Draw a process tree for this code int main int x 0 int y

Draw a process tree for this code.

int main() {

int x = 0;

int y = 0;

... ...

x = fork();

y = fork();

if( x == y)

fork();

... ...

}

Solution

int main() {

int x = 0;

int y = 0;

... ...

x = fork();

y = fork();

if( x == y)

fork();

... ...

}

fork()-A

fork()-B

fork()-C

At level 0,we have only main process.The main will create child c1 and both will continue execution.The children are numbered in increasing order of their creation.

At level 1, we have m and C1 running, and ready to execute fork() – B. (Note that B, C and D named as operands of && and || operators). The initial expression B will be executed in every children and parent process running at this level.

At level 2, due to fork() – B executed by m and C1, we have m and C1 as parents and, C2 and C3 as children.

The return value of fork() – B is non-zero in parent, and zero in child. Since the first operator is &&, because of zero return value, the children C2 and C3 will not execute next expression (fork()- C). Parents processes m and C1 will continue with fork() – C. The children C2 and C3 will directly execute fork() – D, to evaluate value of logical OR operation.