C programming child process is not executing

Question

So this is simple program of creating two process: parent and child. So what I did is have the greeting inside the parent and the name inside the child process. For some reason my child process is not printing despite that I called wait() inside the parent. What should I do?

GOAL OUTPUT: "Hello Sam" OUTPUT I"M GETTING: "Hello"

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

int main()
{
    char *greeting = "Hello";
    char *name;


    pid_t pid;

    pid = fork();

    if(pid == 0)
    {
      name = "Sam";
      exit(0);
    }
    else if(pid == -1)
        printf("Fail\n");
    else{
      wait(0);
     printf("%s %s", greeting, name);
  }

} 

Answer 1

When you make a call to fork(), the child process will receive a copy of the parent process' address space (variables etc...). This means that in the child, "name" is defined, from the parent. However, "name" in the child process is just a copy. So modifying it in the child process does not affect the parent.

To get the behavior that I sense you're after, replace fork() with vfork(). For the purposes of this discussion, the only difference is that:

1. The address space is shared instead of copied. Editing "name" in the child process will be reflected in the parent process
2. The parent process is suspended while the child process executes. I assume that this is OK, because you are already calling wait() in the parent process

Edit: I forgot to add that if you go the vfork route, change exit() to _exit() in the child process

Answer 2

Child process once forked will run independantly of the parent process. So when you set name="Sam" in child it is in different program, than the printf. So you are not able to see the message.

Answer 3

In the child process, you assigned to name and then exit.

if(pid == 0)
{
  name = "Sam";
  exit(0);
}

The printf("%s %s", greeting, name); line is executed only in the else branch, ie, the parent process. It's actually undefined behavior because name is uninitliazed.