【发布时间】:2018-10-31 17:50:35
【问题描述】:
我一直在阅读并遵循 Nick Blundell 编写的关于从头开始编写操作系统的教程,该教程可以在 https://www.cs.bham.ac.uk/~exr/lectures/opsys/10_11/lectures/os-dev.pdf 找到
我已经成功编写了一个可以调用 C 代码的引导加载程序,因此我开始用 C 编写我的内核。我现在正在尝试编写可以在屏幕上打印字符和字符串的函数。当我开始执行 C 代码时,我处于 32 位保护模式,所以我正在尝试从视频内存地址 0xb8000 正确计算内存偏移量。
当我尝试使用计算出的偏移量访问视频内存的特定区域时,我的问题就开始了。由于文本区域是 25 行 x 80 列,我使用公式 ((row * 80) + column) * 2 因为我必须有一个字符字节和一个属性字节。当我设置 row = 0 和 column = 0 时,我尝试打印的 X 不存在。设置row = 0,column = 1,左上角会出现一个X。
从 char* video_memory = 0xb8000 开始并重复发出 video_memory++ 允许我正确访问每个字节并在黑色背景上打印一个空间。
这是我的主要代码:
#include "../drivers/screen.h"
void main() {
//clear_screen();
//print_character('X', 0, 0, 0);
// Helper variables.
int row;
int column;
// We need to point at 0xB8000, where video memory resides.
unsigned char* video_memory = (unsigned char*)0xB8000;
for(row = 0; row < 25; row++) {
for(column = 0; column < 80; column++) {
// Clear the screen by printing a space on a black background.
*video_memory = ' ';
video_memory += 1;
*video_memory = 0x0F;
video_memory += 1;
}
}
// Test the offset calculation by printing at row 0, column 0 (the upper
// left corner of the screen).
row = 0;
column = 0;
// For an 80 by 25 grid. Multiply by 2 to account for the need of two bytes
// to display a character with given attributes.
int offset = ((row * 80) + column) * 2;
// Reset memory location after the loop.
video_memory = (unsigned char*)0xB8000;
// Add the offset to get the desired cell.
// THIS IS WHERE THE PROBLEM IS! Setting column = 1 prints in the first cell
// of video memory instead of the second.
video_memory += offset;
// Set character and its attributes.
*video_memory = 'X';
video_memory++;
*video_memory = 0x0F;
}
这是当 row = 0 和 column = 0 时显示的控制台: The console when row and column are 0. No X appears.
这是行 = 0 列 = 1 时的控制台: The console when row is 0 and column is 1. An X appears.
这是我上面的 kernel.c 文件的 objdump:
kernel.o: file format elf64-x86-64
Disassembly of section .text:
0000000000000000 <main>:
#include "../drivers/screen.h"
void main() {
0: 55 push rbp
1: 48 89 e5 mov rbp,rsp
// Helper variables.
int row;
int column;
// We need to point at 0xB8000, where video memory resides.
unsigned char* video_memory = (unsigned char*)0xB8000;
4: 48 c7 45 f8 00 80 0b mov QWORD PTR [rbp-0x8],0xb8000
b: 00
for(row = 0; row < 25; row++) {
c: c7 45 ec 00 00 00 00 mov DWORD PTR [rbp-0x14],0x0
13: eb 2f jmp 44 <main+0x44>
for(column = 0; column < 80; column++) {
15: c7 45 f0 00 00 00 00 mov DWORD PTR [rbp-0x10],0x0
1c: eb 1c jmp 3a <main+0x3a>
// Clear the screen by printing a space on a black background.
*video_memory = ' ';
1e: 48 8b 45 f8 mov rax,QWORD PTR [rbp-0x8]
22: c6 00 20 mov BYTE PTR [rax],0x20
video_memory += 1;
25: 48 83 45 f8 01 add QWORD PTR [rbp-0x8],0x1
*video_memory = 0x0F;
2a: 48 8b 45 f8 mov rax,QWORD PTR [rbp-0x8]
2e: c6 00 0f mov BYTE PTR [rax],0xf
video_memory += 1;
31: 48 83 45 f8 01 add QWORD PTR [rbp-0x8],0x1
int column;
// We need to point at 0xB8000, where video memory resides.
unsigned char* video_memory = (unsigned char*)0xB8000;
for(row = 0; row < 25; row++) {
for(column = 0; column < 80; column++) {
36: 83 45 f0 01 add DWORD PTR [rbp-0x10],0x1
3a: 83 7d f0 4f cmp DWORD PTR [rbp-0x10],0x4f
3e: 7e de jle 1e <main+0x1e>
int row;
int column;
// We need to point at 0xB8000, where video memory resides.
unsigned char* video_memory = (unsigned char*)0xB8000;
for(row = 0; row < 25; row++) {
40: 83 45 ec 01 add DWORD PTR [rbp-0x14],0x1
44: 83 7d ec 18 cmp DWORD PTR [rbp-0x14],0x18
48: 7e cb jle 15 <main+0x15>
}
}
// Test the offset calculation by printing at row 0, column 0 (the upper
// left corner of the screen).
row = 0;
4a: c7 45 ec 00 00 00 00 mov DWORD PTR [rbp-0x14],0x0
column = 0;
51: c7 45 f0 00 00 00 00 mov DWORD PTR [rbp-0x10],0x0
// For an 80 by 25 grid. Multiply by 2 to account for the need of two bytes
// to display a character with given attributes.
int offset = ((row * 80) + column) * 2;
58: 8b 55 ec mov edx,DWORD PTR [rbp-0x14]
5b: 89 d0 mov eax,edx
5d: c1 e0 02 shl eax,0x2
60: 01 d0 add eax,edx
62: c1 e0 04 shl eax,0x4
65: 89 c2 mov edx,eax
67: 8b 45 f0 mov eax,DWORD PTR [rbp-0x10]
6a: 01 d0 add eax,edx
6c: 01 c0 add eax,eax
6e: 89 45 f4 mov DWORD PTR [rbp-0xc],eax
// Reset memory location after the loop.
video_memory = (unsigned char*)0xB8000;
71: 48 c7 45 f8 00 80 0b mov QWORD PTR [rbp-0x8],0xb8000
78: 00
// Add the offset to get the desired cell.
// THIS IS WHERE THE PROBLEM IS! Setting column = 1 prints in the first cell
// of video memory instead of the second.
video_memory += offset;
79: 8b 45 f4 mov eax,DWORD PTR [rbp-0xc]
7c: 48 98 cdqe
7e: 48 01 45 f8 add QWORD PTR [rbp-0x8],rax
// Set character and its attributes.
*video_memory = 'X';
82: 48 8b 45 f8 mov rax,QWORD PTR [rbp-0x8]
86: c6 00 58 mov BYTE PTR [rax],0x58
video_memory++;
89: 48 83 45 f8 01 add QWORD PTR [rbp-0x8],0x1
*video_memory = 0x0F;
8e: 48 8b 45 f8 mov rax,QWORD PTR [rbp-0x8]
92: c6 00 0f mov BYTE PTR [rax],0xf
}
95: 90 nop
96: 5d pop rbp
97: c3 ret
我已经跟踪并亲手检查了我的偏移计算的实际组装说明,它们似乎是正确的。我怀疑当我尝试将我的偏移量(类型 int)添加到我的视频内存地址(类型 unsigned char*)时会出现问题,但我再次不确定。
另外,我尝试对偏移量的特定数字进行硬编码。使用 video_memory += 0 而不是 video_memory += offset 可以根据需要工作。
【问题讨论】:
-
与您的问题无关,但不要修改
video_memory及其指向的位置。请改用普通数组索引。如video_memory[row * 80 + column] = 'X'; -
你提到这一点很有趣。我的第一次尝试使用数组索引,但无论出于何种原因,结果都不是我所期望的,因此切换到直接修改。我确实打算在解决这个特殊问题后回到数组索引。
-
在这段代码的最后一步写一个'X'和0x0F后,调试器指示的字节值实际占用了0xB8000和0xB80001?
-
你试过写其他值吗?其他角色?其他属性?去其他地方?到 0xB8000、0xB8001、0xB8002、0xB8003?这个想法是建立一个模式。也许您的计算没有关闭,可能是视频内存关闭了一个,因为第一个字节是例如一个全局属性。
-
@selbie 在运行 Bochs 直到我的屏幕被“清除”后,Bochs 报告的值为 0x0f200f20。
标签: c operating-system off-by-one video-memory