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both the kernel and the user space share the same linear address (4 GB) ?

or every task can have each one - or may be each segment in each individual task has 4 GB ????

another question : do i have to own a malloc and free function before start paging ??

when i start paging for the first time should i page all the physical memory so it can be used by the kernel or only the heap for the kernel ???

after enabling task switching every task will have an entry in the page directory or it will have a new page directory and page table ??

in window or linux the linear address of each task is 4 GB ???

mohammed wrote:thanks..by the way your tutorials are great : D
after enabling paging all the addresses that i used in the kernel will be virtual addresses refer to other locations ??
paging doesn't affect the addresses less than 0x100000??
in the boot sector there is a code data and stack segments after loading the GDT is kernel use the segments that i referred to or it still working with the segments that was in the boot sector ?

Normally each task (process, not thread) has it's own page directory, the kernel being mapped identically in all.

ManOfSteel wrote:@JamesM:

Normally each task (process, not thread) has it's own page directory, the kernel being mapped identically in all.

So whenever a user executes a process, the OS paging system allocates all the page tables and pages needed by the process in the newly created page directory, and it also allocates all the page tables and pages needed by the kernel in this same new page directory. So in all processes address space, some addresses will always point to the physical address where the kernel image was first copied at startup. Is that it?

You can have each process' page directory point to the same set of page tables for the kernel. That way, you don't need to keep allocating more pages of memory for kernel page tables just so you can keep the kernel mapped in the same location.

unsigned long *page_directory = (unsigned long *) 0x9C000;
unsigned long *page_table = (unsigned long *) 0x9D000; // the page table comes right after the page directory

unsigned long address=0; // holds the physical address of where a page is
unsigned int i;

// map the first 4MB of memory
for(i=0; i<1024; i++)
{
	page_table[i] = address | 3; // attribute set to: supervisor level, read/write, present(011 in binary)
	address = address + 4096; // 4096 = 4kb
};

it is set up so that what appears to the programmer to be address 0x2250A000 really is 0x2250A000. But if we wanted to, we could make what appears to the programmer to be address 0x2250A000 really be 0x100000.

mohammed wrote: here the page directory points to a physical address - according to intel manual the page directory can point to the physical address if you used 4 MB pages if you used 4 KB the page directory points to the page table why in this code the page directory points to the address directly ?

how he did that i didn't see in the code any thing refer to any linear address he is just mapping "the first 4 MB" ? how he did that although the page directory is pointing to the address 0x9C000??