GATE Overflow Test Series | Operating Systems | Test 1 | Question: 13

Question

A process executes the following segment of code :

if(fork() && fork() || fork()){
    printf("GATE2021");
}

The number of times “$\text{GATE2021}$” will be printed is _______

Answer 1

There are two concepts here, short circuit rule in C and fork in OS.

By short circuit rule in C, if the first operand of logical AND $(\&\&)$ is $0,$ the second operand won't be evaluated. Similarly, if the first operand of logical OR $(||)$ is $1.$

Now, fork(), copies the current process image and creates a child process which starts executing from the next instruction onward. fork() returns the child process id to the parent and 0 to the child process.

Now, let's trace the given code.
 

if((fork1() && fork2()) || fork3()){
    printf("GATE2021");
}

After fork1, we have $2$ processes and only one has return value non zero and only that will execute fork2 creating one new process.

Thus after fork1 and fork2 we will have $3$ processes and two of them will have $0$ as value for the first operand of || and only those $2$ will execute fork3 creating $2$ more processes which will have their return value as $0.$ Thus in total we will have $5$ processes and $3$ of them will have TRUE (non zero) value for "if" and executes the printf.

So, correct answer $3.$

Comments

sir, can you please solve it using the diagram ?

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1,2,3 here means fork1,fork2,fork3 respectively. 1 inside circle is value returned to parent and 0 inside circle is value returned to child.

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Thanks 👍

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In the question the bracket enclosing the operands of the and (&&) operator (given in the solution) is not there.

So, another thing which anyone need to before solving this problem is && has a higher precedence than ||. If the opposite is true then answer would have been 2.