UGC NET CSE | November 2017 | Part 2 | Question: 31

Question

Consider the following program fragment in assembly language:

        mov ax, 0h
        mov cx, 0A h
do loop:
        dec ax
        loop doloop

What is the value of $ax$ and $cx$ registers after the completion of the doloop ?

• A. $ax= FFF5 \: h$ and $cx=0 \:h$
• B. $ax= FFF6 \: h$ and $cx=0 \: h$
• C. $ax= FFF7 \: h$ and $cx=0A \:h$
• D. $ax=FFF5 \: h$ and $cx=0A \: h$

Answer 1

cx is counter register;
so loop will execute until cx!=0 ;
content of cx= 0Ah
A=10 in hex;
statement doloop:
                     dec ax;  decrement value of ax cx times
                    loop doloop;
simply add ( ax + ( -cx ) )
=>  0000 0000 0000 0000 + ( 2's compliment of cx = 1111 1111 1111 0110)
=> 1111 1111 1111 0110= FFF6, cx =0000h

Answer 2

Answer is B.

In 8086, dec instruction is to decrement.

loop instruction decrements the value of cx register by 1 until it becomes 0h.

(ax = 0h or we can say 0000h. I am taking two hexadecimal digits from LSB to simplify calculations. ax = 00h)

Comments

why are you decrementing cx value?

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As I have already mentioned, the loop instruction in 8086 works by decrementing the value of cx register by 1 until it becomes 0h.

As you can see the loop instruction has no stopping condition mentioned explicitly.For this reason I was confused and could not attempt this question in the NET exam.

When I checked later, I learnt that it works by decrementing cx register value until it becomes 0.

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why in 10th iteration, ax is not considered!!!! .

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Because we have to decrement ax only until cx does not become 0.

In the 10th iteration as you can see, cx has taken 00h value so the loop ends here and we do not decrement ax any further.