Homework: intro to xv6

This lecture is the introduction to xv6, our re-implementation of Unix v6. Read the source code in 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 during lecture. Please write up your answers to the exercises below and hand them in to a 6.828 staff member at the beginning of lecture.

Fetch and un-tar the xv6 source:

sh-3.00$ wget http://pdos.csail.mit.edu/6.828/2007/src/xv6-rev1.tar.gz 
sh-3.00$ tar xzvf xv6-rev1.tar.gz
Build xv6:
$ 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>:
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 Bochs, setting a breakpoint at the beginning of main (i.e., the address you just found).

$ make bochs
if [ ! -e .bochsrc ]; then ln -s dot-bochsrc .bochsrc; fi
bochs -q
                       Bochs x86 Emulator 2.2.6
                    (6.828 distribution release 1)
00000000000i[     ] reading configuration from .bochsrc
00000000000i[     ] installing x module as the Bochs GUI
00000000000i[     ] Warning: no rc file specified.
00000000000i[     ] using log file bochsout.txt
Next at t=0
(0) [0xfffffff0] f000:fff0 (unk. ctxt): jmp far f000:e05b         ; ea5be000f0
(1) [0xfffffff0] f000:fff0 (unk. ctxt): jmp far f000:e05b         ; ea5be000f0
Look at the registers and the stack contents:
<bochs> info reg
<bochs> print-stack
Which part of the stack printout is actually the stack? (Hint: not all of it.) Identify all the non-zero values on the stack.

Turn in: the output of print-stack 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.

Now look at kernel.asm for the instructions in main that read:

  10251e:       8b 15 00 78 10 00       mov    0x107800,%edx
  102524:       8d 04 92                lea    (%edx,%edx,4),%eax
  102527:       8d 04 42                lea    (%edx,%eax,2),%eax
  10252a:       c1 e0 04                shl    $0x4,%eax
  10252d:       01 d0                   add    %edx,%eax
  10252f:       8d 04 85 1c ad 10 00    lea    0x10ad1c(,%eax,4),%eax
  102536:       89 c4                   mov    %eax,%esp
(The addresses and constants might be different on your system, and the compiler might use imul instead of the lea,lea,shl,add,lea sequence. Look for the move into %esp).

Which lines in main.c do these instructions correspond to?

Set a breakpoint at the first of those instructions and let the program run until the breakpoint:

<bochs> vb 0x8:0x10251e
<bochs> s
<bochs> c
(0) Breakpoint 2, 0x0010251e (0x0008:0x0010251e)
Next at t=1157430
(0) [0x0010251e] 0008:0x0010251e (unk. ctxt): mov edx, dword ptr ds:0x107800 ; 8b1500781000
(1) [0xfffffff0] f000:fff0 (unk. ctxt): jmp far f000:e05b         ; ea5be000f0
(The first s command is necessary to single-step past the breakpoint at main, otherwise c will not make any progress.)

Inspect the registers and stack again (info reg and print-stack). Then step past those seven instructions (s 7) and inspect them again. Convince yourself that the stack has changed correctly.

Turn in: answers to the following questions. Look at the assembly for the call to lapic_init that occurs after the the stack switch. Where does the bcpu argument come from? What would have happened if main stored bcpu on the stack before those four assembly instructions? Would the code still work? Why or why not?