Re: [Dev-C++] address pointers

[Michael C U. <mu...@cb...>]

Brian,

> int *point = 0x241xff5f;
> cout << *point;

This doesn't work because it changes the value that *point contains.
Remembet that * is the dereference operator. It access the value rather
than the address.

To access the pointer itself, you don't want anything before it. So here
is an example to do what you want:

int *point;                /* declare a pointer */
point = 0x241xff5f;        /* change the address that it points to */
cout < *point;             /* derefernce the poiner and print its value */

Notice that i don't use any operator when assigning the value to point.
That's because I want to change the actual address that point is point at,
rather than change the value that the current address of point contains.

Of course, the above example is only useful if you know that the address
contains an int value. Also, if you try to access a memory address that
doesn't belong to your program, Windows might catch you and kill your
program. So make sure you know that the address you assign to point is
actually "owned" by your program.

Hope this helps,

Mike