I am sorry but it is two years or more since I do not program in C. I probably do not remember the syntax properly, anyway I think that the idea was correct. In addition, when programming in C we have to bear in mind the compiler we are using (there are some subtle differences).

Brian, try the next piece of code in C which represents the same idea:

#include <stdio.h>

int main(){

  int foo[40];
  int *foo2;

  foo2 = (int *)&foo;

  foo[0] = 1;
  foo[1] = 2;
  printf("%d",*foo2);
  foo2 = foo2 + 1;

  printf("%d", *foo2);
  exit(0);
}

Regards

Rafa

-----Original Message-----
From: Brian P Bohnet <bbohnet@(protected)>
To: JSP-INTEREST@(protected)
Date: Wed, 10 Mar 2004 10:30:53 -0600
Subject: Re: thanks rafa

Your code is incorrect, you cannot manipulate an array that way, see
error produced:

: operator + cannot be applied to char[],int
     foo = foo + 1;
            ^

Rafael Tolosana wrote:

> I will try it with two simple examples:
>
> First have a look at this Java code:
> String foo = new String("Example String");
> String foo2 = foo;
>
> Now have a look at this C code:
> char[] foo = "Example String";
> char[] foo2 = foo;
> foo = foo + 1;
>
> In the first example we have two references to the same object String,
> but you cannot manipulate the memory address of the references. On the
> other hand, in the second example we have two pointers to the same
> memory zone and the foo variable has been incremented by one, so that
> it will point to "xample String".
>
> I hope this will help you understand
>
> Rafa

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