Re: [Dev-C++] address pointers

[Per W. <pw...@ia...>]

No, you are confusing the declaration of a pointer variable with the use
of one.

int *point = 0x241xff5f;

does not change what *point contains. It is just an assignment with wrong
type, i.e the hexadecimal constant is not of type 'pointer to int'.

int *point = (*int)0x241xff5f;

would work better - except for the fact that the hexadecimal constant is
broken - where does the second x come from?

The problem with this construct is that since you don't know what the
address 0x241... points to, you don't know if the program has the right to
read the address. If it hasn't, the program will be terminated!

/Per W

On Sat, 31 Jan 2004, Michael C Urban wrote:

> 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
>
>
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