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Question

The program follows to use a shared binary semaphore T :

Process A
int Y;
A1: Y = X*2;
A2: X = Y;
signal(T);

Process B
int Z;
B1: wait(T);
B2: Z = X+1;
X = Z;


T is set to 0 before either process begins execution and, as before, X is set to 5.
Now, how many different values of X are possible after both processes finish executing ?

a.

one

b.

two

c.

three

d.

four

Posted under Operating System

Answer: (a).one

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Q. The program follows to use a shared binary semaphore T : Process A int Y; A1: Y = X*2; A2: X = Y; signal(T); Process B int Z; B1:...

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