Thread: Forking

When you fork and modify a variable, is it modified in the parent process too?

Well, in that case, is there any way to start a new thread sharing the memory with its parent?

Nevermind. Just started reading up on pthreads.

or use "shared memory".
http://www.cs.cf.ac.uk/Dave/C/node27...00000000000000

Originally Posted by Thantos

IIRC no. When the child process is created it gets it's own memory area.

Test program:

Code:

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

int main (void)
{
  int fork_return;
  int x = 5;

  x++;

  fork_return = fork();

  if ( fork_return == 0)
  {
    sleep(10);
    printf("In parent: ");
    x+=10;
  }
  else
  {
    x*= 10;
    printf("In child: ");
  }

  printf(" %d\n", x);

  return 0;
}

Result:
In child: 60
In parent: 16

Shouldn't be like this:

Code:

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

int main (void)
{
  int fork_return;
  int x = 5;

  x++;

  fork_return = fork();

  if ( fork_return == 0)
  {
    sleep(10);
    printf("In child: ");
    x+=10;
  }
  else
  {
    x*= 10;
    printf("In parent: ");
  }

  printf(" %d\n", x);

  return 0;
}

Just a simple question:

Consider this code:

Code:

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

int main(void) {
    int fork_return;
    fork_return = fork();

    if( fork_return == 0 ) {
        sleep(5);
        printf("Child only ");
    }
    else {
        printf("Parent only! ");
    }

    printf("I am touchabled by parent and child");

    return 0;
}

Then this:

Code:

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

int main(void) {
    int fork_return;
    fork_return = fork();

    if( fork_return == 0 ) {
        sleep(5);
        printf("Child only ");
        return 0;
    }
    else {
        printf("Parent only! ");
    }

    printf("Now I am  touchabled by parent only");

    return 0;
}

Ok, I understand this two code. But look at this one:

Code:

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

int main(void) {
    int fork_return;
    fork_return = fork();

    if( fork_return == 0 ) {
        sleep(5);


        //Why does this not show up in console
        printf("Child only ");
        //End of Why


        char *arg[] = {
              "ls",
              NULL
        };
        execvp("ls",arg);
    }
    else {
        printf("Parent only! ");
    }

    printf("Now I am touchabled by parent only");

    return 0;
}

Why?????

Just as reference, I use Slackware 9.1 and GNU C++ compiler version 3.3.3.

> Why?????
1) Try printing a newline with that as well
2) Try calling fflush(stdout) before the exec call

Output is buffered, so my guess is the buffers get trashed along with everything else owned by the program, when the exec happens.
Incomplete (ie unflushed) output is lost.

>>You can only declare variables at the beginning of a block in C
It's a C99 extension that allows this, maybe his compiler has it too. Or maybe it's being compiled as c++

Originally Posted by Salem

> Why?????
1) Try printing a newline with that as well
2) Try calling fflush(stdout) before the exec call

Output is buffered, so my guess is the buffers get trashed along with everything else owned by the program, when the exec happens.
Incomplete (ie unflushed) output is lost.

That works. Thank you.....