C Read stdout from multiple exec called from fork - c

Question

In the code below a process creates one child (fork()) and then the child replaces itself by calling exec(). The stdout of the exec is written in a pipe instead of the shell. Then the parent process reads from the pipe what the exec has written with while (read(pipefd[0], buffer, sizeof(buffer)) != 0)
Can someone tell me how to do the exact same thing as described above but with N number of children processes (who replace themselves with exec as above).
int pipefd[2];
pipe(pipefd);
if (fork() == 0)
{
close(pipefd[0]); // close reading end in the child
dup2(pipefd[1], 1); // send stdout to the pipe
dup2(pipefd[1], 2); // send stderr to the pipe
close(pipefd[1]); // this descriptor is no longer needed
exec(...);
}
else
{
// parent
char buffer[1024];
close(pipefd[1]); // close the write end of the pipe in the parent
while (read(pipefd[0], buffer, sizeof(buffer)) != 0)
{
}
}

Answer 1

I found the answer. I made an array of pipes so that a process does not overwrite the output of another process.
Here is my code. Do you find any mistake?
#include <signal.h>
#include <stdio.h>
#include <sys/types.h>
#include <unistd.h>
#include <stdlib.h>
#include <sys/time.h>
#define N 10
int main(int argc, char *argv[]) {
ssize_t readlen;
int pipefd[N][2];
int i;
for (i = 0; i < N; i++) {
pipe(pipefd[i]);
}
int pid = getpid();
for (i = 0; i < N; i++) {
if (fork() == 0) //The parent process will keep looping
{
close(pipefd[i][0]); // close reading end in the child
dup2(pipefd[i][1], 1); // send stdout to the pipe
dup2(pipefd[i][1], 2); // send stderr to the pipe
close(pipefd[i][1]); // this descriptor is no longer needed
char b[50];
sprintf( b, "%d", i);
execl("/bin/echo", "echo", b,NULL);
}
}
if (pid == getpid()) {
// parent
char buffer[1024];
for (i = 0; i < N; i++) {
close(pipefd[i][1]); // close the write end of the pipe in the parent
while ((readlen=read(pipefd[i][0], buffer, sizeof(buffer))) != 0)
{
buffer[readlen] = '\0';
}
printf("%s\n",buffer);
}
}
}

Answer 2

Maybe this code would do the job:
const int N = 10; //Number of child processes
int pipefd[2];
pipe(pipefd);
int i;
for (i = 0; i < N; i++) {
if (fork() == 0) //The parent process will keep looping
{
close(pipefd[0]); // close reading end in the child
dup2(pipefd[1], 1); // send stdout to the pipe
dup2(pipefd[1], 2); // send stderr to the pipe
close(pipefd[1]); // this descriptor is no longer needed
exec(...);
}
}
// parent
char buffer[1024];
close(pipefd[1]); // close the write end of the pipe in the parent
while (read(pipefd[0], buffer, sizeof(buffer)) != 0)
{
}
WARNING: the output will be mixed. If you want all processes to dump data without being mixed, then you should manage to synchronize processes (by means of public locks, for example).

Answer 3

I think you can create named chanel in any place of the file system (like a local socket) and read all received data to parent process. So child processes must write their getted data to this channel. It will be unix-like architecture.