Hide terminal output from Execve - c

Question

I'm making a C program where it basically reads in a line from the user, interprets it and then tries to execute the command with execve. I'm also forking the execve to a child process if '&' is in the input.
Now I wish to hide any terminal output which comes from the execve command when it's running in the child process.
Is there any relatively easy way to do this?

Answer 1

You can hide the output by redirecting stdout and stderr to /dev/null after forking but before execve(). The idea is to open /dev/null, then make stdout and stderr duplicates of the obtained file descriptor with dup2() (which will also close the originals first). It's almost the same as redirecting to a pipe.
An example (incomplete program, and skipping most error checking):
#include <unistd.h>
#include <fcntl.h>
...
int pid = fork();
if (pid == -1) {
/* fork error */
exit(1);
} else if (pid == 0) {
/* child process */
/* open /dev/null for writing */
int fd = open("/dev/null", O_WRONLY);
dup2(fd, 1); /* make stdout a copy of fd (> /dev/null) */
dup2(fd, 2); /* ...and same with stderr */
close(fd); /* close fd */
/* stdout and stderr now write to /dev/null */
/* ready to call exec */
execve(cmd, args, env);
exit(1);
} else {
/* parent process */
...

Answer 2

I've written simple example, maybe it will help you.
First, try to call it without | echo $1 > /dev/null - it should print files. When you add it, output is empty.
#include <stdio.h>
#include <unistd.h>
int main()
{
int ret;
char *cmd[] = { "ls", "-l", (char *)0 };
char *env[] = {(char *)0 };
ret = execve ("/bin/ls | echo $1 > /dev/null", cmd, env);
return 0;
}