C question

[Chris F.A. Johnson]

On Thu, 20 Jul 2006, Madison Kelly wrote:

> Hi all,
>
>  As something of a follow-up to my last question, I need some C help (a 
> language I am quite 'n00by' in :p ).
>
>  What I want to do is, I think, simple. Take a set of command line options 
> and use them when calling a perl script. From the ANSI C book I have gotten 
> to the point where I can print the commands to STDOUT but I can't figure out 
> how to put them into a variable. I know 'char' is for one byte, but don't 
> know what type to use for a full string or how to concatenate the switches 
> into the string.
>
>  Here's what I've got so far (probably broken from playing with it):
>
> #include <stdio.h>
>
> #define REAL_PATH 
> "/home/digimer/projects/mizu-bu/releases/mizu-bu/cgi-bin/exec-priv.pl"
> main(int argc, char *argv[])

     main() always returns an int:

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

> {
>         setuid(geteuid());
>         setgid(getegid());
> 	 var say;
> 	 say="Hello";

char *say = "hello"; /* pointer to char; string cannot be modified */

char say[] = "hello"; /* array; string can be modified: */
say[2] = x; /* change 3rd char of the string */

>
> 	 int i;
> 	 for (i=1; i<argc; i++ )
> 	 {
> 	 	printf("%s%s", argv[i], (i < argc-1) ? " " : "");
> 	 }
> 	 printf("\n");
> 	 printf("%s", say);
> /*        execv(REAL_PATH, av);*/

     return 0;

> }

-- 
    Chris F.A. Johnson                      <http://cfaj.freeshell.org>
    ===================================================================
    Author:
    Shell Scripting Recipes: A Problem-Solution Approach (2005, Apress)
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