Having trouble understanding command line argument pointers

My professor cited this example in class. Its basically a version of the Unix more command, and I'm unsure about a couple things in it

int main( int ac , char *av[] )
{
  FILE  *fp;

  if ( ac == 1 )
    do_more( stdin );
  else
    while ( --ac )
     if ( (fp = fopen( *++av , "r" )) != NULL )
     {
        do_more(开发者_如何学运维 fp ) ; 
        fclose( fp );
     }
     else
        exit(1);
return 0;
}

I understand that *fp defines a file pointer, and that *av[] is the array of command line arguments. But what does *++av mean in terms of operation?

read *++av like this:

++av // increment the pointer
*av // get the value at the pointer, which will be a char*

in this example, it will open every files passed on the command line.

also:

av[0] // program name
av[1] // parameter 1
av[2] // parameter 2
av[3] // parameter 3
av[ac - 1] // last parameter

Here is a better version of the code that should do the very same thing. Hopefully it will be easier to understand. The names argc and argv are de-facto standard, you should use them to make the code more understandable to other programmers.

int main (int argc, char *argv[])
{
  FILE *fp;
  int i;

  if ( argc == 1 )
  {
    do_more( stdin );
  }
  else
  {
    for(i=1; i<argc; i++) /* skip the name of the executable, start at 1 */
    {
      fp = fopen (argv[i], "r");
      if(fp == NULL)
      {
        /* error message, return etc here */
      }

      do_more( fp ) ;
      fclose( fp );
    }
  }

  return 0;
}