Checkpoint: split alarmtest exercise in two exercises
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@ -10,6 +10,8 @@
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This lab makes you familiar with the implementation of system calls.
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This lab makes you familiar with the implementation of system calls.
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In particular, you will implement a new system call: <tt>alarm</tt>.
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In particular, you will implement a new system call: <tt>alarm</tt>.
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<b>Note: before this lab, it would be good to have recitation section
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on gdb</b>
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<h2>Warmup: system call tracing</h2>
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<h2>Warmup: system call tracing</h2>
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@ -46,6 +48,56 @@ print its output.)
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<p>Optional: print the system call arguments.
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<p>Optional: print the system call arguments.
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<h2>RISC-V assembly</h2>
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<p>For the alarm system call it will be important to understand RISC-V
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assembly. Since in later labs you will also read and write assembly,
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it is important that you familiarize yourself with RISC_V assembly.
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<p>Add a file user/call.c with the following content, modify the
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Makefile to add the program to the user programs, and compile (make
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fs.img). The Makefile also produces a binary and a readable
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assembly a version of the program in the file user/call.asm.
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<pre>
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#include "kernel/param.h"
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#include "kernel/types.h"
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#include "kernel/stat.h"
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#include "user/user.h"
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int g(int x) {
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return x+3;
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}
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int f(int x) {
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return g(x);
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}
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void main(void) {
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printf(1, "%d %d\n", f(8)+1, 13);
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exit();
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}
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</pre>
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<p>Since you will be reading and writing RISC-V assembly code for xv6,
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you should read through call.asm and understand it. The instruction
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manual for RISC-V is in the doc directory (doc/riscv-spec-v2.2.pdf).
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Here are some questions that you should answer for yourself:
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<ul>
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<li>Which registers contain arguments to functions? Which
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register holds 13 in the call to <tt>printf</tt>? Which register
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holds the second one? Which register holds the second one? Etc.
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<li>Where is the function call to <tt>f</tt> and <tt>g</tt>
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in <tt>main</tt>? (Hint: compiler may inline functions.)
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<li>At what address is the function <tt>printf</tt> located?
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<li>What value is in the register <tt>ra</tt> in the <tt>jalr</tt>
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to <tt>printf</tt> in <tt>main</tt>?
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</ul>
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<h2>alarm</h2>
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<h2>alarm</h2>
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<p>
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<p>
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@ -70,25 +122,37 @@ interrupts.
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<p>
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<p>
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You should put the following example program in <tt>user/alarmtest.c</tt>:
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You should put the following example program in <tt>user/alarmtest.c</tt>:
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<b>XXX Insert the final program here</b>
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<pre>
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<pre>
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#include "kernel/param.h"
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#include "kernel/param.h"
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#include "kernel/types.h"
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#include "kernel/types.h"
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#include "kernel/stat.h"
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#include "kernel/stat.h"
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#include "kernel/riscv.h"
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#include "user/user.h"
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#include "user/user.h"
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void test0();
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void test1();
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void periodic();
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void periodic();
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int
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int
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main(int argc, char *argv[])
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main(int argc, char *argv[])
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{
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test0();
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test1();
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exit();
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}
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void test0()
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{
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{
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int i;
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int i;
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printf(1, "alarmtest starting\n");
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printf(1, "test0 start\n");
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alarm(10, periodic);
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alarm(2, periodic);
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for(i = 0; i < 25*500000; i++){
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for(i = 0; i < 1000*500000; i++){
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if((i % 250000) == 0)
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if((i % 250000) == 0)
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write(2, ".", 1);
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write(2, ".", 1);
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}
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}
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exit();
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alarm(0, 0);
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printf(1, "test0 done\n");
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}
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}
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void
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void
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@ -96,13 +160,37 @@ periodic()
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{
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{
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printf(1, "alarm!\n");
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printf(1, "alarm!\n");
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}
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}
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void __attribute__ ((noinline)) foo(int i, int *j) {
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if((i % 2500000) == 0) {
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write(2, ".", 1);
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}
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*j += 1;
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}
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void test1() {
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int i;
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int j;
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printf(1, "test1 start\n");
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j = 0;
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alarm(2, periodic);
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for(i = 0; i < 1000*500000; i++){
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foo(i, &j);
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}
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if(i != j) {
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printf(2, "i %d should = j %d\n", i, j);
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exit();
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}
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printf(1, "test1 done\n");
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}
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</pre>
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</pre>
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The program calls <tt>alarm(10, periodic)</tt> to ask the kernel to
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The program calls <tt>alarm(2, periodic1)</tt> in test0 to ask the kernel to
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force a call to <tt>periodic()</tt> every 10 ticks, and then spins for
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force a call to <tt>periodic()</tt> every 10 ticks, and then spins for
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a while.
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a while.
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After you have implemented the <tt>alarm()</tt> system call in the kernel,
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After you have implemented the <tt>alarm()</tt> system call in the kernel,
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<tt>alarmtest</tt> should produce output like this:
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<tt>alarmtest</tt> should produce output like this for test0:
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<pre>
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<pre>
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$ alarmtest
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$ alarmtest
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@ -125,7 +213,26 @@ alarmtest starting
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(If you only see one "alarm!", try increasing the number of iterations in
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(If you only see one "alarm!", try increasing the number of iterations in
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<tt>alarmtest.c</tt> by 10x.)
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<tt>alarmtest.c</tt> by 10x.)
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Here are some hints:
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<p>The main challenge will be to arrange that the handler is invoked
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when the process's alarm interval expires. In your usertrap, when a
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process's alarm interval expires, you'll want to cause it to execute
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its handler. How can you do that? You will need to understand in
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details how system calls work (i.e., the code in kernel/trampoline.S
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and kernel/trap.c). Which register contains the address where
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systems calls return to?
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<p>Your solution will be few lines of code, but it will be tricky to
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write the right lines of code. Common failure scenarios are: the
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user program crashes or doesn't terminate. You can see the assembly
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code for the alarmtest program in alarmtest.asm, which will be handy
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for debugging.
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<h2>Test0</h2>
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<p>To get started, the best strategy is to first pass test0, which
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will force you to handle the main challenge above. Here are some
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hints how to pass test0:
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<ul>
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<ul>
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<li>You'll need to modify the Makefile to cause <tt>alarmtest.c</tt>
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<li>You'll need to modify the Makefile to cause <tt>alarmtest.c</tt>
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@ -159,19 +266,13 @@ is handled in <tt>usertrap()</tt>; you should add some code here.
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You only want to manipulate a process's alarm ticks if there's a
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You only want to manipulate a process's alarm ticks if there's a
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a timer interrupt; you want something like
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a timer interrupt; you want something like
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<pre>
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<pre>
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if(which_dev == 2) ..
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if(which_dev == 2) ...
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</pre>
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</pre>
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<p>
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<li>Don't invoke the process's alarm function, if the processor
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In your usertrap, when a process's alarm interval expires,
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doesn't have a timer outstanding. Note that the address of the
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you'll want to cause it to execute its handler. How can you do that?
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user's alarm function might be 0 (e.g., in
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alarmtest.asm, <tt>period</tt> is at address 0).
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<li>
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You need to arrange things so that, when the handler returns,
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the process resumes executing where it left off. How can you do that?
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<li>
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You can see the assembly code for the alarmtest program in alarmtest.asm.
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<li>
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<li>
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It will be easier to look at traps with gdb if you tell qemu to use
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It will be easier to look at traps with gdb if you tell qemu to use
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@ -180,17 +281,32 @@ only one CPU, which you can do by running
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make CPUS=1 qemu
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make CPUS=1 qemu
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</pre>
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</pre>
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<li>
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</ul>
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It's OK if your solution doesn't save the caller-saved user registers
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when calling the handler.
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<ul>
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<h2>test1()</h2>
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<p>Test0 doesn't stress whether the handler returns correctly to
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interrupted instruction in test0. If you didn't get this right, it
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is likely that test1 will fail (the program crashes or the program
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goes into an infinite loop).
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<p>A main challenge is to arrange that when the handler returns, it
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returns to the instruction where the program was interrupted. Which
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register contains the return address of a function? When the kernel
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receives an interrupt, which register contains the address of the
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interrupted instruction?
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<p>Your solution is likely to require you to save and restore a
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register. There are several ways to do this. It is ok to change the
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API of alarm() and have an alarm stub in user space that cooperates
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with the kernel.
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<p>
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<p>
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Optional challenges: 1) Save and restore the caller-saved user registers around the
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Optional challenges: Prevent re-entrant calls to the handler----if a
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call to handler. 2) Prevent re-entrant calls to the handler -- if a handler
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handler hasn't returned yet, don't call it again.
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hasn't returned yet, don't call it again. 3) Assuming your code doesn't
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check that <tt>tf->esp</tt> is valid, implement a security attack on the
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kernel that exploits your alarm handler calling code.
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