Question

I have written this code:

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

#define N 10

int fd[2];

void son ()
{
int data=0;
if(read(fd[0], &data, sizeof(int))==sizeof(int))
    fprintf(stderr,"%d\n",data);
else
    fprintf(stderr,"Read failed\n");
}

int main(int argc, char** argv)
{
pid_t sons[N];
pipe(fd);
int data=5;
for(unsigned int i=0; i<N;i++)
{
    sons[i]=fork();
    if(!sons[i])
    {
        son();
        break;
    }
}
write(fd[1], &data, sizeof(int));
data=6;
}

If I try executing it, I get 10 times printed “5”.
What happened? Why all processes have read the same data? Can’t the data be read just once? I thought the reader process was consuming the data read.
I modified the code this way:

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

#define N 10

int fd[2];

void son ()
{
int data=0;
if(read(fd[0], &data, sizeof(int))==sizeof(int))
    fprintf(stderr,"%d\n",data);
else
    fprintf(stderr,"Read failed\n");
if(read(fd[0], &data, sizeof(int))==sizeof(int))
    fprintf(stderr,"%d\n",data);
else
    fprintf(stderr,"Read failed\n");
}

int main(int argc, char** argv)
{
pid_t sons[N];
pipe(fd);
int data=5;
for(unsigned int i=0; i<N;i++)
{
    sons[i]=fork();
    if(!sons[i])
    {
        son();
        break;
    }
}
write(fd[1], &data, sizeof(int));
data=6;
write(fd[1], &data, sizeof(int));
return 0;
}

So I write/read two times, the output I get is “5” and “6”, just once, doesn’t happen that all the 10 processes can read the pipe.Why this?

Answer 1

When you fork() you execute son(), then break out of the loop – but that is not the end of the child process! The child then continues and writes to fd[1].