minishell created in c not working as expected, pipes related - c

Question

im making an UNIX minishell in c, in my OS signature. I only have to improve the shell itself, we have a premade parser for implement yacc and a scanner for implement lex among other files.
The file to modify is msh.c, the main executable file.
Well, the problem begins when I try to implement the pipe secuences. This is what I made:
int executePipeLine (char*** argvv, int bg, char** filev, int n){
int i;
int in = 0;
pid_t pid;
int fd[2];
for (i = 0 ; i < n-1 ; i++){
pipe(fd);
pid_t pid = fork();
if (pid == 0){ //child
if (in != STDIN_FILENO){
dup2(in, STDIN_FILENO);
close(in);
}
if (fd[1] != STDOUT_FILENO){
dup2(fd[1], STDOUT_FILENO);
close(fd[1]);
}
execvp(argvv[i][0], argvv[0]);
}
else { //parent
close(fd[1]);
in = fd[0];
}
}
if(in != STDIN_FILENO){
dup2(in, STDIN_FILENO);
close(in);
}
pid_t lastpid = fork();
if(lastpid == 0){ //child
execvp(argvv[i][0], argvv[0]);
}
if(lastpid == -1){
perror("no se pudo crear el hijo\n");
exit(-1);
}
else { //parent
/* not bg*/
if(!bg ) {
int status;
while (wait(&status) != lastpid); /* wait the child. */
}
else {
/*bg mode*/
printf("pid del proceso last: %d\n", lastpid);
}
return 0;
}
}//end executePipeLine
This seems to work fine. In the main , where we have an extern function called obtain_order(); that returns the number of commands + 1 and save it to ret. If ret is 1 we continue with the prompt, if is 0 means EOF (Control + D keybinding) to end the shell, and if is >1 execute the commands.
Here you have it:
int main(void)
{
char ***argvv;
int command_counter;
int num_commands;
int args_counter;
char *filev[3];
int bg;
int ret;
int reset = 0;
setbuf(stdout, NULL); /* Unbuffered */
setbuf(stdin, NULL);
while (1)
{
fprintf(stderr, "%s", "msh> "); /* Prompt */
ret = obtain_order(&argvv, filev, &bg);
printf("ret: %d\n", ret);
if (ret == 0) break; /* EOF */
if (ret == -1) continue; /* Syntax error */
num_commands = ret - 1; /* Line */
if (num_commands == 0) continue; /* Empty line */
if(num_commands > 1){
executePipeLine(argvv, bg, filev, num_commands);
}
else if (num_commands == 1){
executeCommand(argvv, bg, filev);
}
} //fin while
return 0;
} //end main
All works fine with a simple command. The problem is when I try to execute a pipe line. It show a good result, but i dont know we, after that ret always is 0 in the next iteration, so every time I try to execute a pipe line, it works but close the shell process, and have to execute it again instead of continue with the promt.
You know what is the problem here?
I hope you understand me, my english is not perfec. Thanks

Answer 1

In the parent process (i.e. your mini-shell) you are duping the standard input:
...
else { //parent
close(fd[1]);
in = fd[0];
}
...
if(in != STDIN_FILENO){
dup2(in, STDIN_FILENO);
close(in);
}
When your pipeline terminates, children are dead, pipes are no longer readable => stdin is considered as closed. When you later call obtain_order in your main loop, it returns 0 (EOF) and your program exits.