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<title>Homework: intro to xv6</title>
<html>
<head>
</head>
<body>
<h1>Homework: intro to xv6</h1>
<p>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.
<p>
<b>Hand-In Procedure</b>
<p>
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.
<p>
<p><b>Assignment</b>:
<br>
Fetch and un-tar the xv6 source:
<pre>
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
xv6/
xv6/asm.h
xv6/bio.c
xv6/bootasm.S
xv6/bootmain.c
...
$
</pre>
Build xv6:
<pre>
$ 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
...
$
</pre>
Find the address of the <code>main</code> function by
looking in <code>kernel.asm</code>:
<pre>
% grep main kernel.asm
...
00102454 &lt;mpmain&gt;:
mpmain(void)
001024d0 &lt;main&gt;:
10250d: 79 f1 jns 102500 &lt;main+0x30&gt;
1025f3: 76 6f jbe 102664 &lt;main+0x194&gt;
102611: 74 2f je 102642 &lt;main+0x172&gt;
</pre>
In this case, the address is <code>001024d0</code>.
<p>
Run the kernel inside Bochs, setting a breakpoint
at the beginning of <code>main</code> (i.e., the address
you just found).
<pre>
$ 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
&lt;bochs&gt;
</pre>
Look at the registers and the stack contents:
<pre>
&lt;bochs&gt; info reg
...
&lt;bochs&gt; print-stack
...
&lt;bochs&gt;
</pre>
Which part of the stack printout is actually the stack?
(Hint: not all of it.) Identify all the non-zero values
on the stack.<p>
<b>Turn in:</b> 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.
<p>
Now look at kernel.asm for the instructions in main that read:
<pre>
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
</pre>
(The addresses and constants might be different on your system,
and the compiler might use <code>imul</code> instead of the <code>lea,lea,shl,add,lea</code> sequence.
Look for the move into <code>%esp</code>).
<p>
Which lines in <code>main.c</code> do these instructions correspond to?
<p>
Set a breakpoint at the first of those instructions
and let the program run until the breakpoint:
<pre>
&lt;bochs&gt; vb 0x8:0x10251e
&lt;bochs&gt; s
...
&lt;bochs&gt; 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
&lt;bochs&gt;
</pre>
(The first <code>s</code> command is necessary
to single-step past the breakpoint at main, otherwise <code>c</code>
will not make any progress.)
<p>
Inspect the registers and stack again
(<code>info reg</code> and <code>print-stack</code>).
Then step past those seven instructions
(<code>s 7</code>)
and inspect them again.
Convince yourself that the stack has changed correctly.
<p>
<b>Turn in:</b> answers to the following questions.
Look at the assembly for the call to
<code>lapic_init</code> that occurs after the
the stack switch. Where does the
<code>bcpu</code> argument come from?
What would have happened if <code>main</code>
stored <code>bcpu</code>
on the stack before those four assembly instructions?
Would the code still work? Why or why not?
<p>
</body>
</html>