1. Suppose the Vole memory cells from addresses 0x00 to 0x05 contain the bit patterns given in the following table: Address Contents Ох00 Ox14 Ox01 Ох02 Ox02 Ох34

Systems Architecture
7th Edition
ISBN:9781305080195
Author:Stephen D. Burd
Publisher:Stephen D. Burd
Chapter11: Operating Systems
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1. Suppose the Vole memory cells from addresses 0x00 to 0x05 contain the bit patterns given in the
following table:
Address
Contents
Ох00
Ox14
Ох01
Ох02
Ох02
Ох34
Ox03
Ox17
Ох04
OXCO
Ox05
Ох00
If we start the machine with its program counter containing 0x00, what bit pattern is in the memory cell
whose address is 0x17 when the machine halts?
Transcribed Image Text:1. Suppose the Vole memory cells from addresses 0x00 to 0x05 contain the bit patterns given in the following table: Address Contents Ох00 Ox14 Ох01 Ох02 Ох02 Ох34 Ox03 Ox17 Ох04 OXCO Ox05 Ох00 If we start the machine with its program counter containing 0x00, what bit pattern is in the memory cell whose address is 0x17 when the machine halts?
Expert Solution
Step 1

We must demonstrate the execution of the above-mentioned assembly program in which the vole cell addresses range from 0x00 to 0x05 in this question. As seen in the table, these addresses have a bit of pattern.
Assume the following:- Each cell includes a bit pattern in the form of an address ranging from 0x00 to 0x05.

With counter = 0x00, the programme begins to run.

Now we must calculate the bit pattern of the cell with the address (0x17), which is explained as follows:

- The program counter is set to its beginning position of 0x00 in the first step, and then it begins executing the following instruction based on the memory location.

- The PC (Program counter) is updated in the second phase, and the following instruction at position 0x01 is executed.
As a result of the above procedure, the final fetch instruction (4-byte) is 1402.

- The next instruction, according to the scenario, will be executed at memory location 1402. 1 is a hexadecimal Opcode with a load register of 4 in the given position.


The program counter is increased based on the specified assembly code, and the instruction is executed at address 02. At memory address 0x02, the bit pattern is 34.
The above-mentioned program's execution process is then explained in the form of a flow table, as shown below:

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