the physical memory address.

Note: illustrated with examples in Q 19

Q 19

a )Here the logical addres is specified as

2000: 33BB

The offset is 33BB and 2000 is the value of DS.

To calculate the effective address the following formula is used.

Starting address + offset

Starting address is obtained by appending the contents of segment register with 0H.

Appending 0H to 2000 we get 20000H

Now adding the offset we get 20000 + 33BB that is 233BB becomes the effective address.

b) 0200: AA00

Appending 0H to the segment value ==> 02000H

Effective address = 02000 + AA00 ==>CA00 H

c) 0220:00BC

Starting address = 02200Hb( appending 0H)

Effective address = starting address + offset = 02200+00BC = 22BC H

Q 20

The upper range of the data segment is AFFFF that is A0000+ FFFF.

The low range being A0000 + 0000

The value of the DS after removing 0H being A000 H

venereology answered 1 year ago

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