Xv6 學習紀錄

Lab 連結：Lab: Copy-on-Write Fork for xv6 題目解析 The problem The fork() system call in xv6 copies all of the parent process’s user-space memory into the child. If the parent is large, copying can take a long time. Worse, the work is often largely wasted: fork() is commonly followed by exec() in the child, which discards the copied memory, usually without using most of it. On the other hand, if both parent and child use a copied page, and one or both writes it, the copy is truly needed. ...

課程影片連結： 6.S081 Fall 2020 Lecture 11: Thread Switching 追蹤 Context Switch 的過程 先看一下影片中的範例程式 user/spin.c: #include "kernel/types.h" #include "user/user.h" int main(int argc, char **argv) { int pid; char c; pid = fork(); if (pid == 0) { c = '/'; } else { printf("parent pid is %d, child is %d", getpid(), pid); c = '\\'; } for (int i = 0; ; i++) { if (i % 1000000 == 0) write(2, &c, 1); } exit(0); } 在做的事情基本上是 ...

Lab 連結：Lab: Multithreading Uthread: switching between threads (moderate) In this exercise you will design the context switch mechanism for a user-level threading system, and then implement it. To get you started, your xv6 has two files user/uthread.c and user/uthread_switch.S, and a rule in the Makefile to build a uthread program. uthread.c contains most of a user-level threading package, and code for three simple test threads. The threading package is missing some of the code to create a thread and to switch between threads. ...

課程影片連結： 6.S081 Fall 2020 Lecture 10: Multiprocessors and Locks Deadlock acquire(&l) acquire(&l) release(&l) Locks vs. Performance 越多 lock 雖然可能可以比較安全，但也會出現效能上的不佳 Case study UART kernel/uart.c struct spinlock uart_tx_lock; void uartputc(int c) { acquire(&uart_tx_lock); if(panicked){ for(;;) ; } while(uart_tx_w == uart_tx_r + UART_TX_BUF_SIZE){ // buffer is full. // wait for uartstart() to open up space in the buffer. sleep(&uart_tx_r, &uart_tx_lock); } uart_tx_buf[uart_tx_w % UART_TX_BUF_SIZE] = c; uart_tx_w += 1; uartstart(); release(&uart_tx_lock); } void uartintr(void) { // read and process incoming characters. while(1){ int c = uartgetc(); if(c == -1) break; consoleintr(c); } // send buffered characters. acquire(&uart_tx_lock); uartstart(); release(&uart_tx_lock); } Lock rules protect data structure Lock 的實做 broken lock void acquire(struct spinlock *lk) // does not work! { for(;;) { if(lk->locked == 0) { lk->locked = 1; break; } } } 這是一個錯誤的實做方法，因為可能有兩個 CPU 進入到 if (l->locked == 0) ...

kernel/kalloc.c data structure kmem 與 fun kmem 負責紀錄可使用的 pages，這些 pages 用一個 linked list 做紀錄 struct run { struct run *next; }; struct { struct spinlock lock; struct run *freelist; } kmem; 這個 Linked list run 是一個裡面就只有一個值，指向下一個可用的 page 的 address kalloc() 與 kfree() // Allocate one 4096-byte page of physical memory. // Returns a pointer that the kernel can use. // Returns 0 if the memory cannot be allocated. void * kalloc(void) { struct run *r; acquire(&kmem.lock); r = kmem.freelist; if(r) kmem.freelist = r->next; release(&kmem.lock); if(r) memset((char*)r, 5, PGSIZE); // fill with junk return (void*)r; } kalloc() 中，回傳 freelist 的 head，新的 head 變成 freelist->next Returns 0 if the memory cannot be allocated. 之所以這會成立，是因為在最一開始最一開始的時候，struct run *freelist; 被初始成 null (0)，kinit() 之前的這個唯一的 freelist 最後會變成 linked list 的結尾，這也是 kalloc() 之所以把記憶體用光之後會回傳 0 的原因 // Free the page of physical memory pointed at by pa, // which normally should have been returned by a // call to kalloc(). (The exception is when // initializing the allocator; see kinit above.) void kfree(void *pa) { struct run *r; if(((uint64)pa % PGSIZE) != 0 || (char*)pa < end || (uint64)pa >= PHYSTOP) panic("kfree"); // Fill with junk to catch dangling refs. memset(pa, 1, PGSIZE); r = (struct run*)pa; acquire(&kmem.lock); r->next = kmem.freelist; kmem.freelist = r; release(&kmem.lock); } pa 必須是一個 page 的開頭 kfree() 把這個不要的 pa 放到 freelist 的開頭 freelist 的初始化過程 kernel/main.c void main() { if(cpuid() == 0){ consoleinit(); printfinit(); printf("\n"); printf("xv6 kernel is booting\n"); printf("\n"); kinit(); // physical page allocator <- 這裡進入 kalloc.c kvminit(); // create kernel page table kvminithart(); // turn on paging procinit(); // process table trapinit(); // trap vectors trapinithart(); // install kernel trap vector plicinit(); // set up interrupt controller plicinithart(); // ask PLIC for device interrupts binit(); // buffer cache iinit(); // inode table fileinit(); // file table virtio_disk_init(); // emulated hard disk userinit(); // first user process __sync_synchronize(); started = 1; } else { while(started == 0) ; __sync_synchronize(); printf("hart %d starting\n", cpuid()); kvminithart(); // turn on paging trapinithart(); // install kernel trap vector plicinithart(); // ask PLIC for device interrupts } scheduler(); } kernel/kalloc.c: kinit() void kinit() { initlock(&kmem.lock, "kmem"); freerange(end, (void*)PHYSTOP); } 初始 kmem.lock 使用 freerange() 把可用的 page 都放到 freelist 中 kernel/kalloc.c: freerange() void freerange(void *pa_start, void *pa_end) { char *p; p = (char*)PGROUNDUP((uint64)pa_start); for(; p + PGSIZE <= (char*)pa_end; p += PGSIZE) kfree(p); } 這裡使用先前提到的 kfree() 一次一次的把可用 page 放到 freelist 的前方

這篇主要是為了 lab locks 中的 “Buffer cache” 這一題，要我們先了解 bcache 的一些原理與應用，主要的參考資料是 xv6 book 8.1 ~ 8.3 與 bcache 相關的程式碼 kernel/main.c void main() { if(cpuid() == 0){ consoleinit(); printfinit(); printf("\n"); printf("xv6 kernel is booting\n"); printf("\n"); kinit(); // physical page allocator kvminit(); // create kernel page table kvminithart(); // turn on paging procinit(); // process table trapinit(); // trap vectors trapinithart(); // install kernel trap vector plicinit(); // set up interrupt controller plicinithart(); // ask PLIC for device interrupts binit(); // buffer cache <- 這裡進入 bcache 的初始化過程 iinit(); // inode table fileinit(); // file table virtio_disk_init(); // emulated hard disk userinit(); // first user process __sync_synchronize(); started = 1; } else { while(started == 0) ; __sync_synchronize(); printf("hart %d starting\n", cpuid()); kvminithart(); // turn on paging trapinithart(); // install kernel trap vector plicinithart(); // ask PLIC for device interrupts } scheduler(); } kernel/bio.c: binit() struct { struct spinlock lock; struct buf buf[NBUF]; // Linked list of all buffers, through prev/next. // Sorted by how recently the buffer was used. // head.next is most recent, head.prev is least. struct buf head; } bcache; void binit(void) { struct buf *b; initlock(&bcache.lock, "bcache"); // Create linked list of buffers bcache.head.prev = &bcache.head; bcache.head.next = &bcache.head; for(b = bcache.buf; b < bcache.buf+NBUF; b++){ b->next = bcache.head.next; b->prev = &bcache.head; initsleeplock(&b->lock, "buffer"); bcache.head.next->prev = b; bcache.head.next = b; } } 在 binit() 之後，就不會直接使用 bcache.buf 這個 array，而是使用由 head 開頭的這個環狀 linked-list bread() 與 bwrite() // Return a locked buf with the contents of the indicated block. struct buf* bread(uint dev, uint blockno) { struct buf *b; b = bget(dev, blockno); if(!b->valid) { virtio_disk_rw(b, 0); b->valid = 1; } return b; } 如果 bget() 可以拿到 buf 就直接回傳，如果不行，則使用 virtio_disk_rw() ...

Lab 連結：Lab: locks Memory allocator (moderate) Your job is to implement per-CPU freelists, and stealing when a CPU’s free list is empty. You must give all of your locks names that start with “kmem”. That is, you should call initlock for each of your locks, and pass a name that starts with “kmem”. Run kalloctest to see if your implementation has reduced lock contention. To check that it can still allocate all of memory, run usertests sbrkmuch. Your output will look similar to that shown below, with much-reduced contention in total on kmem locks, although the specific numbers will differ. Make sure all tests in usertests -q pass. make grade should say that the kalloctests pass. ...

// Init fs void fsinit(int dev) { readsb(dev, &sb); if(sb.magic != FSMAGIC) panic("invalid file system"); initlog(dev, &sb); } // Zero a block. static void bzero(int dev, int bno) { struct buf *bp; bp = bread(dev, bno); memset(bp->data, 0, BSIZE); log_write(bp); brelse(bp); } balloc() 與 bfree() // Blocks. // Allocate a zeroed disk block. // returns 0 if out of disk space. static uint balloc(uint dev) { int b, bi, m; struct buf *bp; bp = 0; // 掃過 super block for(b = 0; b < sb.size; b += BPB){ bp = bread(dev, BBLOCK(b, sb)); for(bi = 0; bi < BPB && b + bi < sb.size; bi++){ m = 1 << (bi % 8); if((bp->data[bi/8] & m) == 0){ // Is block free? bp->data[bi/8] |= m; // Mark block in use. log_write(bp); brelse(bp); bzero(dev, b + bi); return b + bi; } } brelse(bp); } printf("balloc: out of blocks\n"); return 0; } balloc() 是 block allocate 的意思 ...

課程連結：6.S081 Fall 2020 Lecture 14: File Systems

Lab 連結：Lab: file system Large files (moderate) Modify bmap() so that it implements a doubly-indirect block, in addition to direct blocks and a singly-indirect block. You’ll have to have only 11 direct blocks, rather than 12, to make room for your new doubly-indirect block; you’re not allowed to change the size of an on-disk inode. The first 11 elements of ip->addrs[] should be direct blocks; the 12th should be a singly-indirect block (just like the current one); the 13th should be your new doubly-indirect block. You are done with this exercise when bigfile writes 65803 blocks and usertests runs successfully: ...