interrupt

Question

Answer 1

Instruction Code	Memory Address
I0	2000
I1	2001-2002
I2	2003-2004
I3	2005
I4	2006-2009
I5	2010-2011
I6	2012

when interrupt occur during execution of I5 . Interrupt will be handled after execution of current statement. 

Interrupt size = 2 Word = 64 bit = 8 Byte so Interrupt will be handled from 2012 to 2019 . So , 2019 will be the last address upto which interrput will be handled.

Comments

@nice explanations @ Anup patel

Pls Explain also this

in 22.2 2.23 size 32 bits

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In 2.22 , Memory size is byte addressable and of 32 bit and instruction size is in words so in this question 

1 word = 4 byte

Note that When HALT instruction will occur then program counter will store the starting address of that instruction.

Here i am naming instruction as I1,I2,I3,I4,HALT (Just for easy writing)

Instruction	Memory Address
I1	1000-1007
I2	1008-1011
I3	1012-1015
I4	1016-1023
HALT	1024-1027

So , during execution of HALT statement PC will have address 1024 and when interrupt will occur at this stage then program counter will Save this address in stack and return address 1024 after servicing of Interrupt.

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And in 2.23 for word addressble memory address will be changel like : 

Instruction	Memory Address
I1	1000-1001
I2	1002
I3	1003
I4	1004-1005
HALT	1006

Here interrupt has occured during execution of I3 so PC Will store 1004