How to make script in linux use my interpteter and work? (#!)

Question

I've made a simple shell for linux. It's reading line by line with getline() until ctrl+d (eof/-1) is entered into standard input.

While entering into stdin line by line code like that:

ls -al &
ls -a -l

My shell works pretty well.

I've tried to run script through my shell, but it's not working. When I execute script, my shell is automatically executed (1st line) but the shell do not interprete other lines.

#!/home/arbuz/Patryk/projekt/a.out
ls -al &
ls -a -l

What could cause it? I have to say that I'm very beginner in linuxes and teacher didnt say anything about all that stuff. Just a homework. I've done some researches but that's all I've found.

Here's code of my Shell. I've added shell path into etc/shells but its still not working

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

int main()
{

    ssize_t bufer_size = 0;
    char* line = NULL;
    int line_size;

    while ((line_size = getline(&line, &bufer_size, stdin)) != -1) // while end of file
    {
        char** words_array;
        words_array = (char**)malloc(200 * sizeof(char*));

        int words_count = 0;
        int i;
        int j = 0;
        int words_length = 0;
        char word[100];
        for (i = 0; i < line_size; i++)
        {
            if (line[i] == ' ' || line[i] == '\n')
            {
                words_array[words_count] = (char*)malloc(words_length * sizeof(char));
                int b;
                for (b = 0; b < words_length; b++)
                {
                    words_array[words_count][b] = word[b];
                }
                j = 0;
                words_count++;
                words_length = 0;
            }
            else
            {
                word[j] = line[i];
                j++;
                words_length++;
            }
        }

        bool run_in_background = false;

        if (words_array[words_count - 1][0] == '&')
        {
            run_in_background = true;
            words_array[words_count - 1] = NULL;
        }

        int a = fork();

        if (a == 0) // child process
        {
            execvp(words_array[0], words_array);
        }
        else       // parent process
        {
            if (run_in_background == true)
            {
                printf("\n ---- running in background. \n");
            }
            else
            {
                printf("\n ---- running normal \n");
                wait(NULL);
            }
        }
    }

    return 0;
}

Answer 1

Your shell must accept command line arguments. In this case, your program will be called like this:

/home/arbuz/Patryk/projekt/a.out your_script

So you'll need a main() of this signature:

int main(int argc, char* argv[])

and then parse the arguments. argc contains the amount of arguments. The script's filename is passed in argv[1]. You'll need to open it (using fopen()) and read commands from it instead of stdin. You should probably make sure that your shell ignores the first line of a file if it starts with a #.

If your script is called without an absolute path (a path that doesn't start with a /), then the filename is relative to the current directory. You can get that from the environment or programmatically with getcwd().

Answer 2

The problem is that your shell reads from standard input, while a she-bang #! causes the script to passed as a command-line argument. So your shell is called as

/home/arbuz/Patryk/projekt/a.out <script>

... ignores the command line argument and waits for commands on standard input. You have to read the script from argv[1].