What does this statement mean?

Legend986 · #1 (**permalink**) 11-17-2007

If I declare a structure as:

Code:

typedef struct
	{
		int id;
		void* ptr;
		int size;
	} kpage_t;

And then use this to do some initialization...

Code:

*((kpage_t**)page->ptr) = page;

What exactly is being done? I have some trouble with pointers here...

And if there's a macro BASEADDRESS(x) defined as

Code:

((void*)(((int) (x)) & ~(PAGESIZE-1)))

What exactly is happening here assuming that PAGESIZE is a power of two number...

uvrakesh · #2 (**permalink**) 11-17-2007

*((kpage_t**)page->ptr) = page;

This means that page is holding an address of the structure kpage_t(ie kpage_t *)

ptr is of type void * and thus can be made to point to any address, which can be address of any type no matter its an address of another address storing variable

thus page->ptr is an address which points to another address and that address is an address where a variable of type kpage_t is residing

Then regarding the macro question
((void*)(((int) (x)) & ~(PAGESIZE-1)))

This would yeild 0 if the PAGESIZE is a in 2^n

Regards,
RUV

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