164 lines
4.8 KiB
HTML
164 lines
4.8 KiB
HTML
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<title>Homework: intro to xv6</title>
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<html>
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<head>
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</head>
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<body>
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<h1>Homework: intro to xv6</h1>
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<p>This lecture is the introduction to xv6, our re-implementation of
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Unix v6. Read the source code in the assigned files. You won't have
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to understand the details yet; we will focus on how the first
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user-level process comes into existence after the computer is turned
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on.
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<p>
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<b>Hand-In Procedure</b>
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<p>
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You are to turn in this homework during lecture. Please
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write up your answers to the exercises below and hand them in to a
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6.828 staff member at the beginning of lecture.
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<p>
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<p><b>Assignment</b>:
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<br>
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Fetch and un-tar the xv6 source:
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<pre>
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sh-3.00$ wget http://pdos.csail.mit.edu/6.828/2007/src/xv6-rev1.tar.gz
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sh-3.00$ tar xzvf xv6-rev1.tar.gz
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xv6/
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xv6/asm.h
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xv6/bio.c
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xv6/bootasm.S
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xv6/bootmain.c
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...
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$
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</pre>
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Build xv6:
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<pre>
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$ cd xv6
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$ make
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gcc -O -nostdinc -I. -c bootmain.c
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gcc -nostdinc -I. -c bootasm.S
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ld -N -e start -Ttext 0x7C00 -o bootblock.o bootasm.o bootmain.o
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objdump -S bootblock.o > bootblock.asm
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objcopy -S -O binary bootblock.o bootblock
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...
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$
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</pre>
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Find the address of the <code>main</code> function by
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looking in <code>kernel.asm</code>:
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<pre>
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% grep main kernel.asm
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...
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00102454 <mpmain>:
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mpmain(void)
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001024d0 <main>:
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10250d: 79 f1 jns 102500 <main+0x30>
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1025f3: 76 6f jbe 102664 <main+0x194>
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102611: 74 2f je 102642 <main+0x172>
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</pre>
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In this case, the address is <code>001024d0</code>.
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<p>
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Run the kernel inside Bochs, setting a breakpoint
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at the beginning of <code>main</code> (i.e., the address
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you just found).
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<pre>
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$ make bochs
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if [ ! -e .bochsrc ]; then ln -s dot-bochsrc .bochsrc; fi
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bochs -q
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========================================================================
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Bochs x86 Emulator 2.2.6
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(6.828 distribution release 1)
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========================================================================
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00000000000i[ ] reading configuration from .bochsrc
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00000000000i[ ] installing x module as the Bochs GUI
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00000000000i[ ] Warning: no rc file specified.
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00000000000i[ ] using log file bochsout.txt
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Next at t=0
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(0) [0xfffffff0] f000:fff0 (unk. ctxt): jmp far f000:e05b ; ea5be000f0
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(1) [0xfffffff0] f000:fff0 (unk. ctxt): jmp far f000:e05b ; ea5be000f0
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<bochs>
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</pre>
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Look at the registers and the stack contents:
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<pre>
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<bochs> info reg
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...
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<bochs> print-stack
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...
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<bochs>
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</pre>
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Which part of the stack printout is actually the stack?
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(Hint: not all of it.) Identify all the non-zero values
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on the stack.<p>
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<b>Turn in:</b> the output of print-stack with
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the valid part of the stack marked. Write a short (3-5 word)
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comment next to each non-zero value explaining what it is.
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<p>
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Now look at kernel.asm for the instructions in main that read:
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<pre>
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10251e: 8b 15 00 78 10 00 mov 0x107800,%edx
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102524: 8d 04 92 lea (%edx,%edx,4),%eax
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102527: 8d 04 42 lea (%edx,%eax,2),%eax
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10252a: c1 e0 04 shl $0x4,%eax
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10252d: 01 d0 add %edx,%eax
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10252f: 8d 04 85 1c ad 10 00 lea 0x10ad1c(,%eax,4),%eax
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102536: 89 c4 mov %eax,%esp
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</pre>
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(The addresses and constants might be different on your system,
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and the compiler might use <code>imul</code> instead of the <code>lea,lea,shl,add,lea</code> sequence.
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Look for the move into <code>%esp</code>).
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<p>
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Which lines in <code>main.c</code> do these instructions correspond to?
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<p>
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Set a breakpoint at the first of those instructions
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and let the program run until the breakpoint:
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<pre>
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<bochs> vb 0x8:0x10251e
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<bochs> s
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...
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<bochs> c
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(0) Breakpoint 2, 0x0010251e (0x0008:0x0010251e)
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Next at t=1157430
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(0) [0x0010251e] 0008:0x0010251e (unk. ctxt): mov edx, dword ptr ds:0x107800 ; 8b1500781000
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(1) [0xfffffff0] f000:fff0 (unk. ctxt): jmp far f000:e05b ; ea5be000f0
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<bochs>
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</pre>
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(The first <code>s</code> command is necessary
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to single-step past the breakpoint at main, otherwise <code>c</code>
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will not make any progress.)
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<p>
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Inspect the registers and stack again
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(<code>info reg</code> and <code>print-stack</code>).
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Then step past those seven instructions
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(<code>s 7</code>)
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and inspect them again.
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Convince yourself that the stack has changed correctly.
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<p>
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<b>Turn in:</b> answers to the following questions.
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Look at the assembly for the call to
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<code>lapic_init</code> that occurs after the
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the stack switch. Where does the
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<code>bcpu</code> argument come from?
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What would have happened if <code>main</code>
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stored <code>bcpu</code>
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on the stack before those four assembly instructions?
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Would the code still work? Why or why not?
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<p>
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</body>
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</html>
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