Hello,
I'm new to UNIX programming (and UNIX in general) and am trying to
learn it to improve my career.  I have written the following program
but do not understand why I get the duplicate line printed out:

Here's the output that I've received and you'll note duplicate "Value
of PID..." output:

#include <stdio.h>

#include <unistd.h>
using std::pipe;

#define SIZE 2048

int main(int argc, char **argv)
{
int pipefd[2];
int what_we_read;
int process_id;
char buffer_p[SIZE];

if (pipe(pipefd) == -1)
{
perror("no pipe");
exit(1);
}
if ((process_id = fork()) < 0)
{
perror("no fork");
exit(1);
}

printf("\nValue of process_id before if statement: %i\n",
process_id);

if (process_id == 0)
{
printf("\nEntering child process.\n");
close(pipefd[1]);
while ((what_we_read = read(pipefd[0], buffer_p, SIZE)) != 0)
printf("from the pipe the forked child has read %s\n", buffer_p);
close(pipefd[0]);
} else {
printf("\nEntering parent.\n");
close(pipefd[0]);
strcpy(buffer_p, "howdyyall");
write(pipefd[1], buffer_p, strlen(buffer_p)+1);
close(pipefd[1]);
}
exit(0);
}

Here's the output and you'll note that duplicate "Value of
process_id..." line:

Value of process_id before if statement: 0

Entering child process.

Value of process_id before if statement: 17706

Entering parent.
from the pipe the forked child has read howdyyall

End of program output.

I can't understand why I get the duplicate line.  I clearly have it
above the if statement starting the child/parent interaction.   I
expected it only to appear once, I don't have a good explanation as to
why the duplicate line (I know it's not an exact match of course).  The
child and parent execute and then exit, no need for the printf
statement to repeat.

I looked through the man pages for pipe and fork and didn't see any
rationale as to what I see this behaviour.

Unfortunately, I do not have access to a UNIX workstation and have to
compile and run this using Windows telnet session, so I can't debug it
as I normally would using MSVC.  I would like to use ddd but it's not
possible to do that from a Windows machine.

Kim.

[ Post a follow-up to this message ]

In article <1159572846.387311.63640@h48g2000cwc.googlegroups.com>,
"kimi" <testcpp@gmail.com> wrote:

> Hello,
> I'm new to UNIX programming (and UNIX in general) and am trying to
> learn it to improve my career.  I have written the following program
> but do not understand why I get the duplicate line printed out:
>
> Here's the output that I've received and you'll note duplicate "Value
> of PID..." output:
>
> #include <stdio.h>
>
> #include <unistd.h>
> using std::pipe;
>
> #define SIZE 2048
>
> int main(int argc, char **argv)
> {
>   int pipefd[2];
>   int what_we_read;
>   int process_id;
>   char buffer_p[SIZE];
>
>   if (pipe(pipefd) == -1)
>   {
>     perror("no pipe");
>     exit(1);
>   }
>   if ((process_id = fork()) < 0)
>   {
>     perror("no fork");
>     exit(1);
>   }
>
>   printf("\nValue of process_id before if statement: %i\n",
> process_id);
>
>   if (process_id == 0)
>   {
>     printf("\nEntering child process.\n");
>     close(pipefd[1]);
>     while ((what_we_read = read(pipefd[0], buffer_p, SIZE)) != 0)
>       printf("from the pipe the forked child has read %s\n", buffer_p);
>     close(pipefd[0]);
>   } else {
>     printf("\nEntering parent.\n");
>     close(pipefd[0]);
>     strcpy(buffer_p, "howdyyall");
>     write(pipefd[1], buffer_p, strlen(buffer_p)+1);
>     close(pipefd[1]);
>   }
>   exit(0);
> }
>
> Here's the output and you'll note that duplicate "Value of
> process_id..." line:

I don't see any duplicated lines.  I see one line that says 0, and
another that says 17706.  The first line is printed by the child, the
second one is printed by the parent.

>
> Value of process_id before if statement: 0
>
> Entering child process.
>
> Value of process_id before if statement: 17706
>
> Entering parent.
> from the pipe the forked child has read howdyyall
>
> End of program output.
>
> I can't understand why I get the duplicate line.  I clearly have it
> above the if statement starting the child/parent interaction.   I
> expected it only to appear once, I don't have a good explanation as to
> why the duplicate line (I know it's not an exact match of course).  The
> child and parent execute and then exit, no need for the printf
> statement to repeat.

After fork, you have two processes executing the same program; the only
difference is the value of process_id.  They will both execute all the
statements preceding the if(), then they will diverge because the
condition in the if() will be true in the child, false in the parent.

Since the printf() is before the if, they both execute it, so they each
print their value of process_id.

--
Barry Margolin, barmar@alum.mit.edu
Arlington, MA
*** PLEASE post questions in newsgroups, not directly to me ***
*** PLEASE don't copy me on replies, I'll read them in the group ***

[ Post a follow-up to this message ]

Barry,
Thank you for your reply.

I wouldn't have thought the process would start a separate copy of the
same program without having to make a call to that program, but I know
now, will have to reread the man section again on fork.  I'm sure
there's some good books out there I could purchase.

Kim

Barry Margolin wrote:
> In article <1159572846.387311.63640@h48g2000cwc.googlegroups.com>,
>  "kimi" <testcpp@gmail.com> wrote:
> 
>
> I don't see any duplicated lines.  I see one line that says 0, and
> another that says 17706.  The first line is printed by the child, the
> second one is printed by the parent.
> 
>
> After fork, you have two processes executing the same program; the only
> difference is the value of process_id.  They will both execute all the
> statements preceding the if(), then they will diverge because the
> condition in the if() will be true in the child, false in the parent.
>
> Since the printf() is before the if, they both execute it, so they each
> print their value of process_id.
>
> --
> Barry Margolin, barmar@alum.mit.edu
> Arlington, MA
> *** PLEASE post questions in newsgroups, not directly to me ***
> *** PLEASE don't copy me on replies, I'll read them in the group ***

[ Post a follow-up to this message ]