shared memory problem

I am supposed to create shared variables in which multiple processes (p1 to p8) update it. Say processes p1-p4 update shared variable x1 and processes p5-p8 update shared variable x2.

how do it do it ?

is it like ?

Code:

main 

create shared variables x1 and x2

for (i=1;i<=4;i++)
{pid=fork()
if(pid==0)
update x1;
}

for (j=1;j<=4;j++)
{pid=fork()
if(pid==0)
update x2;
}

I doubt in it because. I think second fork will create some problems as it will fork children which has childrens generated in earlier for loop..

I am confused how to solve it...

--------------------------------------------------------------------------------

Code:

isChild = 0;
for ( group = 0 ; group < 2 && !isChild; group++ ) {
  for ( i = 0 ; i < 4 && !isChild ; i++ ) {
    if ( fork() == 0 ) { isChild = 1; }
  }
}

// At this point in the code, you will have 9 processes running in parallel
// the parent will have isChild = 0, the rest will have isChild = 1
// 4 of the children will have group = 0
// 4 of the children will have group = 1

ohhh....thanks a lot Salem, I will try it out and let you know if i fall into any trap...

Thanks a lot ...

say if i use only like this

for (i=0;i<2;i++)
fork()

then in that case, will it go like this ??

when i =0

child1 , similar to parent created

when i=1;

child 2 similar to child1 (which is again similar to parent) created.

when i=2;
child3 , similar to child2 (which is again similar to child1, which is similar to parent) is created..

so,

child1 will be like parent
child 2 will be something like parent, child1 and other child
child3 will be something like parent,child1,child2 and child3

in total we will have 6 childs..
am i correct ??

But when I wrote simple prog. and tested, it produced 8 children...

why ??

Trace is as follows:

Code:

i = 0;
parent forks()....
We now have 2 processes.  1 parent and 1 child

i = 1;
parent forks()....
We now have 3 processes.  1 parent and 2 children
child forks()....
We now have 4 processes.  1 parent and 2 children and 1 grandchild

i = 2;
Loop ends.
Total processes: 4

I wrote the following bad code really quickly:

Code:

#include <stdio.h>
#include <wait.h>
                        
int main()
{
        int i, iReturn;
        pid_t pid;
        for(i=0;i<2;i++)
        {
                pid = fork();
                printf("%d reporting for duty\n",getpid());
                if(pid > 0) 
                {
                        pid = wait(&iReturn);
                }
        }
        return 0;
}

Output:

Code:

17244 reporting for duty
17245 reporting for duty
17245 reporting for duty
17246 reporting for duty
17244 reporting for duty
17247 reporting for duty

Now you'll get 6 entries, but notice the identical PID's? That's because you're in a loop and printing notices multiple times by some of the processes.

Taking out the duplicate entries:

Code:

17244 reporting for duty
17245 reporting for duty
17246 reporting for duty
17247 reporting for duty

Surprise! Only 4 processes.