QUESTION 2.16:
Solution:
Each word is of size: 32 bits = 32/8 Bytes = 4 Bytes.
Program has been loaded starting from memory location :1000.
Instruction Size |
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2Words=8Bytes |
1000 |
1001 |
1002 |
1003 |
1004 |
1005 |
1006 |
1007 |
1Word=4Bytes |
1008 |
1009 |
1010 |
1011 |
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1Word=4Bytes |
1012 |
1013 |
1014 |
1015 |
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2Words=8Bytes |
1016 |
1017 |
1018 |
1019 |
1020 |
1021 |
1022 |
1023 |
1Word=4Bytes |
1024 |
1025 |
1026 |
1027 |
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After halt instruction memory location will be : 1027+1 = 1028.
Ans: 1028 Option D.
Hope it helps..!