GATE2003-48

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:

If the initial value of register A is A0 the value of register B after the program execution will be

• A. the number of 0 bits in $A_0$
• B. the number of 1 bits in $A_0$
• C. $A_0$
• D. 8

Comments

can anyone plz explain this ques. solution with example.

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Content of A0 is copied to A, we are rotating register to right with Carry, every 1 present in the register A carry will be set to 1, and everytime carry flag is set to 1, B will be incremented.

A contains the content of A0, it means when this program terminates, B will contain the number of 1's in the A (or) A0. there will be 8 iterations of the loop as value of C is initially 8 and with every iteration it is decremented by 1.

Answer 1

Option $(B)$. The code is counting the number of 1 bits in $A_0$. When a 1 is moved to carry, B is incremented.

Comments

Dear Arjun sir

I have a very stupid doubt here. Please pardon my ignorance.

First time the value of C will be 8 ==> 1000

Now , after SUB C,#1 , value of C will be 7 ==> 0111

Now , will the RRC be 1011 ? so , B will be 0+1 = 1

After this C will be 7 ( CMP C,#0 will be zero flag false ) , now C <- C-1 , so C is 6 , 0110 ,

Now , RRC will be 0011 , so B will not be incremented.

Will this program flow happen in this way ?

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@ Arjun sir, In 49 , why rotation will happen 8 times ?

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@coolcoder are you rotating C? Question is rotating A and decrementing C.

@Amsar Because C is initialized to 8.

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Arjun sir,

yes yes , sorry ... 

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@Arjun SIR 
Can u pls explain what is happening in line

RRC A, #1

Carry Flad =0 Regiser A = 10010
After RRC  010010
When does carry flag set? 

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when $a_0=1$. This is mentioned in question.

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Sir what will be contents of A after this code. Will it be 00000000, bcoz of CMP imstruction, as it will reset carry and zero flag when C !=0.

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please provide the steps and iterated solution with one example.....

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Sir what is the initial value of the register A.....??  Not mentioned anywhere!!

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@ Preety singhal 

   you can assume any initial value of register content A  (i.e: the content of Ao )

like...suppose A = 00101101  or whatever you want can assume.and at last when we execute this program the register content of B will be the total number of 1's which are present in register A.( in this case at last content of register B will be 4, because 4,  1's are present in register A )

And one more thing u can also assume initial carry either 0 or 1.

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8 bit register along with 1 carry bit->9 bits are rotated right 8 times.

So, In Each rotation $i$, bit $a_{i-1}$ comes into carry flag and depending on whether it's 1 or 0, CY flag is set or reset.

8 times rotation, bits $a_0-a_7$ each time come in CY flag.

After $8^{th}$ rotation, register would contain $a_6a_5a_4a_3a_2a_1a_0c_0$ and CY flag $a_7$