Homework: intro to xv6

This lecture continues understanding xv6, our re-implementation of Unix v6. Read the source code in the assigned files; The assigned chapter for today provides a commentary on the assigned files.

You won't have to understand the details yet; we will focus on how the first user-level process comes into existence after the computer is turned on.

Hand-In Procedure

You are to turn in this homework before lecture. Please email your answers to 6.828-homework@pdos.csail.mit.edu, preferably in plain text.

Assignment

Fetch and un-tar the xv6 source:

sh-3.00$ wget http://pdos.csail.mit.edu/6.828/2010/src/xv6-rev4.tar.gz
sh-3.00$ tar xzvf xv6-rev4.tar.gz
xv6/
xv6/asm.h
xv6/bio.c
xv6/bootasm.S
xv6/bootmain.c
...
$

Build xv6:

$ add -f 6.828
$ cd xv6
$ make
gcc -O -nostdinc -I. -c bootmain.c
gcc -nostdinc -I. -c bootasm.S
ld -N -e start -Ttext 0x7C00 -o bootblock.o bootasm.o bootmain.o
objdump -S bootblock.o > bootblock.asm
objcopy -S -O binary bootblock.o bootblock
...
$

Find the address of the main function by looking in kernel.asm:

% grep main kernel.asm
...
00102454 <mpmain>:
mpmain(void)
001024d0 <main>:
  10250d:       79 f1                   jns    102500 <main+0x30>
  1025f3:       76 6f                   jbe    102664 <main+0x194>
  102611:       74 2f                   je     102642 <main+0x172>

In this case, the address is 001024d0.

Run the kernel inside QEMU GDB, setting a breakpoint at the beginning of main (i.e., the address you just found). (Note: If you're running your own copy of QEMU and have installed the QEMU accelerator kernel module (kqemu), you'll need to disable kqemu for these exercises using the -no-kqemu argument to QEMU in order to work around a bug in its implementation of breakpoints.)

$ make qemu-gdb
...
$ gdb
GNU gdb 6.8-debian
Copyright (C) 2008 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later 
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.  Type "show copying"
and "show warranty" for details.
This GDB was configured as "x86_64-linux-gnu".
+ target remote localhost:26000
[New Thread 1]
The target architecture is assumed to be i8086
[f000:fff0] 0xffff0:    ljmp   $0xf000,$0xe05b
0x0000fff0 in ?? ()
(gdb) br * 0x001024d0
Breakpoint 1 at 0x1024d0
(gdb) c
Continuing.
The target architecture is assumed to be i386
0x1024d0:       lea    0x4(%esp),%ecx

Breakpoint 1, 0x001024d0 in ?? ()

The details of what you see are likely to differ from the above output. Look at the registers and the stack contents:

(gdb) info reg
...
(gdb) x/24x $esp
...
(gdb)

Which part of the stack printout is actually the stack? (Hint: not all of it.) Identify all the non-zero values on the stack.

Submit: The output of x/16x $esp with the valid part of the stack marked. Write a short (3-5 word) comment next to each non-zero value explaining what it is.