Could someone Help me with this code please?
#include <stdio.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <string.h>
#include <fcntl.h>
#define SIZE_B 256 /*buffer's size */
#define NUM_ARG 20 /* max number of args for any command */
int main()
{
char *arg1[NUM_ARG];
char *arg2[NUM_ARG];
int pippa;
pid_t pid1,pid2;
arg1[0]="ls"; /*first command*/
arg1[1]="-l";
arg1[2]=0;
arg2[0]="grep"; /*second command*/
arg2[1]="mishell";
arg2[2]=0;
mkfifo("./mishell_pipa",S_IRWXU); /*make fifo*/
pippa=open("./mishell_pipa",O_RDWR|O_CREAT|O_TRUNC|O_SYNC); /*get fd*/
unlink("./mishell_pipa"); /*unlink file*/
if(!(pid1=vfork())) /*first process*/
{
dup2(pippa,STDOUT_FILENO);
execvp(arg1[0],arg1);
_exit(0);
}
else
{
waitpid(pid1,NULL,0); /*wait first process*/
if(!(pid2=vfork())) /*second process*/
{
dup2(pippa,STDIN_FILENO);
execvp(arg2[0],arg2);
_exit(0);
}
else
waitpid(pid2,NULL,0); /*wait second process*/
return 0;
}
}
I have to create a pipe between 2 process such as the output of the first process
is the input of the second.
In practice I have to do the same thing that the bash does using the follow command:
ls -l|grep mishell
The program works well, but after that grep gives me its output it doesn't come back to
the shell, but stays blocked.
I have tried also to use the mode O_NONBLOCK in the open function, but so the first problem leaves
its place to another. The programm works well, but at the end grep gives me this error:
grep: (standard input): Resource temporarily unavailable
I think that the problem is that grep attend a particular kind of input to terminate, and ls doesn't
give it this input. I've tried to write a EOF in the fifo after the execution of ls, but there
aren't changes.
What can I do????
Help me please!
If you're going to use a named pipe you would need to open it twice. But you should just use a pipe. Your first process will not create your third process until after the second process finishes. But the second process would be waiting for the third process to read its data. This is a deadlock. You only need two processes anyway. And don't use vfork() anymore, use fork().
#include <stdio.h>
#include <unistd.h>
#include <sys/types.h>
#define NUM_ARG 20 /* max number of args for any command */
int main()
{
char *arg1[NUM_ARG];
char *arg2[NUM_ARG];
pid_t pid1,pid2;
int pipefd[2];
arg1[0]="ls"; /*first command*/
arg1[1]="-l";
arg1[2]=0;
arg2[0]="grep"; /*second command*/
arg2[1]="mishell";
arg2[2]=0;
pipe(pipefd);
if(pid1=fork())
{
dup2(pipefd[1],STDOUT_FILENO);
close(pipefd[0]);
execvp(arg1[0],arg1);
}
else
{
dup2(pipefd[0],STDIN_FILENO);
close(pipefd[1]);
execvp(arg2[0],arg2);
}
}
Thanx Perderabo, I have solved my problem using as pipe as fifo thanks your suggestion of open
the pipe twice.
I have modifyed your source because your line:
if(pid1=fork())
was incorrect.
I hope that you don't judge me presumptuos, but the lack of the ! was harmful for the program. 
I tested it before I posted. I just retested and cannot get my code to fail. What problem did you see? What version of unix are you using?
The prolems are 2.
1) the lack of the ! into the (pid1=fork())
2) the lack of a wait or waitpid function after the else.
The output of your program is correct, but sometimes the first output line is wrote on the shell line, and
always the program waits an enter to terminate.
I have tested your code with a Linux Mandrake 8.2 .
This is your code with my changes:
#include <stdio.h>
#include <unistd.h>
#include <sys/types.h>
#define NUM_ARG 20 /* max number of args for any command */
int main()
{
char *arg1[NUM_ARG];
char *arg2[NUM_ARG];
pid_t pid1,pid2;
int pipefd[2];
arg1[0]="ls"; /*first command*/
arg1[1]="-l";
arg1[2]=0;
arg2[0]="grep"; /*second command*/
arg2[1]="mishell";
arg2[2]=0;
pipe(pipefd);
if(!(pid1=fork()))
{
dup2(pipefd[1],STDOUT_FILENO);
close(pipefd[0]);
execvp(arg1[0],arg1);
exit(0);
}
else
{
waitpid(pid1,NULL,0);
dup2(pipefd[0],STDIN_FILENO);
close(pipefd[1]);
execvp(arg2[0],arg2);
exit(0);
}
}