Child process not exiting with fork()

Question

I am creating a simple Linux command shell in C. I am having trouble understanding where my code is having problems. "commands" is a list of strings of Linux commands that I want to be executed concurrently as the children processes of one parent. When all are done executing, I want the parent to exit the function. However, when I call exit(0), the for loop continues and parses the current command again, causing the args to be executed again in execvp. Am I using fork() and wait() correctly here? I have tried using waitpid() as well with no luck.

void executeShell(char** commands){
    char **arr = commands;
    char *c;
    pid_t pid, wpid;
    int status = 0;

    for (c = *arr; c; c=*++arr){
        // printf("%d-\n", strcmp(command, "exit"));

        if (strcmp(c, "exit") == 0){
            EXIT = 1;
            return;
        }

        printf("Running command \'%s\'...\n", c);

        char** args = parseStringToTokenArray(c, " ");
        free(args);

        /* fork and execute the command */
        pid = fork();
        if(pid < 0){
            perror("fork() error\n");
            return;
        }
        /* child process executes command */
        else if (pid == 0){
            /* 'cd' is not a part of /bin library so handle it manually */
            if (strcmp(args[0], "cd") == 0){
                changeDirectory(args[1]);
            }
            else if (strcmp(args[0], "sdir") == 0){
                searchDirectory(args[1]);
            }else{
                /* parse commands with arguments */
                execvp(args[0], args);//execute the command
            }
            exit(0);// <-----command is finished, so why no exit?
        }                                     
    }                                          
    /* wait for children to complete */          
    while((wpid = wait(&status)) > 0);         
}

As @nissimabehcera mentioned, looks like the father execution is still going on back to the loop. — Miguel Ruivo, Feb 25 '19 at 10:35
Why do you `free(args);` right after creating it ? Shouldn't you wait until after you used it (or no longer need it) ? — Sander De Dycker, Feb 25 '19 at 10:54
What commands are you forking off as children? Do they ever finish or do they run indefinitely? — ilkkachu, Feb 25 '19 at 11:04

JeremyP · Accepted Answer · 2019-02-25T19:33:31.587

1

If execvp succeeds, the entire child process address space is replaced by the program invoked by execvp(). This means that the exit(0) will only ever be invoked following your two special cases i.e. cd and sdir. As far as your code is concerned execvp() should never return, unless there is an error.

A further problem is that you free args immediately after allocating it and then go on to use it in your child processes. This is undefined behaviour.

The only problem I see with your wait code is that, if any of the children block waiting for user input, the parent will block waiting for the child to exit.

The cd code, has no effect on any process except the child in which it is executed. The parent's current directory is not affected. As you state in the comments, this can bet fixed by handling cd in the parent without forking.

edited Feb 25 '19 at 19:33

answered Feb 25 '19 at 10:54

JeremyP

84,577
15
123
161

The problem was the free() de-allocation before use. Also, I had to move the 'cd' case outside the loop because when exit(0) is called the updated working directory would be lost. Wow, such a simple error to cause such a massive headache! thank you :) – M. Rosenlof Feb 25 '19 at 19:02
Yes, I debated with myself about whether to mention the `cd` issue. I might add it to the answer for completeness. – JeremyP Feb 25 '19 at 19:30
@M.Rosenlof : re. simple error -> massive headache : that's [undefined behavior](https://stackoverflow.com/questions/2397984/undefined-unspecified-and-implementation-defined-behavior) for you ! – Sander De Dycker Feb 26 '19 at 08:51

Child process not exiting with fork()

1 Answers1