Question

the question in the exam was :

write the output of the following program :

int i = 2 ;
int main () {

    int j = 10, p ;
    while (i-- && p == fork())
        if ( p < 0 ) exit(1);
    j += 2;
    if (p == 0) {
        i *= 3;
        j *= 3;
    }
    else {
        i *= 2;
        j *= 2;
    }

    printf("i = %d, j = %d \n",i,j);

    return 0;
}

Output from the console using xcode with include this lines before int i = 2; :

    #include <stdio.h>
    #include <stdlib.h>
    #include <unistd.h>

Output:

i = 3, j = 36 
i = 0, j = 36 
i = -3, j = 36 

Note: I notice that the output is different if we use the Ubuntu.

i think this is the output of the Ubuntu :

i = 2 , j = 24
i = 2 , j = 24

any brief explain or trace would be great Thanks

Answer 1

Assuming that was a typo, and while (i-- && p == fork()) was really while (i-- && (p = fork())), then the output depends on the OS scheduler.

The main process forks off processes with

• i=1 p=0 <– process A, parent.p=A
• i=0 p=0 <– process B, parent.p=B
• i=-1 p=B <– parent process, parent.p = B

Processes A and B do not continue the loop, because the p=fork() evaluates as false for them.

Each process adds 2 to j (which might as well be j=12). In summary:

A: i=+1 p=0    j=12
B: i= 0 p=0    j=12
P: i=-1 p=*B*  j=12

The cases where p=0, have i and j multiplied by 3, the ones with p!=0 (the parent process) have them multiplied by 2. This yields the following perfectly reasonable output for me:

i = -2, j = 24
i = 3, j = 36
i = 0, j = 36

(The order is going to be somewhat random)

As sharptooth pointed out, your code as written is simply producing random results, depending on a piece of uninitialized memory.