Pipe commands with fork and dup2 - c

Question

I wrote the following code in order to pipe two commands:
#include <stdlib.h>
#include <unistd.h>
char *program_1[3] = {"/bin/cat", "/dev/random", NULL};
char *program_2[2] = {"/bin/ls", NULL};
char *program_3[2] = {"/usr/bin/sort", NULL};
int main(void)
{
int fd[2];
int pid;
pipe(fd);
if ((pid = fork()) == 0) //Child process
{
dup2(fd[1], STDOUT_FILENO);
close(fd[0]);
execve(program_3[0], program_3, NULL);
}
else if (pid > 0) //Parent process
{
dup2(fd[0], STDIN_FILENO);
close(fd[1]);
execve(program_2[0], program_2, NULL);
}
return (EXIT_SUCCESS);
}
Each pair of program_x / program_y where x != y works fine, except this one.
When i pipe sort into ls, ls well prints its output on stdout, but then, sort throw this error: sort: Input/output error.
When I type sort | ls into bash, it prints ls result as my program, but then waits for input.
Am I doing someting wrong ?
edit: I'm trying to reimplement the shell's behaviour

Answer 1

The problem is that when ls finishes, the parent process will exit which will close the read-end of the pipe, which will lead to an error being propagated to the write-end of the pipe which is detected by sort and it write the error message.
That it doesn't happen in the shell is because shells handle pipes differently than your simple example program, and it keeps the right-hand side of the pipe open and running (possibly in the background) until you pass EOF (Ctrl-D) to the sort program.

Answer 2

Your program isn't quite equivalent to what a shell typically does.
You're replacing the parent with ls; whereas shell would create who child processes and connect them and wait for them to finish.
It's more like:
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
char *program_2[2] = {"/bin/ls", NULL};
char *program_3[2] = {"/usr/bin/sort", NULL};
int main(void)
{
int fd[2];
pid_t pid;
pid_t pid2;
pipe(fd);
if ((pid = fork()) == 0) //Child process
{
dup2(fd[1], STDOUT_FILENO);
close(fd[0]);
execve(program_3[0], program_3, NULL);
}
else if (pid > 0) //Parent process
{
if ( (pid2 = fork()) == 0) {
dup2(fd[0], STDIN_FILENO);
close(fd[1]);
execve(program_2[0], program_2, NULL);
}
}
waitpid(pid, 0, 0);
waitpid(pid2, 0, 0);
return (EXIT_SUCCESS);
}

Answer 3

I finally found the solution, we were close to:
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
char *cat[3] = {"/bin/cat", "/dev/random", NULL};
char *ls[2] = {"/bin/ls", NULL};
char *sort[2] = {"/usr/bin/sort", NULL};
int main(void)
{
int fd[2];
pid_t pid;
pid_t pid2;
pipe(fd);
if ((pid = fork()) == 0)
{
dup2(fd[1], STDOUT_FILENO);
close(fd[0]);
execve(cat[0], cat, NULL);
}
else if (pid > 0)
{
if ( (pid2 = fork()) == 0)
{
dup2(fd[0], STDIN_FILENO);
close(fd[1]);
execve(ls[0], ls, NULL);
}
waitpid(pid2, 0, 0);
close(fd[0]);
}
waitpid(pid, 0, 0);
return (EXIT_SUCCESS);
}
We need to close the read end of the pipe once the last process ends, this way, if the first process tries to write on the pipe, an error will be throwed and the process will exit, else if it only reads from stdin as sort, it will keep reading as stdin is still open