我一直在阅读并遵循Nick Blundell从零开始编写操作系统的教程，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，因此我开始在C中编写我的内核。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，所以我试图正确计算视频内存地址0xb8000的内存偏移量。
当我尝试使用计算的偏移量访问视频内存的特定区域时，我的问题就出现了。由于文本区域是25行乘80列，我使用公式（（行* 80）+列）* 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，我正在尝试打印的X不存在。设置row = 0和column = 1，X出现在左上角。
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空格。
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;
}
这是在row = 0和column = 0时显示的控制台： The console when row and column are 0. No X appears.
当row = 0且column = 1时，这是控制台： 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
我已经跟踪并亲自检查了我的偏移计算的实际装配说明，它们看起来是正确的。我怀疑当我尝试将我的偏移量（类型为int）添加到我的视频内存地址（类型unsigned char *）时会出现问题，但我不能完全确定。
另外，我尝试对特定数字进行硬编码以进行偏移。使用video_memory + = 0而不是video_memory + = offset可以根据需要使用。
答案 0 :(得分：3)
经过更多搜索，我在ARM Information Center上发现了一篇文章，描述了使用C指针访问内存映射I / O设备的特定地址。
将我的video_memory指针变量声明为＆＃39; volatile＆＃39;确保编译器始终执行内存访问，而不是优化它们...＆＃34;。显然，根据this answer on Quora，编译器可以生成指令，在将数据刷新到内存之前覆盖写缓冲区中的数据，这就是我的问题发生的地方。
因此将我的变量声明为volatile unsigned char* video_memory = 0xB8000;会产生预期的结果。