char *vidmem=(char *)0xb8000

[slacker]

why do you need the "(char *)" in front of the address cause shouldnt you be able to leave out this part?

[Therx]

You can leave it out but GCC (with -Wall) will warn that you are making a pointer from integor. For a clean build you don't want this clogging the display

[slacker]

why would you be making a pointer from an integer?

[Nairou]

Because 0xb8000 is a number, an integer, while *vidmem is a pointer. Integers and pointers are not the same thing, so the compiler will give you a warning about it. In this case it would still work, but its best to eliminate all warnings, both for cleanliness and stability (you never know when a warning will be the cause of a bug you're trying to find). By putting the (char*) in front of the integer, you are telling the compiler that you really do know what you're doing, and you really do want to put the integer into the pointer.

[slacker]

but by writing char *vidmem your saying that vidmem will be a pointer to a char. you already told the compiler that your pointing at a char...why do you need to tell it again?

[Therx]

If you consider setting it to a lower int

char *ptr;
ptr = 100;

The compiler is warning that you may have meant.

char *ptr;
*ptr = 100;

these have completely different results. The (char *) simply tells the compiler that you definately want the first. I know that once my whole device manager would not work for the sake of one * amongst over 250 lines of code. Unfortunately it was passing an argument to a function so I didn't get a warning and I eventually had to post it here to find the bug.

[slacker]

k now im really confused.

char *pntr;
*pntr=0xb8000; <----means that the pointer will be referencing memory at that specific location.

now if you say:
pntr='a';

that means that the value at that location is 'a';

i thought this is how it works?

[Therx]

No

ptr = 0x80000
Sets to pointer to 0x80000

Then
*ptr = 'a'
would set the byte at 0x80000 to 'a'

[slacker]

im pretty sure its the other way around...

[slacker]

yea i found a site..your right

[slacker]

my compiler wont let me do this:

char *vidmem;
vidmem=0xb8000;

why?

[Tim]

Because you're trying to make a pointer from an integer .

vidmem is of type char*, and 0xb8000 is of type int. You can't assign an int to a char* -- they're just different types.

Now, it happens that, on the x86 CPU in 32-bit pmode, ints and char*s are the same size. So what you need to do is tell the compiler, "I know what I'm doing, leave me alone", and put in the (char*) cast.

[df]

So what you need to do is tell the compiler, "I know what I'm doing, leave me alone"

heheh i like that. I'm not casting! I'm bossing the compiler around!

[slacker]

but the pointer is the size of an int. the char says the pointer is pointing to a char at a certain address. doesnt the pointer know that the only type of memory address it can accept to point to is an int?

[Tim]

No. The only values you can assign a pointer are: another pointer of the same type, a void pointer (but not in C++), or the value 0 (NULL). Assigning an int to a pointer is a hack which will only work on machines with a flat address space, so you need the cast to let you do it.