GATE CSE 2004 | Question: 63

Question

Consider the following program segment for a hypothetical CPU having three user registers $R_1, R_2$ and $R_3.$ $$\begin{array}{|l|l|c|} \hline \text {Instruction}  &  \text{Operation }& \text{Instruction size} \\&& \text{(in words)} \\\hline \text{MOV $R_1,5000$} & \text{$R_1$} \leftarrow  \text{Memory$[5000]$}& \text{$2$} \\\hline\text{MOV $R2,(R1)$} & \text{$R2$} \leftarrow  \text{Memory$[(R_1)]$}& \text{$1$} \\\hline \text{ADD $R_2,R_3$} & \text{$R2$} \leftarrow  \text{$R_2 + R_3$} & \text{$1$} \\\hline \text{MOV $6000,R_2$} & \text{Memory$[6000]$} \leftarrow  \text{$R_2$} & \text{$2$} \\\hline \text{HALT} & \text{Machine Halts} & \text{$1$} \\\hline \end{array}$$Consider that the memory is byte addressable with size $32$ bits, and the program has been loaded starting from memory location $1000$ (decimal). If an interrupt occurs while the CPU has been halted after executing the HALT instruction, the return address (in decimal) saved in the stack will be

• A. $1007$
• B. $1020$
• C. $1024$
• D. $1028$

Comments

If (C) is the PC value, what happens after Interrupt is handled and now it's time to start executing instructions again.. PC will again point to HALT (address 1024). So, how will you escape?

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@Raizel 

What is the loop here?

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32 b / 8  =  4 B
= 4*2  +   4*1  +  4*1  +   4*2   +   4*1
=  8 + 4 + 4 + 8 + 4  = 28
1000 + 28

Answer 1

Option is D.

Word size is $32$ $bits$ ($4$ $bytes$). Interrupt occurs after execution of HALT instruction NOT during,  So address of next instruction will be saved on to the stack which is $1028$.  

(We have $5$ instructions starting from address $1000$, each of size $2, 1, 1, 2, 1$ totaling $7$ words $= 7 *4 =28$ $bytes$). 

$1000+ 28 = 1028$,

$1028$ is the starting address of NEXT Instruction .

After HALT instruction CPU enters a HALT state and if an interrupt happens the return address will be that of the instruction after the HALT. 

 References :

1.  https://x86.puri.sm/html/file_module_x86_id_134.html [ X86 Instructors Manual ]
2. http://electronics.stackexchange.com/questions/277735/what-happens-if-the-interrupt-occurs-during-the-execution-of-halt-instruction
3. https://en.wikipedia.org/wiki/HLT_(x86_instruction)

Comments

I have a doubt. Memory starting address is given in decimal and size of total instruction is in bytes,then how decimal and bytes addition is taking place?

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Refer my comment HERE for several standard resources supporting the answer 1028. NOT a Single Standard Resource supports 1024. Students who are giving answer are following some poor coaching notes, not following the standard resources.

Halt Instruction Complete Explanation & Misconceptions: https://youtu.be/D6mjFhYkgNg?feature=shared

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SIR CAN U ANSWER HOW TO DO SIR

Answer 2

answer should be 1024 because consider the case where halt statement is succeeded by some AlU instructions now if you'll store the next intruction as the return address when an interrupt occur while executing halt then in that case after executing the interrupt program will execute those alu instruction which should not happen as program is already halted

Comments

True it is a valid case.

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@Sneha negi Ma'am, I understood from your answer that after HALT is executed no other instruction should be executed. But the below statement in the question 

after executing the HALT instruction, the return address will be...

Doesn’t it means now HALT has been executed and now its time for another instruction ?

Thanks in advance ! 

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Pogram Counter (PC) is incremented before decoding the instruction.

 

Answer 3

It is clearly mentioned that HALT instruction has executed completely.

Now some other instruction just began its fetch at 1028, interrupt came, so it is bit OBVIOUS that 1028 will be the answer,

1007, 1020 are useless options

 

D is the corect answer

Comments

No, HALT instruction changes the Program counter value to 1024, so that program remains halted until it is restarted again, so the PC value will remain 1024 which will be pushed onto the stack

Answer 4

Working of HALT:

HALT instruction stops the execution of a program by changing the PC value to the current instruction program counter value (1028 is replaced with 1024 in this case) so when the HLT instruction is completed, the CPU fetches the instruction from the location specified in PC (PC contains 1024, ie HLT instruction) and increments the PC value to 1028 and does the same thing.

Now the fetched HALT instruction does the same thing again and again ( ie Changing the PC value to 1024 from 1028). It halts the program only from the user point of view, but HALT is simply busy waiting from the CPU point of view. 

So ans is 1024

 

PS: From Chottu’s comment "Interrupt comes during the halt instruction" and "Interrupt comes after halt instruction", both are the same. Because interrupt is served only after execution of current instruction.

Answer 5

Instruction1

Instruction2

Instruction3

Instruction4

Instruction5

1000        1008        1012           1016          1024          1028

After HALT Instruction PC will contain 1028. So,1028 will be pushed into the top of the stack so that it can execute this Program after servicing the Interrupt.

 

Comments

not after HALT instruction,

during the fetch operation of the HALT instruction itself, the PC will get updated to 1028

Then excution of HALT instruction takes place then operation will get stopped and stack will store the 1028 memory address !