Word Addressable (GATE 2004)

Question

 

Comments

Please confirm the answer of Question 17 it's option B right ?

I didn't get this question on GO

In word addressable every word is given and address so During Add instruction the address of next instruction I.e. 1004 will be stored .
Am I right ?

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Anybody help here please :/

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if i assume that 1 word is 1 byte then should be 1016?

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But in word addressable every Word is given an address. How u got 1016 ?

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Its Like : -

Address          Bytes

     0                B B B B

     1                B B B B

     2                B B B B

and so on

Here not every byte is given a unique address unlike in byte addressable memory

Correct me if i am wrong Brother :)

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Ok Sry I haven't noticed that instruction size is given in words. and word size is given 32 bits(4 bytes) yes then answer should be 1004 i was assuming 1 word is 1 byte.

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My above interpretation is right na Brother ?

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yes according to me you are right brother..

Answer 1

1005 , because program counter will point to 1001 and  + 4 = 1005

Answer 2

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
2Words=8Bytes	1000	1001	1002	1003	1004	1005	1006	1007
1Word=4Bytes	1008	1009	1010	1011
1Word=4Bytes	1012	1013	1014	1015
2Words=8Bytes	1016	1017	1018	1019	1020	1021	1022	1023
1Word=4Bytes	1024	1025	1026	1027

After halt instruction memory location will be : 1027+1 = 1028.

Ans: 1028 Option D.

 

Hope it helps..!