Lab1
歷時2天，完成了LAB1，完整代碼倉庫可點擊：https://github.com/Elio-yang/MIT6.828
partA 練習
- exercise3
gdb指令：
x/Ni addr :反匯編addr處的N條指令
x/Nx addr:列印N位元組addr處的記憶體
b *addr:在addr處設定斷點
readsect(): 0x7c7c
bootmain():0x7d25
回圈結束的第一條指令是0x7d81處的call *0x10018,利用gdb，0x10018記憶體處的值為0x10000c，故第一條指令是call 0x10000c，這個地址就是kernel的entry，
At what point does the processor start executing 32-bit code? What exactly causes the switch from 16- to 32-bit mode?
從ljmp $PROT_MODE_CSEG,$protcseg這條指令后開始執行32位代碼，真正造成切換的，是CR0的PE位被置為，進入了保護模式，
What is the last instruction of the boot loader executed, and what is the first instruction of the kernel it just loaded?
last：
call *0x10018
first：
f010000c <entry>:
f010000c: 66 c7 05 72 04 00 00 movw $0x1234,0x472
Where is the first instruction of the kernel?
很顯然在0x10000c，
How does the boot loader decide how many sectors it must read in order to fetch the entire kernel from disk? Where does it find this information?
都是通過ELF header得知的，
Loading the kernel
第一個需要注意的是代碼的鏈接地址和裝載地址
使用命令
objdump -h <file.o>
# -x Display all available header information
# -f Display entry point
# 更多用法 man objdump 即可
在kernel中這兩者是不同的，但是在之前的boot中
二者是一致的，在kern/entry.S中有這樣一段代碼
# Turn on paging.
movl %cr0, %eax
orl $(CR0_PE|CR0_PG|CR0_WP), %eax
movl %eax, %cr0
這便開啟了地址映射，在此之前kernel的VMA和LMA地址處的記憶體一般是不同的，但是開啟分頁之后，LMA映射到了VMA，
The Kernel
第一個值得注意的是：開啟分頁模式，將虛擬地址[0, 4MB)映射到物理地址[0, 4MB)，[0xF0000000, 0xF0000000+4MB)映射到[0, 4MB）（/kern/entry.S）
分頁模式下的尋址，在Intel手冊中也有給出
開啟這個模式的代碼如下
# Load the physical address of entry_pgdir into cr3. entry_pgdir
# is defined in entrypgdir.c.
movl $(RELOC(entry_pgdir)), %eax
movl %eax, %cr3
# Turn on paging.
movl %cr0, %eax
orl $(CR0_PE|CR0_PG|CR0_WP), %eax
movl %eax, %cr0
關于地址的映射在kern/entrypgdir.c有代碼實作
__attribute__((__aligned__(PGSIZE)))
pde_t entry_pgdir[NPDENTRIES] = {
// Map VA's [0, 4MB) to PA's [0, 4MB)
[0]
= ((uintptr_t)entry_pgtable - KERNBASE) + PTE_P,
// Map VA's [KERNBASE, KERNBASE+4MB) to PA's [0, 4MB)
[KERNBASE>>PDXSHIFT]
= ((uintptr_t)entry_pgtable - KERNBASE) + PTE_P + PTE_W
};
編譯器分配的空間是強制性4kB頁對齊的，pgdir是一個1024項的陣列，這里可以不用詳細了解原理For now, you don't have to understand the details of how this works, just the effect that it accomplishes.
- exercise7
在開啟分頁之前，這兩個地址內容是不一致的，后面經過了地址映射，這兩者的內容一致，注釋掉
movl %eax, %cr0程式會崩潰，
Formated Printing to the Console
首先是幾個函式的呼叫關系
然后練習題
- exercise8
這個檔案就是lib/printfmt.c
// (unsigned) octal
case 'o':
// Replace this with your code.
num=getuint(&ap,lflag);
base=8;
goto number;
對照背景關系很容易補全，
下面是回答問題：
Explain the interface between
printf.candconsole.c. Specifically, what function doesconsole.cexport? How is this function used byprintf.c?
對照上文呼叫關系圖即可
Explain the following from console.c
if (crt_pos >= CRT_SIZE) { int i; memmove(crt_buf, crt_buf + CRT_COLS, (CRT_SIZE - CRT_COLS) *sizeof(uint16_t)); for (i = CRT_SIZE - CRT_COLS; i < CRT_SIZE; i++) crt_buf[i] = 0x0700 | ' '; crt_pos -= CRT_COLS; }
首先文本模式最多能顯示25*80個字符，即25行每行80個，此處
// console.h
#define CRT_ROWS 25
#define CRT_COLS 80
#define CRT_SIZE (CRT_ROWS * CRT_COLS)
因此，這一段處理的是超出一螢屏以后的做法：即舍棄最上面一行，整體上移一行，
For the following questions you might wish to consult the notes for Lecture 2. These notes cover GCC's calling convention on the x86.
Trace the execution of the following code step-by-step:
int x = 1, y = 3, z = 4; cprintf("x %d, y %x, z %d\n", x, y, z);
- In the call to
cprintf(), to what doesfmtpoint? To what doesappoint?- List (in order of execution) each call to
cons_putc,va_arg, andvcprintf. Forcons_putc, list its argument as well. Forva_arg, list whatappoints to before and after the call. Forvcprintflist the values of its two arguments.
GCC 函式呼叫約定是引數從右往左入堆疊，此處fmt指向的就是第一個引數的位置，而ap指向第一個可變引數，也就是第二個引數x的位置，關于變引數，JOS使用的是GCC builtin來實作的，其實作可以用如下代碼進行大致說明（不是嚴謹的完整實作）：
#define va_start(list,param_1st) ( list = (va_list)¶m1+ sizeof(param_1st) )
#define va_arg(list,type) ( (type *) ( list += sizeof(type) ) )[-1]
#define va_end(list) ( list = (va_list)0 )
因此：
va_list：即char*
va_start:獲取第一個可變引數的地址
va_arg：回傳指向下一個引數的指標
va_end：清空引數串列
Run the following code.
unsigned int i = 0x00646c72; cprintf("H%x Wo%s", 57616, &i);What is the output? Explain how this output is arrived at in the step-by-step manner of the previous exercise.
Here's an ASCII tablethat maps bytes to characters.
The output depends on that fact that the x86 is little-endian. If the x86 were instead big-endian what would you set
ito in order to yield the same output? Would you need to change57616to a different value?Here's a description of little- and big-endian and a more whimsical description.
把這段代碼加入init.c中，運行make qemu,結果如下
0xe110=57616這很好解釋，查閱ASCII表，得知
00(\0) 64(d) 6c(l) 72(r)
顯然這是由于小端模式而使用的一個數，為了證明這一點，可以輸出&i記憶體處的位元組，將下面這段代碼放在上面列印代碼的后面
cprintf("addr of i: %p\n",&i);
char *p=(char*)&i;
for(int i=0;i<4;i++){
cprintf("[%x]",*p);
p++;
}
輸出結果如下：
In the following code, what is going to be printed after
'y='? (note: the answer is not a specific value.) Why does this happen?cprintf("x=%d y=%d", 3);
運行結果如下
顯然y的值并不一定固定，他就是把記憶體中那個位置的數拿來充當了第二個引數，
Let's say that GCC changed its calling convention so that it pushed arguments on the stack in declaration order, so that the last argument is pushed last. How would you have to change
cprintfor its interface so that it would still be possible to pass it a variable number of arguments?
更改了入堆疊方式，相應地更改va_start和va_start即可，
The Stack
先看這個練習
- exercise9
在entry.S中可以找到如下代碼
# where the stack is set.
# Clear the frame pointer register (EBP)
# so that once we get into debugging C code,
# stack backtraces will be terminated properly.
movl $0x0,%ebp # nuke frame pointer
# Set the stack pointer
movl $(bootstacktop),%esp
# now to C code
call i386_init
利用gdb得知，movl $(bootstacktop),%esp會被編譯為movl $0xf0110000,%esp，因此堆疊何時初始化，堆疊放在哪兒都清楚了，繼續看代碼
###################################################################
# boot stack
###################################################################
.p2align PGSHIFT # force page alignment
.globl bootstack
bootstack:
.space KSTKSIZE
.globl bootstacktop
這便開辟了堆疊的大小，即32KB，堆疊由高地址向低地址增長，
下面，關于函式的呼叫程序，做一個總結，可以參考[CSAPP，p164]，
這是從課件ppt截取的兩頁
關于函式的呼叫，一般有如下的動作發生：
- 函式呼叫者(caller)將引數入堆疊，按照從右到左的順序入堆疊
- call指令會自動將當前%eip(指向call的后面一條指令)入堆疊，ret指令將自動從堆疊中彈出該值到eip暫存器
- 被呼叫函式(callee)負責：將%ebp入堆疊，%esp的值賦給%ebp，
因此函式開頭都會是類似的兩條指令
push %ebp
mov %esp,%ebp
因此整個呼叫鏈差不多可以描述成如下形式
來到下一個練習
- exercise10
每次call之后會干什么，上文已經分析了，至于每次遞回入堆疊的字，偽代碼可以表示為
push %eip
push %ebp
push %esi
push %ebx
共計0x10B，
- exercise11
需要我們更改mom_backtrace()函式，達到的效果如下：
題目中已經說明，獲得%ebp的函式就是read_ebp()，那么編碼作業應該很好完成了(利用呼叫鏈中%ebp的鏈)
int
mon_backtrace(int argc, char **argv, struct Trapframe *tf)
{
// Your code here.
uint32_t *ebp=(uint32_t*)read_ebp();
while(ebp!=NULL){
cprintf("ebp %8x eip %8x args %08x %08x %08x %08x %08x\n",
ebp,ebp[1],ebp[2],ebp[3],ebp[4],ebp[5],ebp[6]);
ebp=(uint32_t *)(*ebp);
}
return 0;
}
運行結果如下
- exercise12
練習12的任務有三個：
- 搞清楚
__STAB_* - 添加命令
backtrace - 完善
mon_backtrace
-
任務一
根據提示，查看這幾個檔案，首先是
kernel.ld，.stab : { PROVIDE(__STAB_BEGIN__ = .); *(.stab); PROVIDE(__STAB_END__ = .); BYTE(0) /* Force the linker to allocate space for this section */ } .stabstr : { PROVIDE(__STABSTR_BEGIN__ = .); *(.stabstr); PROVIDE(__STABSTR_END__ = .); BYTE(0) /* Force the linker to allocate space for this section */ }可以知道
.stab和.stabstr應該是兩個段，接著
objdump -h obj/kern/kernel
然后是``objdump -G obj/kern/kernel``
執行后面的操作以后，大致可以知道這是一個段，包含了除錯資訊(符號表)，細節可以不用太了解，接著找到``stab.h``，其中
這兩項便是后文編碼尋找行號時需要的，下面開始任務二和三
-
任務二
題目中提示了需要使用
debuginfo_eip,查找這個函式發現，他會將需要的資訊存到型別為struct Eipdebuginfo的結構體中，查看該結構體定義(kern/kdebebug.h)// Debug information about a particular instruction pointer struct Eipdebuginfo { const char *eip_file; // Source code filename for EIP int eip_line; // Source code linenumber for EIP const char *eip_fn_name; // Name of function containing EIP // - Note: not null terminated! int eip_fn_namelen; // Length of function name uintptr_t eip_fn_addr; // Address of start of function int eip_fn_narg; // Number of function arguments };因此只需要使用
debuginfo_eip填充該結構體，再輸出資訊即可，static struct Command commands[] = { { "help", "Display this list of commands", mon_help }, { "kerninfo", "Display information about the kernel", mon_kerninfo }, { "backtrace", "Show stack backtrace",mon_stacktrace} }; //...... int for_stack(int argc,char **argv,struct Trapframe *tf) { uint32_t *ebp=(uint32_t*)read_ebp(); while(ebp!=NULL){ struct Eipdebuginfo info; uint32_t eip = ebp[1]; debuginfo_eip((int)eip, &info); cprintf(" ebp %8x eip %8x args %08x %08x %08x %08x %08x\n", ebp,ebp[1],ebp[2],ebp[3],ebp[4],ebp[5],ebp[6]); const char* filename=(&info)->eip_file; int line = (&info)->eip_line; const char * not_null_ter_fname=(&info)->eip_fn_name; int offset = (int)(eip)-(int)((&info)->eip_fn_addr); cprintf(" %s:%d: %.*s+%d\n",filename,line,info.eip_fn_namelen,not_null_ter_fname,offset); ebp=(uint32_t *)(*ebp); } return 0; } int mon_stacktrace(int argc,char **argv,struct Trapframe *tf) { cprintf("Stack backtrace:\n"); return for_stack(argc,argv,tf); }其中關于檔案行號的查找實作，對照背景關系就能實作，注意
N_SLINE這就是之前說stab時提到的一個有用的屬性，// Search within [lline, rline] for the line number stab. // If found, set info->eip_line to the right line number. // If not found, return -1. // // Hint: // There's a particular stabs type used for line numbers. // Look at the STABS documentation and <inc/stab.h> to find // which one. // Your code here. stab_binsearch(stabs, &lline, &rline, N_SLINE, addr); if(lline<=rline){ info->eip_line=stabs[lline].n_desc; }else{ return -1; }運行結果如下：
之后運行評分程式
至此，Lab1完結，完整代碼倉庫可點擊：https://github.com/Elio-yang/MIT6.828
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