多个前叉语句

Question

This code creates 3 processes in addition to the the original one. So in total 4 processes exist. As far as i know, this code should print 8 statements. However the result is just 4 statements. What am i missing here?

#include <stdio.h>
#include <unistd.h>
#include <sys/types.h>

// error checking is omitted 

int main(int argc, char const *argv[])
{
    pid_t pid, pid2;

    fflush(stdout);// used to clear buffers before forking
    pid = fork();
    fflush(stdout);
    pid2 = fork();

    if(pid == 0) {
        printf("%d is the first generation child from first fork\n", getpid());
    }

    else if(pid > 0) {
        printf("%d is the original process\n", getpid());
        wait();
    }

    else if(pid2 == 0) {
        printf("%d is the 2nd generation child from the second fork by the first generation child  \n", getpid());
    }

    else if(pid2 > 0) {
        printf("%d is the first generation younger child from the 2nd fork by the original\n", getpid() );
        wait();
    }

    return 0;
}

output

4014 is the original process 4016 is the original process 4015 is the first generation child from first fork 4017 is the first generation child from first fork

Answer 1

It's because of else if, each process can only print one line and 4 process means 4 lines.

You should replace:

else if(pid2 == 0) {

By:

 if(pid2 == 0) {

Both tests on pid and pid2 must be done to print 2 lines per process.