lesson 3, boot sector with memory addressing

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Carlos Fenollosa 10 years ago
parent 81e9f2c974
commit bdca5e65da

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TBD
*Concepts you may want to Google beforehand: memory offsets, pointers*
Our new boot sector will refer to memory addresses and labels
The only goal of this lesson is to learn where the boot sector is stored
Please open page 14 [of this document](
http://www.cs.bham.ac.uk/~exr/lectures/opsys/10_11/lectures/os-dev.pdf)<sup>*</sup>
and look at the figure with the memory layout.
I could just go ahead and tell you that it starts at `0x7C00`, but it's
better with an example.
We want to print an X on screen. We will try 4 different strategies
and see which ones work and why.
First, we will define the X as data, with a label:
```nasm
the_secret:
db "X"
```
Then we will try to access `the_secret` in many different ways:
1. `mov al, the_secret`
2. `mov al, [the_secret]`
3. `mov al, the_secret + 0x7C00`
4. `mov al, 2d + 0x7C00`, where `2d` is the actual position of the X in the binary
Take a look at the code and read the comments.
Compile and run the code. You should see a string similar to `1[2¢3X4X`, where
the bytes following 1 and 2 are just random garbage.
If you add or remove instructions, remember to compute the new offset of the X
by counting the bytes, and replace `0x2d` with the new one.
~~~~~
This whole tutorial is heavily inspired on that document. Please read the
root-level README for more information on that.

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mov ah, 0x0e
; attempt 1
; Fails because it tries to print the memory address (i.e. pointer)
; not its actual contents
mov al, "1"
int 0x10
mov al, the_secret
int 0x10
; attempt 2
; It tries to print the memory address of 'the_secret' which is the correct approach.
; However, BIOS starts loading at address 0x7c00
; so we need to add that padding beforehand. We'll do that in attempt 3
mov al, "2"
int 0x10
mov al, [the_secret]
int 0x10
; attempt 3
; Add the BIOS starting offset 0x7c00 to the memory address of the X
; and then dereference the contents of that pointer
mov al, "3"
int 0x10
mov bx, the_secret
add bx, 0x7c00
mov al, [bx]
int 0x10
; attempt 4
; We try a shortcut since we know that the X is stored at byte 0x2d in our binary
mov al, "4"
int 0x10
mov al, [0x7c2d]
int 0x10
jmp $
the_secret:
; ASCII code 0x58 is stored just before the zero-padding
; on this code that is at byte 0x2d (check it out using xdd)
db "X"
times 510-($-$$) db 0
dw 0xaa55
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