Addressing Modes- Q4

+2 votes

Which of the following address in decimal can not be the starting address of the word

In a system with word size of 8 byte, If word alignment is implemented ?

A. 1024
B. 2048
C. 1004
D. 2008

asked Jun 4 in Computer Organization by gbmentor (63,050 points)
reshown Jun 5 by gbmentor

2 Answers

+1 vote
 
Best answer
The starting address should be divisible by 8,if we look at every option,we will notice 1004 is not divisible by 8,hence it is the right option

Option C
answered Jun 5 by soumyadeeppro20 (1,750 points)
selected Jun 5 by vinayakjhapro18
0 votes
Answer is not  multiple of 8 so 1004 c is correct one.
answered Jun 5 by tsnikhilsharmagate2018 (23,380 points)
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