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 <[EMAIL PROTECTED]> To: [EMAIL 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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