Question

The output of the following code is "11 0" But I am not sure why, can someone please explain to me clearly and concisely why this is the case?

main(){
int n=1;
if(fork())
wait(&n);
else
n=n+10;
printf("%d\n", n);
exit(0);
}

Answer 1

main(){
int n=1;
if(fork())
wait(&n);
else
n=n+10;
printf("%d\n", n);
exit(0);
}

fork() is a system call to create a process. When the fork() method is called a duplicate copy of itself is created called as child process. In the above code when fork() method is called a duplicate is created and this child process starts executing the lines return after calling fork method.

fork() method returns a negative value if no child process is created. It returns a positive pid number to the parent process if a child process is created. And it returns 0 to the child process if child process is created.

So in short if a child process is created then 0 is returned to child process and pid number of child is returned to parent process.

Now the above code after fork() is executed the code runs two times(one by parent process and another by child process).

• For child process the fork() method has returned 0 and hence it moves to else part and computes n = n+10 and n becomes 11.
• For the parent process the fork() method has returned a pid number of child and hence it enters the if condition and wait(&n) is executed.

There is a difference between wait() and wait(&n).

• wait() or wait(null) - This will only wait for any child to terminate.
• wait(&n) or wait(&status) - This also will wait for any child to terminate but with additional information that is the status code of the termination which is the exit code. Since its exit(0) its printing 0 as n becomes 0 since n is used in wait(&n).

Summary of the output 11 and 0 :

• 11 is printed by the child process which enters else condition and n = n+10 is executed. and n becomes 11. And in the next print statement n is executed as 11.
• 0 is printed by the parent process as wait(&status) gets the status code from the exit(0) as 0 and again executes the print status as n which is now 0. If you change the exit(2) then the output will be 11 and 512(status code of exit(2)).
• Else if wait(&n) is changed to different variable such a s, then the output would be 11 printed by child process and 1 as n would not be changed and would remain as 1, printed by parent process.

Please leave a comment if u need more explanation.