Sorry, I thought we were speaking java only, thanks for the response.

Rafael Tolosana wrote:

>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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