not really ...
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struct pgEntry {
unsigned present : 1;
unsigned flags:11;
unsigned frame:20;
} __attribute__((packed));
struct pgEntry *pgDirectory=(struct pgEntry*) 0xFFFFF000;
struct pgEntry *pgTables=(struct pgEntry*) 0xFFC00000;
All the content of the virtual memory starting from tables[] are defined by the content of directory[]. It uses the last 4MB of your virtual address space. if pgDirectory
.present==0, then pgTables[i*1024 .. ((i+1)*1024)-1] are not valid addresses.
if you want to access page directory entry related to address X, you'll use PDE=pgDirectory[X>>22]: the last 10 bits are the entry index in the directory (see Intel manuals) and C will gracefully replace this by
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mov ebx,[pgDirectory]
mov esi, <X>
shr esi,22
mov eax,[ebx+esi*4]
mov <PTE>,eax
Say X=0x12345678, the PDE is at FFFFF000|((0x12345678>>22)*4) == FFFFF480, which is translated by the MMU as:
PDE[3FF](uses the page directory as page table)->PTE[3FF](use the page directory as the page) ->offset[480..483]
If you want to access the page table entry for address X, you should first read the page directory entry and check its present bit is 1, at least if you're in a page fault handler. If not, you might try to read an invalid page table, which would result in weird error conditions (including Exception loop)
As we are sure page tables are contiguous in the pgTables[...] structure, we don't have to do the PDE lookup to know where the page table actually is: it is immediate
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PTE=pgTables[X>>12];
/* once again, there is a "*4" implied in this C construct
* because sizeof(pgEntry) == 4
*/
if X=0x12345678, you have PTE = 0xFFC00000 + 12345*4 == 0xFFC0000 | 0x00048D14 = 0xFFC48d14, which will be interpreted by the MMU as
PDE[3FF](uses the page directory as page table)--> PTE[048](uses page table for 0x120xxxxx as page) --> offset [d14 .. d17] (read the entry for 0x12345xxx)
Hope i made it clearer.
seems there's something wrong thus in the previous picture: 0xffd20000 should be replaced by 0xffc48000.
).