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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.
$$\begin{array}{lll} & \text{MOV B, #0}&& \text{;} & \text{$B \leftarrow 0$} \\& \text{MOV C, #8} && \text{;}& \text{$C \leftarrow 8$} \\
\text{Z:} & \text{CMP C, #0} &&\text{;}& \text{compare C with 0} \\ 
& \text{JZ X} && \text{;}& \text{jump to X if zero flag is set}  \\
& \text{SUB C, #1} && \text{;}& \text{$C \gets C-1$}  \\ 
& \text{RRC A, #1} && \text{;}& \text{right rotate A through carry by one bit. Thus:}  \\  & \text{} && \text{;}& \text{If the initial values of A and the carry flag are $a_7\ldots a_0$ and}  \\ 
& \text{} && \text{;}& \text{$c_0$ respectively, their values after the execution of this}  \\ 
& \text{} && \text{;}& \text{instruction will be $c_0a_7\ldots a_1$ and $a_0$ respectively.}  \\ 
& \text{JC Y} && \text{;}& \text{jump to Y if carry flag is set}  \\
& \text{JMP Z} && \text{;}& \text{jump to Z}  \\
\text{Y:} & \text{ADD B, #1} && \text{;}& \text{$B \gets B+1$}  \\ 
& \text{JMP Z} && \text{;}& \text{jump to Z}  \\ 
\text{X:}& \text{} && \text{;}& \text{}  \\ \end{array}$$
If the initial value of register A is A0 the value of register B after the program execution will be

  1. the number of $0$ bits in $A_0$
  2. the number of $1$ bits in $A_0$
  3. $A_0$
  4. $8$
in CO and Architecture by Veteran (52.1k points)
edited by | 3.9k views
0
can anyone plz explain this ques. solution with example.
+2
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.

1 Answer

+22 votes
Best answer
Option $(B)$. The code is counting the number of 1 bits in $A_0$. When a 1 is moved to carry, B is incremented.
by Veteran (416k points)
edited by
0

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 ?

0
@ Arjun sir, In 49 , why rotation will happen 8 times ?
0
@coolcoder are you rotating C? Question is rotating A and decrementing C.

@Amsar Because C is initialized to 8.
0

Arjun sir,

yes yes , sorry ... 

0

@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? 

0
when $a_0=1$. This is mentioned in question.
0
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.
+1
please provide the steps and iterated solution with one example.....
0
Sir what is the initial value of the register A.....??  Not mentioned anywhere!!
+1

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.

+1
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 $a_7$
0
Answer:

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