Synchronization

Question

Ans is B) ...but after z...y should not printed since b would become 0 then....so II should not be a possible output...

Plz explain......

Answer 1

When b becomes 1, the while loop in P2 breaks. Now the next instruction is printf("y"). But before this instruction gets executed, the instructions
b=0;
printf("z");
CAN be executed by P1. (There is a possibility for that). So, B is the correct answer.

Comments

In the question it is told P1 and P2 are executed once.

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Thanks.

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@Arjun Sir It means if a  process has instructions 1 2 3 ,can it happen that 3 executes before 1?