Lab lazy: Lazy allocation

One of the many neat tricks an OS can play with page table hardware is lazy allocation of heap memory. Xv6 applications ask the kernel for heap memory using the sbrk system call. In the kernel we’ve given you, sbrk allocates physical memory and maps it into the process’s virtual address space. There are programs that allocate memory but never use it, for example to implement large sparse arrays. Sophisticated kernels delay allocation of each page of memory until the application tries to use that page—as signaled by a page fault. You’ll add this lazy allocation feature to xv6 in this lab.

To start the lab, update your repository and create a new branch for your solution:

$ git fetch origin
$ git checkout -b lazy origin/xv6-21au
$ make clean

Printing the page table

To help you visualize RISC-V page tables, and perhaps to aid future debugging, your first task is to write a function that prints the contents of a page table.

Exercise

Define a function called vmprint. It should take a pagetable_t argument, and print that pagetable in the format described below. Insert if(p->pid==1) vmprint(p->pagetable) in exec.c just before the return argc, to print the first process’s page table.

Now when you start xv6 it should print output like this, describing the page table of the first process at the point when it has just finished exec()ing init:

page table 0x0000000087f6e000
 ..0: pte 0x0000000021fda801 pa 0x0000000087f6a000
 .. ..0: pte 0x0000000021fda401 pa 0x0000000087f69000
 .. .. ..0: pte 0x0000000021fdac1f pa 0x0000000087f6b000
 .. .. ..1: pte 0x0000000021fda00f pa 0x0000000087f68000
 .. .. ..2: pte 0x0000000021fd9c1f pa 0x0000000087f67000
 ..255: pte 0x0000000021fdb401 pa 0x0000000087f6d000
 .. ..511: pte 0x0000000021fdb001 pa 0x0000000087f6c000
 .. .. ..510: pte 0x0000000021fdd807 pa 0x0000000087f76000
 .. .. ..511: pte 0x000000002000200b pa 0x0000000080007000

The first line displays the argument to vmprint. After that there is a line for each PTE, including PTEs that refer to page-table pages deeper in the tree. Each PTE line is indented by a number of “ ..” that indicates its depth in the tree. Each PTE line shows the PTE index in its page-table page, the pte bits, and the physical address extracted from the PTE. Don’t print PTEs that are not valid. In the above example, the top-level page-table page has mappings for entries 0 and 255. The next level down for entry 0 has only index 0 mapped, and the bottom-level for that index 0 has entries 0, 1, and 2 mapped.

Your code might emit different physical addresses than those shown above. The number of entries and the virtual addresses should be the same.

Some hints:

Question

Explain the output of vmprint. What does page 0 contain? What is in page 2? When running in user mode, could the process read/write the memory mapped by page 1? What do last two pages contain?

You don’t need to submit answers to the questions in this lab. Do answer them for yourself though!

Eliminating allocation from sbrk

Exercise

Delete page allocation from the sbrk(n) system call implementation, which is the function sys_sbrk() in sysproc.c. The sbrk(n) system call grows the process’s memory size by n bytes, and then returns the start of the newly allocated region (i.e., the old size). Your new sbrk(n) should just increment the process’s size (myproc()->sz) by n and return the old size. It should not allocate memory—so you should delete the call to growproc() (but you still need to increase the process’s size!).

Try to guess what the result of this modification will be: what will break?

Make this modification, boot xv6, and type echo hi to the shell. You should see something like this:

init: starting sh
$ echo hi
usertrap(): unexpected scause 0x000000000000000f pid=3
            sepc=0x00000000000011dc stval=0x0000000000004008
va=0x0000000000004000 pte=0x0000000000000000
panic: uvmunmap: not mapped

The “usertrap(): …” message is from the user trap handler in trap.c; it has caught an exception that it does not know how to handle. Make sure you understand why this page fault occurs. The “stval=0x0..04008” indicates that the virtual address that caused the page fault is 0x4008.

Lazy allocation

Exercise

Modify the code in trap.c to respond to a page fault from user space by mapping a newly allocated page of physical memory at the faulting address, and then returning back to user space to let the process continue executing. You should add your code just before the printf call that produced the “usertrap(): …” message. Your solution is acceptable if it passes both lazytests and usertests.

Hints:

If all goes well, your lazy allocation code should result in echo hi working. You should get at least one page fault (and thus lazy allocation) in the shell, and perhaps two.

If you have the basics working, now turn your implementation into one that handles the corner cases too:

Your solution is acceptable if your kernel passes both lazytests and usertests:

$ lazytests
lazytests starting
running test lazy alloc
test lazy alloc: OK
running test lazy unmap...
usertrap(): ...
test lazy unmap: OK
running test out of memory
usertrap(): ...
test out of memory: OK
ALL TESTS PASSED
$ usertests
...
ALL TESTS PASSED
$

You may run make grade to ensure that your code passes all of the tests.

This completes the lab. In the lab directory, commit your changes, type make tarball, and submit the tarball through Canvas.