Question

I’m currently trying to have my C program read Unix arguments from the user. I’ve so far searched this site but I haven’t been able to figure out exactly what I’m doing wrong – though admittedly my pointer implementation skills are rather limited.

The following is how I have the code now; I’ve been messing around with the pointers with no luck. The errors are also saying that I need to use const *char, but I’ve seen in other examples the *char can be input by the user.

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>

main()
{
    char args[128];
    //User input
    printf("> ");
    fgets(args, 128, stdin);
    execvp(args[0], *args[0]);
}

The error I get is as follows:

smallshellfile.c: In function ‘main’:
smallshellfile.c:13:21: error: invalid type argument of unary ‘*’ (have ‘int’)
smallshellfile.c:13:5: warning: passing argument 1 of ‘execvp’ makes pointer from integer without a cast [enabled by default]
/usr/include/unistd.h:575:12: note: expected ‘const char *’ but argument is of type ‘char’

Does anyone know what the problem may be?

Answer 1

You have several problems:

1. *args[0] is meaningless. args is array. args[0] is char. what is *args[0]?

2. You have to create a NULL-terminated array of char*, to pass as 2nd argument.

3. args[0] is the first char in args. you should pass the whole string (just args), not only its first char.

Try something like:

char *argv[]={args,NULL};
execvp(args,argv);