Xv6 homework 4

Since this has gone missing on the internet...

Lecture 4 homework

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 for Sep 18

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.

Assignment: 
Download xv6 and expand the tar ball:

sh-3.00$ wget -q http://pdos.csail.mit.edu/6.828/2006/src/xv6-rev0.tar.gz 
sh-3.00$ tar xzvf xv6-rev0.tar.gz
xv6/
xv6/8253pit.c
xv6/asm.h
xv6/bio.c
...
$
Build xv6:
$ cd xv6
$ make
cc -o mkfs mkfs.c
gcc -fno-builtin -O2 -Wall -MD   -c -o usertests.o usertests.c
gcc -fno-builtin -O2 -Wall -MD   -c -o ulib.o ulib.c
...
$ 
Find the address of the main0 function by looking in kernel.asm:
sh-3.00$ grep main0 kernel.asm
001015b0 <main0>:
  1015f0:	7e ee                	jle    1015e0 <main0+0x30>
  1016a7:	74 22                	je     1016cb <main0+0x11b>
  1016d0:	eb dc                	jmp    1016ae <main0+0xfe>
sh-3.00$ 
In this case, the address is 001015b0. Note that this address may be different on 
Run the kernel inside Bochs, setting a breakpoint at the beginning of main0 (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
              Build from CVS snapshot on January 29, 2006
========================================================================
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
<bochs> vb 0x8:0x1015b0
<bochs> c
(0) Breakpoint 1, 0x001015b0 (0x0008:0x001015b0)
Next at t=901856
(0) [0x001015b0] 0008:0x001015b0 (unk. ctxt): push ebp                  ; 55
(1) [0xfffffff0] f000:fff0 (unk. ctxt): jmp far f000:e05b         ; ea5be000f0
<bochs>
Look at the registers and the stack contents:
<bochs> info reg
...
<bochs> print-stack
...
<bochs>
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 main0 that read:

  1015fc:	ba 7c a6 10 00       	mov    $0x10a67c,%edx
  101601:	89 d4                	mov    %edx,%esp
  101603:	ba 9c a6 10 00       	mov    $0x10a69c,%edx
  101608:	89 d5                	mov    %edx,%ebp
(The addresses and constants might be different for you. Look for the moves into %esp and %ebp).
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 0x8:0x1015fc
<bochs> s
Next at t=901858
(0) [0x00102ea8] 0008:0x00102ea8 (unk. ctxt): jnz .+0xfffffff7          ; 75f7
(1) [0xfffffff0] f000:fff0 (unk. ctxt): jmp far f000:e05b         ; ea5be000f0
<bochs> c
(0) Breakpoint 2, 0x001015fc (0x0008:0x001015fc)
Next at t=1191513
(0) [0x001015fc] 0008:0x001015fc (unk. ctxt): mov edx, 0x0010a67c       ; ba7ca61000
(1) [0xfffffff0] f000:fff0 (unk. ctxt): jmp far f000:e05b         ; ea5be000f0
<bochs> 
(The first s command is necessary to single-step past the breakpoint, or else c will not make any progress.)
Inspect the registers and stack again (info reg and print-stack). Then step past those four instructions (s 4) 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 immediately follows the stack switch. Where does the bcpu argument come from? What would have happened if the compiler had instead chosen to save bcpu on the stack before those four assembly instructions? Would the code still work? Why or why not?

(You can test your answer to the last two questions by running

$ make clean
$ make 'CFLAGS=-fno-builtin -Wall -MD'
to build a kernel without optimizations (the default CFLAGS in the Makefile also says -O2). Without optimization, the compiler will use the stack for every variable reference. Be sure to run make clean once you're finished experimenting.