Question

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

int main()
{
    int a = 1, b = 1;
    int rval, pid;

    pid = fork();
    switch(pid)
    {
        case -1:
            printf("I am bad.\n");

        case 0:
            printf("I am in child.\n");
            rval = a + b;
            printf("leaving child.\n");
        default:
            wait();
            printf("I am back to parent.\n");
            //wait();
            printf("%d \n", rval);
            printf("leaving parent.\n");
    }
    return 0;
}

I was expecting child statements first, then parents. But there shouldn’t be any reptition. Instead,

>> ./demo
I am in child.
leaving child.
I am back to parent.
2 
leaving parent.
I am back to parent.
134513584 
leaving parent.

Parent is repeated twice. Why is that?
Furthermore, how did parent get 2 from the child? Child’s has its own virtual memory for a and b so how did we transfer that result to parent?
I am confused.

---

Further investigation
When I insert return 0 into the case 0, it will not repeat. It looks like the switch statement continues. Did I miss something about switch statement?

Answer 1

In C, switch statements fall through cases by default; you have to put a break; (or return) statement at the end of each case, otherwise, e.g., the child will execute both the child and the parent code.