Question

When I do something like:

./foo -uxw --bar something

Does the shell automatically parse these commands, or does each program have to do the parsing itself?

Answer 1

No, the shell doesn’t parse it for you. Each program has to parse it on its own. The following code should make it clear what is going on.

#include <stdio.h>

int main(int argc, char **argv)
{
    int i;
    printf("argc: %d\n", argc);
    for (i = 0; i < argc; i++) {
        printf("argv[%d] = %s\n", i, argv[i]);
    }
    return 0;
}

Let us compile this program.

susam@nifty:~$ gcc args.c -o args

Now let us run this and see the output:

.susam@nifty:~$ ./args
argc: 1
argv[0] = ./args
susam@nifty:~$ ./args foo bar
argc: 3
argv[0] = ./args
argv[1] = foo
argv[2] = bar
susam@nifty:~$ ./args -a foo --b bar
argc: 5
argv[0] = ./args
argv[1] = -a
argv[2] = foo
argv[3] = --b
argv[4] = bar

The only thing the shell does is pass each argument you specify in the command line to your program. While it would pass foo bar as two separate arguments to your program, it would pass "foo bar" or 'foo bar‘ as a single argument to your program. Yes, so the shell does some sort of parsing of the arguments before passing it to your program. It considers quoted strings as a single argument. Here is a demonstration:

susam@nifty:~$ ./args -a foo bar
argc: 4
argv[0] = ./args
argv[1] = -a
argv[2] = foo
argv[3] = bar
susam@nifty:~$ ./args -a "foo bar"
argc: 3
argv[0] = ./args
argv[1] = -a
argv[2] = foo bar
susam@nifty:~$ ./args -a 'foo bar'
argc: 3
argv[0] = ./args
argv[1] = -a
argv[2] = foo bar
susam@nifty:~$ ./args -a "foo bar" 'car tar war'
argc: 4
argv[0] = ./args
argv[1] = -a
argv[2] = foo bar
argv[3] = car tar war