In executing the following code:MOV CX, 0002START:MOV SI, 0300MOV DI, 0302LOOP:XOR (DI), ALINC SIMOV AL, (SI)INC DIDEC CXJNZ LOOPHLTwith the following byte contents:(0300H) = 0AH, (0301H) = OBH, (0302H) = 11H, (0303H) = 21H.The result will be:Select one:a. (0300H) = 0AH, (0301H) = OBH, (0302H) = 1AH, (0303H) = 2BH(0300H) = 0AH, (0301H) = OBH, (0302H) = 10H, (0303H) = 20Hb.c. (0300H) = 0AH, (0301H) = OBH, (0302H) = 1AH, (0303H) = 20Hd.None of the options given heree. (0300H) = 0AH, (0301H) = OBH, (0302H) = 1BH, (0303H) = 2AH

The correct answer for the above question is option C.

Answered: In executing the following code: MOV…

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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