Test and Set

Question

What if lock=True initially and process P1, P2 ,P3 are trying to enter into critical section 

If already lock is true then again in Test - set Lock is becoming true

I didn't get how Test- set is ensuring Mutual Exclusion

Answer 1

First of all let me tell you that you are absolutely right that, if lock variable is initially true then it will never go forward. It blocks all the three process there only. 

If you notice then test-set looks like a function, but it's not a function, it's an instruction. Check this,

So, Now question is that what is the meaning of having TestAndSet as instruction? The advantage of having TestAndSet, as instruction is that, When it executes then it executes completely, It never get preempted in between. 

Now Consider Lock = False

P1 will go for enter, P1 is going for execution, it gets value false in rv, it sets lock as True, and return false. Due to the false, condition is failed, hence P1 goes for executing critical section. 

Now suppose while P1 is in critical section, P2 goes for execution, its gets Lock value true in rv, assigning true in lock and return true, condition is true, it will wait there only. 

Similar for P3.

Hence Mutual exclusion preserved. 

Comments

Is Lock value initially fixed to false ??

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It will be mentioned in the question. If it is not then take it as false, otherwise it will not go anywhere.

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Sir if it is true how can we move forward so we cant ensure ME

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First, I am not a sir. I m in the final year right now. Second, If it is not moving forward, it does not mean that it's not satisfying ME, while it means that It's in deadlock.

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Ya Ok I Understood and Thank You

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Mark it best :P

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And one more

https://gateoverflow.in//35693/ipc-synchronization-is-deadlock-possible-here

I have doubt in this too  

here if initially flag1=true/false and flag2=false then in process i starts execution and in while() --- false so both can enter CS ?? 

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let me see.

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By considering lock=TRUE, which states that the critical section is already occupied, even though we are not visualizing with three processes.By meaning no other process should enter critical section. So no process will be allowed to enter the Critical Section.