GATE Overflow Test Series | Operating Systems | Test 1 | Question: 12

Question

A process executes the following segment of code :

#include <stdio.h>
#include <unistd.h>
int main()
{
    fork();
    fork() && (fork() || fork());
    printf("GO 2020");
}

The number of times “$\text{GO 2020}$” 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.

#include <stdio.h>
#include <unistd.h>
int main()
{
    fork(); //2 processes after this
    fork1() && (fork2() || fork3());

    printf("GO 2020");
}

fork1 will create $4$ processes in total from the $2$ processes executing them.

Only 2 of these will execute the RHS of $\&\&$ as for the other $2$ fork1 calls return $0.$

Now, fork2 will create $2$ child processes from the $2$ parent ones executing it. These $2$ child processes will have return value non zero and so skip the fork3 call. The two parents will have the return value $0$, and only those $2$ will execute fork3 creating $2$ new processes. Thus we get in total $4+2+2=8$ processes which all will print "GO 2020" once each. So, the correct answer is $8.$

Comments

Wouldn’t the expression inside brackets be evaluated first(acc to rule of precedence)?

although the answer would still be 8

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can anyone make a diagram of it ?..please

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@herenow https://www.geeksforgeeks.org/short-circuiting-in-c-and-linux/

Because expressions will be evaluated from left to right.

Answer 2

short circuit is considered*