madeeasytestseries

Question

Consider the following segment:

int count = 0;

void tally( )

{

for(int i=1;i<=5;i++)

count=count+1;

}

main()

{

parbegin

tally();

tally();

parend

}

 Note: Assume the count = count + 1; will execute in ‘3’ different instructions. I. Load Ri, m[count] II. INC Ri III. Store m[count], Ri Pre­emption can occur while executing the above instructions. After completion of both functions of tally( ), The minimum final values of count is.....?

Comments

Can someone explain how the answer is 2?

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Lets say when tally() is called for the first time count =count +1 will be executed for 5 times which have 3 micro instruction.
In each microinstruction value of count is loaded in to register Ri , incremented and stored. There can be preemption as operations are not atomic. Lets say for first iteration value of 0 is loaded into Ri and thread performing this operation is preempted. Now other threads execute and set count = count + 1 to 4 but first thread which got preempted comes back read its register value i.e 0 and increment to 1 and update value to 1 for count over writing value of 4. count value is 1 at end of first tally() call.

Same thing happens during second call setting minimum value to 2.

Answer 1

Chronology samajhiye ....