GO Classes Test Series 2024 | Mock GATE | Test 14 | Question: 37

Question

Suppose that an operating system provides two functions, block() which puts the calling process on the blocked queue, and wakeup(P) which moves process P to the ready queue if it is currently on the blocked queue. wakeup(P) does not perform any action if the process P is NOT in the blocked queue.

Consider two processes A and B running the code given below.


Which of the following is TRUE regarding the behaviour of the system when we run A and B concurrently on the uni-processor system?

• A. It is possible to have deadlock in the system.
• B. A and B will always run forever with strick alternation printing ABABAB….
• C. A and B will always run forever with strick alternation printing BABABA….
• D. None of these

Comments

@Sachin Mittal 1 sir, here in question shouldn't it be mentioned what wakeup(A) does when A is not in block state.

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@gaurav_kumar Yes thats right but expectedly it will not do anything.
We will mention in question

Answer 1

IT IS POSSIBLE TO HAVE DEADLOCK WHEN B STARTS FIRST AND PRINTS "B" AND THEN WAKEUP A AND BLOCKS ITSELF , THEN WHENEVER A STARTS IT PROCESS FROM NOW IT WILL FIRST BLOCK ITSELF.

AS BOTH PROCESSES ARE IN BLOCKED STATE AND BOTH LOOKING FOR EACHOTHER TO WAKEUP IT LEADS TO A DEAD STATE 

OPTION C ALSO POSSIBLE BUT IT IS CORRECT IF THE WORD "ALWAYS " IS NOT THERE. 

CORRECT ANSWER: "A"

Comments

I think Option C will be correct if the Option is like
"A and B may run forever with strict alternation printing "BABABA..." ".

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@Prabhas  I think your sequence is little bit wrong 

after printing first " B"  , run the process till wakeup(A) code.

then A will be getting released from suspended list and printing "A" and then continue wakeup(B) and again starting execution of void A() function and getting blocked.

in that time we came back to void B() and trying to execute from the last preempted point after the "wakeup(A)" line and getting blocked . 

thus deadlock occurred.