Nigel
b4cff3e667
- 1 block = 4096 bytes : because this will make page fault handling possibly somewhat easier - 1 byte in the bitmap = 8 blocks of physical memory unsure if the allocation is perfect ... guess i'll find out some day if this is actually correct. The bitmap needs 16kb to keep track of 2gb of physical memory. Seems a decent percentage to me.
31 lines
827 B
C++
31 lines
827 B
C++
#pragma once
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#include <stdint.h>
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#include "./memory.h"
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#include "./../tty/kterm.h"
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#define KB 1024
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typedef uintptr_t address_t;
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static const int MAX_PAGES = 1024 * KB; // 4GB , 4kB/page
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static volatile address_t pmem_stack[MAX_PAGES];
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static volatile address_t pmem_stack_top = MAX_PAGES; // top down allocation
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extern "C" void loadPageDirectory (uint32_t* addr );
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extern "C" void enablePaging();
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struct page_directory_entry {};
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struct page_table_entry{};
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class PageDirectory {
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public:
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void enable ();
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void MapPhysicalToVirtualAddress ( address_t PAddress , address_t VAddress, uint32_t size );
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private:
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uint32_t page_directory[1024] __attribute__((aligned(4096))); // align on 4 kiloByte pages
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uint32_t first_page_table[1024] __attribute__((aligned(4096))); // align on 4 kiloByte pages
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}; |