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

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

why would you be making a pointer from an integer?

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.

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?

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.

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?

No

ptr = 0x80000
Sets to pointer to 0x80000

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

im pretty sure its the other way around...

yea i found a site..your right

my compiler wont let me do this:

char *vidmem;
vidmem=0xb8000;

why?

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.

Tim Robinson wrote: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!

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?

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.