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Question

  • Part- I

If the program below, the number of times the FIRST and SECOND JNZ instruction cause the control to be transferred to LOOP respectively
MVI H,02H
MVI L,05H
LOOP : DCR L
FIRST : JNZ LOOP
DCR H
SECOND : JNZ LOOP

A. 5 and 2
B. 21 and 1
C. 259 and 1
D. 260 and 1

   

  • Part- II 

If In place of DCR L instruction, if we use SUB L,#1.

The number of times the FIRST and SECOND JNZ instruction cause the control to be transferred to LOOP respectively

                                                                                                                                                                                   

  1.  5 and 2
  2.  260 and 1
  3.  259 and 1
  4.  6 and 1
in CO and Architecture by (67 points)
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1 Answer

+6 votes
Best answer

MVI L,05H
LOOP : DCR L      ///////////  L==> 04
FIRST : JNZ LOOP            CONTROL transfered 1st time

LOOP : DCR L      ///////////  L==> 03
FIRST : JNZ LOOP            CONTROL transfered 2nd time

LOOP : DCR L      ///////////  L==> 02
FIRST : JNZ LOOP            CONTROL transfered 3rd time

LOOP : DCR L      ///////////  L==> 01
FIRST : JNZ LOOP            CONTROL transfered 4th time

LOOP : DCR L      ///////////  L==> 00
FIRST : JNZ LOOP            CONTROL transfered will not be transfred condition failed  now execute  next instruction

LOOP : DCR L
FIRST : JNZ LOOP
DCR H                                  /////////////h==>01
SECOND : JNZ LOOP         /////////////// CONTROL transfered 1st time

LOOP : DCR L
FIRST : JNZ LOOP
DCR H                                  /////////////h==>00
SECOND : JNZ LOOP         /////////////// CONTROL  will not be transfred condition failed

part 1 first = 4 ,,second ==1/////////////all options given are wrong

part---2 beginswink

MVI L,05H
LOOP : SUB L , #1       ///////////  L==> 04
FIRST : JNZ LOOP            CONTROL transfered 1st time

......will be same as above till  L==00//////////////means it ran 4times upto now

                           LOOP : SUB L , #1 
                           FIRST :JNZ LOOP
  next           DCR H              ///h==>01
                  SECOND : JNZ LOOP   //////CONTROL transfered 1st time

 

         LOOP :      SUB L , #1     //////  now this will be again executed SUB actually invokes carry flag but DCR doesn't

             

                         00000000 (carry flag will be generated )

subtract(--)     1111111 

                   11111111===>255  

               
                           FIRST :JNZ LOOP///////////////////so  first will again run 255 times

                  after that
                   DCR H              ///h==>00
                  SECOND : JNZ LOOP   //////CONTROL not transfered condition failed

so part2 first= 255+4 ==259 and second ==1...
           

by Boss (10k points)
selected by
0
thanks fr the xplanation
+6

Both the case ANS will be 259,1

Part1:

It will go first L DEC 4 time(L=0) first then H DEC 1 time (H=0)and transfer control goto to LOOP.in this case L is again dec then it will become 255(11111111) 

During the exicution of decrement instruction value 1 is subtract from the register content. When it will satisfied the minimum limit.then all 00000000 raising up to 11111111

00000000

-               1


11111111

So both the case PART1  and PART2 ans will be same but syntax is differ but exicution is same.

Ans:.  259 OR 1

ANS

0
But how can we say so surely that when L=0 and a decrement operation is performed then L will be rolled back to 1111 1111 (decimal-255)?
It is not written anywhere that the computer uses signed or unsigned data.By default, All computers use signed 2's complement representation.
0
-1 representation in 2's complement form is 1111 1111
0
Sir.... I dont understand in the first case what happen to DCR L when second transefer to the loop for the first time... According to your answer it seems like nothing happens and DCR H make H=00 and transfer fails..
Answer:

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