Thread: Process get's killed when invoking bash interactive with oneliner

Hi,

I was coding a shell for fun and I discovered a weird bahaviour when invoking bash like this :

Code:

bash -ic <some_command>

Basically instead of looping through, the program gets killed and I don't really know why. Could someone have an explanation ?

Here is the basic code :

Code:

#include<stdio.h>
#include<stdlib.h>
#include<unistd.h>

void loop_pipe (char***cmd)
{
     pid_tpid,wpid;
     intstatus;

     pid=fork();
     if(pid==0)
     {
     // Child process
     if(execvp((*cmd)[0],*cmd)==-1)
         perror("Command error");
     exit(EXIT_FAILURE);
     }
     elseif(pid<0)
     {
         // Error forking
         perror("Fork error");
     }
     else
     {
         // Parent process
         do{
             wpid=waitpid(pid,&status,WUNTRACED);
         }while(!WIFEXITED(status)&&!WIFSIGNALED(status));
     }

}

int main()
{
     char*ls[]={"bash","-ic","uname",NULL};
     char**cmd[]={ls,NULL};

     while(1)
     {
         loop_pipe(cmd);
     }
}

Thanks !

Please edit your post and copy/paste your code as plain text.

Presumably your problem is that you are forcing the shell to run interactively with -i. Try getting rid of that.

BTW, you're missing <sys/wait.h> for waitpid.

I just did that, I don't know where this weird colored formatting thing came from.

The thing is that I'm coding a shell to I have to prevent this from happening. But how does running bash in interactive mode completely kills the parent process that executed it ? I'm forking the process so again how does bash gets to have my shell killed ?

Not to mention that running bash with only the -i option doesn't to anything, it's rather combination or -ic that completely wrecks my shell or this code above

EDIT :

I think I got confused. My process wasn't being killed. It was suspended due to the interactive options you were right. But now I want to simulate the behavior of bash. If I run bash -ic, my previous bash doesn't get suspend. Whereas my shell does and I want to avoid that bu I don't know how then.

Looks like you might have just copy/pasted your code again from this website. Notice how all the spacing is gone? Look at the first two lines in loop_pipe.
Anyway, obviously if you are making your own shell you can't invoke bash. That would be pointless! Your shell will be a replacement for the bash shell.

EDIT: just saw your edit.
Yeah, I was going to say that for me it's suspended.

If I run bash -ic, my previous bash doesn't get suspend.

Why would it? The bash you ran with bash -ic executes its command and exits. It's that bash (not the "previous" one) that is getting suspended in your program. I'm not entirely sure why, but I suspect you cannot combine -i and -c since the definition of -c is to run the given command and exit.

EDIT2:
It doesn't get suspended if you comment out the while(1) in main, therefore only calling loop_pipe once. Although it is killed, i.e., it doesn't just drop you into a second invocation of bash.

Originally Posted by john.c

Looks like you might have just copy/pasted your code again from this website. Notice how all the spacing is gone? Look at the first two lines in loop_pipe.
Anyway, obviously if you are making your own shell you can't invoke bash. That would be pointless! Your shell will be a replacement for the bash shell.

I just saw that it's now impossible for me to edit the main post. So if a mod could do it for me that would be awesome thanks.

Originally Posted by john.c

Why would it? The bash you ran with bash -ic executes its command and exits. It's that bash (not the "previous" one) that is getting suspended in your program. I'm not entirely sure why, but I suspect you cannot combine -i and -c since the definition of -c is to run the given command and exit.

I mean, if it was bash that is running with -ic that got suspended, why does the program stop to execute then ? I'm still a bit confused here.
I get the point where bash executes it's command, but then when it finishes, I should be putted back to my shell or in this case, I should be able to run again the same command, that is bash -ic uname. But somehow it gets suspended and I have to manually tell bash to resume the program, effectively iterate the infinite loop manually

Originally Posted by john.c

EDIT2:
It doesn't get suspended if you comment out the while(1) in main, therefore only calling loop_pipe once. Although it is killed, i.e., it doesn't just drop you into a second invocation of bash.

Well that would solve the question reading my example for sure. But how about my shell then ? I'm using the exact same "logic", get input from user, fork, execute command given by the user, wait on it to finish and repeat the whole process. So if the user passes bash -ic <command> well my shell gets suspended and bash takes the lead. I'm still unsure what's going on I'm already sorry for my slow understanding

I'm still a bit confused here.

I'm confused too. I have no idea why it's being suspended.
I see what you mean that if you made an actual shell it would have trouble running a (possibly malformed) command like 'bash -ic command' and would apparently get suspended.
Clearly bash does not get suspended when you run that command.

BTW, here's a fixed-up version of your original code (slightly reorganized but the same) so someone can more easily copy it to run it.

Code:

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/wait.h>
 
void loop_pipe(char***cmd)
{
    pid_t pid = fork();
    if (pid < 0)
    {
        perror("fork you");
        exit(EXIT_FAILURE);
    }
 
    if (pid == 0)
    {
        // Child process
        execvp((*cmd)[0], *cmd); // does not return on success
        perror("execvp error");
        exit(EXIT_FAILURE);
    }
 
    // Parent process
    int status;
    do {
        waitpid(pid, &status, WUNTRACED);
    } while (!WIFEXITED(status) && !WIFSIGNALED(status));
}
 
int main()
{
    char *ls[] = {"bash", "-ic", "uname", NULL};
    char **cmd[] = {ls, NULL};
 
    while (1)
    {
        loop_pipe(cmd);
    }
    return 0;
}