GATE CSE 2003 | Question: 49

Question

Consider the following assembly language program for a hypothetical processor $A, B,$ and $C$ are $8$ bit registers. The meanings of various instructions are shown as comments.

	MOV B, #0	;	$B \leftarrow 0$
	MOV C, #8	;	$C \leftarrow 8$
Z:	CMP C, #0	;	compare C with 0
	JZ X	;	jump to X if zero flag is set
	SUB C, #1	;	$C \gets C-1$
	RRC A, #1	;	right rotate A through carry by one bit. Thus:
		;	If the initial values of A and the carry flag are $a_7..a_0$ and
		;	$c_0$ respectively, their values after the execution of this
		;	instruction will be $c_0a_7..a_1$ and $a_0$ respectively.
	JC Y	;	jump to Y if carry flag is set
	JMP Z	;	jump to Z
Y:	ADD B, #1	;	$B \gets B+1$
	JMP Z	;	jump to Z
X:

Which of the following instructions when inserted at location $X$ will ensure that the value of the register $A$ after program execution is as same as its initial value?

• A. $\text{RRC A}, \#1$
• B. $\text{NOP} ;$ no operation
• C. $\text{LRC A,} \#1; $  left rotate $A$ through carry flag by one bit
• D. $\text{ADD A,} \#1$

Comments

Can anyone explain why CMP C, #0 is not resetting the carry flag? 

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neel19

see this , https://gateoverflow.in/938/gate-cse-2003-question-48?show=9784#a9784

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For the given question, NO option is correct. Detailed Video Solution with Complete Analysis of this question is HERE.

$ADD,SUB, CMP$ instructions affect all the status flags, like carry flag, zero flag etc.

$SUB$ instruction sets carry flag when there is need of borrow into the MSB i.e. when we subtract bigger number from smaller number. $ADD$ instruction sets carry flag when there is carry out of MSB. $CMP$ instruction subtracts second operand from first for flags only. If first operand is $\geq$ second operand, then carry flag is reset, else set. 

NOTE that just before $RRC$ instruction, the $SUB$ instruction will also affect the Carry Flag, & in each iteration, since there is no requirement of borrow for the subtraction operation, Carry Flag will be reset just before the $RRC$ instruction. 

Hence, WHATEVER be the value of $A$ in the beginning, the final value of $A$ will ALWAYS be $0.$ 

NOTE: In question, “SUB C, #1” should be replaced with “DEC C” to prevent modification of carry flag. Also, “ADD B, #1” should be replaced with “INC B” to prevent modification of carry flag. Also, it should be mentioned that $CMP$ instruction, somehow, doesn't affect carry flag. With these 3 changes, answer becomes Option A.

Detailed Video Solution with Complete Analysis: https://youtu.be/5aC2H2u_VgI 

Both Question 48 & Q 49 are analyzed in the above lecture. 

Answer 1

Option $(A) \text{RRC}\; a, \#1.$  As the 8 bit register is rotated via carry $8$ times.

• $a_7a_6a_5a_4a_3a_2a_1a_0$
• $c_0a_7a_6a_5a_4a_3a_2a_1$, now $a_0$ is the new carry. So, after next rotation,
• $a_0c_0a_7a_6a_5a_4a_3a_2$

So, after $8$ rotations, $a_6a_5a_4a_3a_2a_1a_0c_0$ and carry is $a_7$.

Now, one more rotation will restore the original value of $A_0$.

PS: In question, “ADD B, #1” should be replaced with “INC B” to prevent modification of carry flag. 

Comments

@ArjunSir Then Where  SUB C#1 in this program is used?

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@Surya Pratap Singh S if we take your example then after first 1 execution will jump to Y and carry bit is 1 and then we are performing an arithmetic operation i.e we are adding 1 to B so it is arithmetic operation so this will modify carry bit to 0 so the first 1 became 0 because of this

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not only add B,#1. isn't there other instructions also which might change the carry flag thus loosing the data bit in it, such as sub c,#1?

Answer 2

If A has n bits with 1 carry bit ... then number of right rotations through carry after which we will get the same A is (n+1) .. Here this code is doing right rotation for n times .. So at the last step of program we do one more rotation ...

Answer 3

$Let A=7(00000111)$
$|B=0|C=8|carry=0,zero=0|zero\neq1 |C=7|00000011\ \ 1|B=1$

$|carry=0,zero=0|zero\neq1 |C=6|10000001\ \ 1|B=2$

$|carry=0,zero=0|zero\neq1 |C=5|11000000\ \ 1|B=3$

$|carry=0,zero=0|zero\neq1 |C=4|11100000\ \ 0|$

$|carry=0,zero=0|zero\neq1 |C=3|01110000\ \ 0|$

$|carry=0,zero=0|zero\neq1 |C=2|00111000\ \ 0|$

$|carry=0,zero=0|zero\neq1 |C=1|00011100\ \ 0|$

$|carry=0,zero=0|zero\neq1 |C=0|00001110\ \ 0|$

$|carry=0,zero=1|zero=1:Jump\ to\ X|$

Which of the following instructions when inserted at location X will ensure that the value of the register A after program execution is as same as its initial value?

$If\ X:RRC \ A,\#1$

$|00000111\ \ 0|$

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$Ans:A$

Comments

1. “zero” is denoted as zero flag here. So, when you write zero=0, it means value of zero flag is zero and  zero != 1 shows only that value of zero flag is still zero, it just shows that we don’t have to go to “X” now. Values in zero flag is either 0 or 1.

2. In Assembly language, CMP A,B instruction means processor compares the two arguments $A$ and $B$ and when they are equal then zero flag is set which means value of the zero flag register is 1 i.e. zero=1. Here, at the end, $C=0$ and so, when processor goes to CMP C, #0 instruction, it finds that both arguments are equal and so, it sets the zero flag i.e zero=1.

3. Carry flag is set means value of the carry flag is 1. 

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@ankitgupta.1729 

Thank you 

one last doubt , you have written :->

 “Here, at the end, C=0 and so, when processor goes to CMP C, #0 instruction, it finds that both arguments are equal and so, it sets the zero flag i.e zero=1.”

It is not mentioned to make zero =1 in the program , does it automatically happen after the condition became true.

 

Thank you very much I understand the solution today(after 3 day😄).

 

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Yes, it is not mentioned here because the person who made this question might be expecting to know the functionality of “CMP” instruction from the candidates for this exam.