90 lines
No EOL
4.5 KiB
NASM
90 lines
No EOL
4.5 KiB
NASM
; Declare constants for the multiboot header.
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MBALIGN equ 1 << 0 ; align loaded modules on page boundaries
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MEMINFO equ 1 << 1 ; provide memory map
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FLAGS equ MBALIGN | MEMINFO ; this is the Multiboot 'flag' field
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MAGIC equ 0x1BADB002 ; 'magic number' lets bootloader find the header
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CHECKSUM equ -(MAGIC + FLAGS) ; checksum of above, to prove we are multiboot
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; Declare a multiboot header that marks the program as a kernel. These are magic
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; values that are documented in the multiboot standard. The bootloader will
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; search for this signature in the first 8 KiB of the kernel file, aligned at a
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; 32-bit boundary. The signature is in its own section so the header can be
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; forced to be within the first 8 KiB of the kernel file.
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section .multiboot
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align 4
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dd MAGIC
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dd FLAGS
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dd CHECKSUM
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; The multiboot standard does not define the value of the stack pointer register
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; (esp) and it is up to the kernel to provide a stack. This allocates room for a
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; small stack by creating a symbol at the bottom of it, then allocating 16384
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; bytes for it, and finally creating a symbol at the top. The stack grows
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; downwards on x86. The stack is in its own section so it can be marked nobits,
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; which means the kernel file is smaller because it does not contain an
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; uninitialized stack. The stack on x86 must be 16-byte aligned according to the
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; System V ABI standard and de-facto extensions. The compiler will assume the
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; stack is properly aligned and failure to align the stack will result in
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; undefined behavior.
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section .bss
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align 16
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stack_bottom:
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resb 16384 ; 16 KiB
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stack_top:
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; The linker script specifies _start as the entry point to the kernel and the
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; bootloader will jump to this position once the kernel has been loaded. It
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; doesn't make sense to return from this function as the bootloader is gone.
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; Declare _start as a function symbol with the given symbol size.
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section .text
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global _start:function (_start.end - _start)
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_start:
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; The bootloader has loaded us into 32-bit protected mode on a x86
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; machine. Interrupts are disabled. Paging is disabled. The processor
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; state is as defined in the multiboot standard. The kernel has full
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; control of the CPU. The kernel can only make use of hardware features
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; and any code it provides as part of itself. There's no printf
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; function, unless the kernel provides its own <stdio.h> header and a
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; printf implementation. There are no security restrictions, no
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; safeguards, no debugging mechanisms, only what the kernel provides
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; itself. It has absolute and complete power over the
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; machine.
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; To set up a stack, we set the esp register to point to the top of our
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; stack (as it grows downwards on x86 systems). This is necessarily done
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; in assembly as languages such as C cannot function without a stack.
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mov esp, stack_top
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; This is a good place to initialize crucial processor state before the
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; high-level kernel is entered. It's best to minimize the early
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; environment where crucial features are offline. Note that the
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; processor is not fully initialized yet: Features such as floating
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; point instructions and instruction set extensions are not initialized
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; yet. The GDT should be loaded here. Paging should be enabled here.
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; C++ features such as global constructors and exceptions will require
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; runtime support to work as well.
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; Enter the high-level kernel. The ABI requires the stack is 16-byte
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; aligned at the time of the call instruction (which afterwards pushes
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; the return pointer of size 4 bytes). The stack was originally 16-byte
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; aligned above and we've since pushed a multiple of 16 bytes to the
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; stack since (pushed 0 bytes so far) and the alignment is thus
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; preserved and the call is well defined.
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; note, that if you are building on Windows, C functions may have "_" prefix in assembly: _kernel_main
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extern kernel_main
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call kernel_main
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; If the system has nothing more to do, put the computer into an
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; infinite loop. To do that:
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; 1) Disable interrupts with cli (clear interrupt enable in eflags).
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; They are already disabled by the bootloader, so this is not needed.
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; Mind that you might later enable interrupts and return from
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; kernel_main (which is sort of nonsensical to do).
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; 2) Wait for the next interrupt to arrive with hlt (halt instruction).
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; Since they are disabled, this will lock up the computer.
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; 3) Jump to the hlt instruction if it ever wakes up due to a
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; non-maskable interrupt occurring or due to system management mode.
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cli
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.hang: hlt
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jmp .hang
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.end: |