MIT 6.S081学习笔记（第七章）
〇、前言
本文主要完成MIT 6.S081 实验七：locks。
开始之前，切换分支：
$ git fetch
$ git checkout lock
$ make clean
一、Memory allocator (moderate)
Question requirements
The program user/kalloctest stresses xv6’s memory allocator: three processes grow and shrink their address spaces, resulting in many calls to
kallocandkfree.kallocandkfreeobtainkmem.lock. kalloctest prints (as “#fetch-and-add”) the number of loop iterations in acquire due to attempts to acquire a lock that another core already holds, for thekmem lockand a few other locks. The number of loop iterations in acquire is a rough measure of lock contention.
Some hints
- You can use the constant NCPU from
kernel/param.h; - Let freerange give all free memory to the CPU running freerange;
- The function cpuid returns the current core number, but it’s only safe to call it and use its result when interrupts are turned off. You should use
push_off()andpop_off()to turn interrupts off and on. - Have a look at the
snprintffunction inkernel/sprintf.cfor string formatting ideas. It is OK to just name all locks “kmem” though.
Answer
这个实验比较简单，目标就是尽量避免这样一种情况：当线程 A 拿到锁时，B 尽量不要再去尝试获取锁。因此，我们要重新设计（改进）原来的内存分配机制，为每一个CPU 都创建一个内存链表，这样每个核心都不会竞争别的 CPU 的内存链表，它们有自己的链表。
但是我们不得不承认，当 CPU0 的内存链表为空时，它无法为跑在本 CPU0 上的线程分配内存。因此，它必须去和其它的 CPU 去沟通，让其它的 CPU 给它分配一些内存。
来看看原来的设计：
// kernel/kalloc.c
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);
return (void*)r;
}
可以看到，原来的设计中，由于修改链表必须当做原子化操作，因此每次获取内存时，都要占用锁。由于并发性，其它 CPU 在尝试获取内存时，大概率会 aquire() 失败。
因此，我们需要重新设计该机制。
// kernel/kalloc.c
struct {
struct spinlock lock;
struct run *freelist;
} kmem[NCPU];
char *kmem_lock_names[] = {
"kmem_cpu_0",
"kmem_cpu_1",
"kmem_cpu_2",
"kmem_cpu_3",
"kmem_cpu_4",
"kmem_cpu_5",
"kmem_cpu_6",
"kmem_cpu_7",
};
void
kinit()
{
for(int i=0;i<NCPU;i++) { // 初始化锁，这是题目要求
initlock(&kmem[i].lock, kmem_lock_names[i]);
}
freerange(end, (void*)PHYSTOP);
}
释放内存：
// kernel/kalloc.c
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;
push_off(); // 题目说，调用 cpuid()必须关中断，关了之后就立即打开
int cpu = cpuid();
pop_off();
acquire(&kmem[cpu].lock);
// 头插法
r->next = kmem[cpu].freelist;
kmem[cpu].freelist = r;
release(&kmem[cpu].lock);
}
内存分配：
void *
kalloc(void)
{
struct run *r;
push_off();
int cpu = cpuid();
pop_off();
acquire(&kmem[cpu].lock);
if(!kmem[cpu].freelist) {
int steal_left = 64;
for(int i=0;i<NCPU;i++) {
if(i == cpu) continue;
acquire(&kmem[i].lock);
struct run *rr = kmem[i].freelist;
while(rr && steal_left) {
kmem[i].freelist = rr->next;
// 头插法
rr->next = kmem[cpu].freelist;
kmem[cpu].freelist = rr;
rr = kmem[i].freelist;
steal_left--;
}
release(&kmem[i].lock);
if(steal_left == 0) break; // 转移完成
}
}
r = kmem[cpu].freelist;
if(r)
kmem[cpu].freelist = r->next;
release(&kmem[cpu].lock); // 释放锁
if(r)
memset((char*)r, 5, PGSIZE); // fill with junk
return (void*)r;
}
至此，这个实验就完成了：
== Test running kalloctest ==
$ make qemu-gdb
(186.2s)
== Test kalloctest: test1 ==
kalloctest: test1: OK
== Test kalloctest: test2 ==
kalloctest: test2: OK
== Test kalloctest: test3 ==
kalloctest: test3: OK
但是这个解决方案会导致死锁，比如：当 cpu0拿到本锁，之后，尝试获取 cpu1 的锁来进行偷取。同时，cpu1拿到本锁之后，尝试拿到 cpu0 的锁进行偷取。这样就导致了循环等待，发生了死锁。
解决方法也很简单，当 cpu0 拿到本锁之后，当它确认要偷取 cpu1时，可以释放自己的本锁。这样就避免了循环等待。但是，我们仍然必须保证窃取过程是原子化操作的，怎么办呢？我们可以考虑再加一把大一点的锁。
这样就得到了不会死锁的终极代码：
struct {
struct spinlock stealing_lock;
struct spinlock lock;
struct run *freelist;
} kmem[NCPU]; // 为每一个 cpu 分配一个 freelist
char *kmem_lock_names[] = {
"kmem_cpu_0",
"kmem_cpu_1",
"kmem_cpu_2",
"kmem_cpu_3",
"kmem_cpu_4",
"kmem_cpu_5",
"kmem_cpu_6",
"kmem_cpu_7",
};
char *kmem_stealing_lock_names[] = {
"kmem_cpu_00",
"kmem_cpu_11",
"kmem_cpu_22",
"kmem_cpu_33",
"kmem_cpu_44",
"kmem_cpu_55",
"kmem_cpu_66",
"kmem_cpu_77",
};
初始化：
void
kinit()
{
for(int i = 0; i < NCPU; i++){ // 修改初始化代码
initlock(&kmem[i].lock,kmem_lock_names[i]);
initlock(&kmem[i].stealing_lock,kmem_stealing_lock_names[i]);
}
freerange(end, (void*)PHYSTOP);
}
内存分配：
void *
kalloc(void)
{
struct run *r;
push_off();
int cpu = cpuid();
pop_off();
acquire(&kmem[cpu].lock);
if(!kmem[cpu].freelist){ // 如果链表为空,就偷
// 获取大锁
acquire(&kmem[cpu].stealing_lock);
// 释放本锁
release(&kmem[cpu].lock);
int steal_left = 64;
for(int i = 0; i < NCPU; i++){
if(i == cpu) continue; // 不要偷自己的内存
acquire(&kmem[i].lock);
struct run *rr = kmem[i].freelist;
while(rr && steal_left){
kmem[i].freelist = rr->next; // 取出一块内存
rr->next = kmem[cpu].freelist; // 插入
kmem[cpu].freelist = rr;
rr = kmem[i].freelist;
steal_left--;
}
release(&kmem[i].lock);
if(steal_left == 0) break; // 转移完成
}
acquire(&kmem[cpu].lock);
release(&kmem[cpu].stealing_lock);
}
r = kmem[cpu].freelist;
if(r)
kmem[cpu].freelist = r->next;
release(&kmem[cpu].lock);
if(r)
memset((char*)r, 5, PGSIZE); // fill with junk
return (void*)r;
}
我们再看看是否会有死锁。当 cpu0 拿到本锁后，先拿到本steal 锁，开始窃取 cpu1 的内存时，获取 cpu1本锁，然后修改cpu1 链表。cpu1 如果此时持有本锁，那么窃取等待。而 cpu1 拿着本锁，就会轻松窃取 cpu0 的内存，而没有任何阻碍。完美解决了死锁问题！
但是还有一个非常关键的问题！那就是当 cpu0 完全放弃本锁后，在修改 cpu0 链表的过程中，发生了进程调度，比如 kfree() 被调用，那么就会发生错误！这如何解决？
答案是不用解决，因为 xv6 保证了，只要某一个线程还没有释放完所有的锁，进程就不会调度。这就是那个大锁 stealing_lock的妙处，它使得整个操作变得原子化。详见关于这一个问题的讨论。
修改后，自然通过了测试：
== Test running kalloctest ==
$ make qemu-gdb
(195.1s)
== Test kalloctest: test1 ==
kalloctest: test1: OK
== Test kalloctest: test2 ==
kalloctest: test2: OK
== Test kalloctest: test3 ==
kalloctest: test3: OK
二、Buffer cache (hard)
Question requirements
If multiple processes use the file system intensively, they will likely contend for
bcache.lock, which protects the disk block cache inkernel/bio.c.bcachetestcreates several processes that repeatedly read different files in order to generate contention onbcache.lock.
You will likely see different output, but the number of acquire loop iterations for the
bcache lockwill be high. If you look at the code inkernel/bio.c, you’ll see thatbcache.lockprotects the list of cached block buffers, the reference count (b->refcnt) in each block buffer, and the identities of the cached blocks (b->devandb->blockno).
Modify the
block cacheso that the number of acquire loop iterations for all locks in the bcache is close to zero when runningbcachetest. Ideally the sum of the counts for all locks involved in the block cache should be zero, but it’s OK if the sum is less than 500. Modifybgetandbrelseso that concurrent lookups and releases for different blocks that are in the bcache are unlikely to conflict on locks (e.g., don’t all have to wait for bcache.lock). You must maintain the invariant that at most one copy of each block is cached.
Some hints
- It is OK to use a fixed number of buckets and not resize the hash table dynamically. Use a prime number of buckets (e.g., 13) to reduce the likelihood of hashing conflicts.
- Searching in the hash table for a buffer and allocating an entry for that buffer when the buffer is not found must be atomic.
- Remove the list of all buffers (bcache.head etc.) and instead time-stamp buffers using the time of their last use (i.e., using ticks in
kernel/trap.c). With this changebrelsedoesn’t need to acquire the bcache lock, andbgetcan select the least-recently used block based on the time-stamps.- It is OK to serialize eviction in
bget(i.e., the part ofbgetthat selects a buffer to re-use when a lookup misses in the cache).- Your solution might need to hold two locks in some cases; for example, during eviction you may need to hold the bcache lock and a lock per bucket. Make sure you avoid deadlock.
- When replacing a block, you might move a
struct buffrom one bucket to another bucket, because the new block hashes to a different bucket. You might have a tricky case: the new block might hash to the same bucket as the old block. Make sure you avoid deadlock in that case.- Some debugging tips: implement bucket locks but leave the global bcache.lock acquire/release at the beginning/end of
bgetto serialize the code. Once you are sure it is correct without race conditions, remove the global locks and deal with concurrency issues. You can also run make CPUS=1 qemu to test with one core.
在一个缓存系统中，特别是像 bcache 这样的共享缓存，多个进程（或者多个 CPU）都可以同时访问和操作相同的区块。
在这种情况下，为每个 CPU 预先分配专属的页可能并不可行，因为这些页会被多个进程或 CPU 共享，无法简单地将其分配给单个处理器使用。这需要更复杂的管理和同步机制来确保多个进程或 CPU 可以安全地访问和使用这些共享的区块缓存。
这里锁竞争优化就不能简单降低分享了（比如上一个实验就是降低分享，将链表私有化），只能通过减少在关键区中停留的时间（对应“大锁化小锁”，降低锁的粒度）来优化了。
源码分析
先来看看原始的方案：
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;
这里通过一个链表，将所有的 buf 连接起来。head 是一个头结点，head.next 是最近使用的数据块，head.prev 是最久未使用的数据块。
初始化：
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;
}
}
这个函数初始化了整个 buffer，形成了一个双向链表结构。
最关键的是这个函数：
// Look through buffer cache for block on device dev.
// If not found, allocate a buffer.
// In either case, return locked buffer.
static struct buf*
bget(uint dev, uint blockno)
{
struct buf *b;
acquire(&bcache.lock);
// Is the block already cached?
for(b = bcache.head.next; b != &bcache.head; b = b->next){
if(b->dev == dev && b->blockno == blockno){
b->refcnt++;
release(&bcache.lock);
acquiresleep(&b->lock);
return b;
}
}
// Not cached.
// Recycle the least recently used (LRU) unused buffer.
for(b = bcache.head.prev; b != &bcache.head; b = b->prev){
if(b->refcnt == 0) {
b->dev = dev;
b->blockno = blockno;
b->valid = 0;
b->refcnt = 1;
release(&bcache.lock);
acquiresleep(&b->lock);
return b;
}
}
panic("bget: no buffers");
}
首先获取整个 buffer 的锁，然后依次遍历，当要查询的数据块吻合时，修改引用计数，并且释放 buffer 的锁，获取这个数据块的锁（要准备写入数据）。
如果没有命中，就向前遍历，这也是最久未使用的块。找到 b->refcnt == 0 的块，然后设置一些信息，释放大锁获取小锁，执行下一步对数据块的操作。
如果所有的缓冲都被占用了，怎么办？它是通过直接中断程序来结束的。直接 panic()，这显然有点不负责任。
// Write b's contents to disk. Must be locked.
void
bwrite(struct buf *b)
{
if(!holdingsleep(&b->lock))
panic("bwrite");
virtio_disk_rw(b, 1);
}
把这个块(b)拿到以后，先进行安全检查，然后就执行写入操作了。
// Release a locked buffer.
// Move to the head of the most-recently-used list.
void
brelse(struct buf *b)
{
if(!holdingsleep(&b->lock))
panic("brelse");
releasesleep(&b->lock);
acquire(&bcache.lock);
b->refcnt--;
if (b->refcnt == 0) {
// no one is waiting for it.
b->next->prev = b->prev;
b->prev->next = b->next;
b->next = bcache.head.next;
b->prev = &bcache.head;
bcache.head.next->prev = b;
bcache.head.next = b;
}
release(&bcache.lock);
}
这是释放快的操作，第一步就是安全检查，然后释放掉这个块的锁（因为我们不对它内部进行操作了），但是要对整个 buffer 进行修改，所以要上大锁，并将引用计数自减。
当它的引用计数为 0 时，就该释放掉了。这里为了保持最近使用到最近最久未使用的序列，先将 b 踢除，然后把它又插在了 head->next的位置，也就是最近未使用的位置。最后将大锁释放掉，就完成了释放 b 快的操作。
void
bpin(struct buf *b) {
acquire(&bcache.lock);
b->refcnt++;
release(&bcache.lock);
}
void
bunpin(struct buf *b) {
acquire(&bcache.lock);
b->refcnt--;
release(&bcache.lock);
}
这两个函数很简单，就是封装的引用计数自增自减函数。
可以看到整个结构是，当要操作某个 buf 块的属性以及整个 buffer 时，需要大锁；当要往 buf 块中写数据时，需要小锁，理解锁的层次结构很重要。
源码重新设计
原版 xv6 的设计中，使用双向链表存储所有的区块缓存，每次尝试获取一个区块 blockno 的时候，会遍历链表，如果目标区块已经存在缓存中则直接返回，如果不存在则选取一个最近最久未使用的，且引用计数为 0 的 buf 块作为其区块缓存，并返回。
新的改进方案，可以建立一个从 blockno 到 buf 的哈希表，并为每个桶单独加锁。仅有在两个进程同时访问的区块同时哈希到同一个桶的时候，才会发生锁竞争。当桶中的空闲buf 不足的时候，从其他的桶中获取 buf。
首先重新设计 bcache 结构体：
struct {
struct buf buf[NBUF];
struct spinlock eviction_lock; // 大锁
struct buf bufmap[NBUFMAP_BUCKET]; // 桶
struct spinlock bufmap_locks[NBUFMAP_BUCKET]; // 每一个桶的锁
} bcache;
这个结构体定义了一个大锁，并定义了桶组，还有桶锁组。
void
binit(void)
{
// 初始化缓冲桶s
for(int i = 0; i < NBUFMAP_BUCKET; i++){
initlock(&bcache.bufmap_locks[i],"bcache_bufmap");
bcache.buf[i].next = 0;
}
initlock(&bcache.eviction_lock, "bcache_eviction");
// 初始化缓冲区
for(int i = 0; i < NBUF; i++){
struct buf *b = &bcache.buf[i];
initsleeplock(&b->lock, "buffer");
b->lastuse = 0;
b->refcnt = 0;
// 将所有的缓冲区全部放进 bcache.bufmap[0]
b->next = bcache.bufmap[0].next;
bcache.bufmap[0].next = b;
}
}
将所有的桶的 next指针 悬空，再将所有的缓冲区链到 桶0 中。
接下来就是 bget 函数的改写了，先用伪代码写出框架：
// 由设备号、块号得到桶key(哈希生成函数)
// 拿到 桶key 的锁
// 循环遍历 桶key:
for(b = bcache.bufmap[key].next;b;b = b->next){
if(b->dev == dev && b->blockno == blockno){
b->refcnt++;
release(&bcache.bufmap_locks[key]);
acquiresleep(&b->lock); // 获取睡眠锁
return b;
}
}
// 释放 桶key 的锁，防止持有锁时又获取其它桶的锁
// 获取大锁
// 由于中途放弃过锁,可能被其它 CPU 已经将该要映射的块映射过,防止重复映射，再检查一次
for(b = bcache.bufmap[key].next;b;b = b->next){
if(b->dev == dev && b->blockno == blockno){
acquire(&bcache.bufmap_locks[key]);
b->refcnt++;
release(&bcache.bufmap_locks[key]);
release(&bcache.eviction_lock); // 释放大锁
acquiresleep(&b->lock); // 获取睡眠锁
return b;
}
}
// 如果依然没有命中，循环循环找目标 buf
// 定义一个最久未使用的前一个结点：struct buf *before_least = 0;
uint holding_bucket = -1;
for(int i = 0; i < NBUFMAP_BUCKET; i++){
// 获取小锁（此时持有大锁）
acquire(&bcache.bufmap_locks[i]);
int newfound = 0;
// 选中一个最久未使用的（ticks大的）且引用计数为 0 的 buf块
for(b = &bcache.bufmap[i]; b->next; b = b->next) {
if(b->next->refcnt == 0 && (!before_least || b->next->lastuse < before_least->next->lastuse)) {
before_least = b;
newfound = 1;
}
}
if(!newfound) {
release(&bcache.bufmap_locks[i]);
} else {
// 如果选中的桶更新了，就释放前一个桶的锁，继续持有更新的桶的锁,这养还能防止其它 cpu 对这个桶引用数+1，从而导致驱逐错误
if(holding_bucket != -1) release(&bcache.bufmap_locks[holding_bucket]);
holding_bucket = i;
// keep holding this bucket's lock....
}
}
// 如果遍历完了所有的桶，发现没有任何 buff 块可用，继续 panic().
// b 就是我们要找的缓冲块
b = before_least->next;
// 我们此时找到了一个桶，这个桶的编号并不是我们映射到的 key，那么将这个桶的 buf 转移到 key 桶中就好
if(holding_bucket != key) {
// 从原来的桶中转移
before_least->next = b->next;
// 现在释放
release(&bcache.bufmap_locks[holding_bucket]);
// 拿到 key锁，头插法直接插入
acquire(&bcache.bufmap_locks[key]);
b->next = bcache.bufmap[key].next;
bcache.bufmap[key].next = b;
}
// 设置
b->dev = dev;
b->blockno = blockno;
b->refcnt = 1;
b->valid = 0;
release(&bcache.bufmap_locks[key]);
release(&bcache.eviction_lock); // 释放大锁
acquiresleep(&b->lock);
return b;
以上就是伪代码和注释，以下为修改好的完整代码以及其它几个函数的代码。
kernel/bio.c：
// Buffer cache.
//
// The buffer cache is a linked list of buf structures holding
// cached copies of disk block contents. Caching disk blocks
// in memory reduces the number of disk reads and also provides
// a synchronization point for disk blocks used by multiple processes.
//
// Interface:
// * To get a buffer for a particular disk block, call bread.
// * After changing buffer data, call bwrite to write it to disk.
// * When done with the buffer, call brelse.
// * Do not use the buffer after calling brelse.
// * Only one process at a time can use a buffer,
// so do not keep them longer than necessary.
#include "types.h"
#include "param.h"
#include "spinlock.h"
#include "sleeplock.h"
#include "riscv.h"
#include "defs.h"
#include "fs.h"
#include "buf.h"
// bucket number for bufmap
#define NBUFMAP_BUCKET 13
// hash function for bufmap
#define BUFMAP_HASH(dev, blockno) ((((dev)<<27)|(blockno))%NBUFMAP_BUCKET)
struct {
struct buf buf[NBUF];
struct spinlock eviction_lock; // 大锁
struct buf bufmap[NBUFMAP_BUCKET]; // 桶s
struct spinlock bufmap_locks[NBUFMAP_BUCKET]; // 桶锁s
} bcache;
void
binit(void)
{
for(int i = 0; i < NBUFMAP_BUCKET; i++){
initlock(&bcache.bufmap_locks[i],"bcache_bufmap");
bcache.buf[i].next = 0;
}
initlock(&bcache.eviction_lock, "bcache_eviction");
for(int i = 0; i < NBUF; i++){
struct buf *b = &bcache.buf[i];
initsleeplock(&b->lock, "buffer");
b->lastuse = 0;
b->refcnt = 0;
b->next = bcache.bufmap[0].next;
bcache.bufmap[0].next = b;
}
}
// Look through buffer cache for block on device dev.
// If not found, allocate a buffer.
// In either case, return locked buffer.
static struct buf*
bget(uint dev, uint blockno)
{
struct buf *b;
uint key = BUFMAP_HASH(dev,blockno);
acquire(&bcache.bufmap_locks[key]);
// hit cache
for(b = bcache.bufmap[key].next;b;b = b->next){
if(b->dev == dev && b->blockno == blockno){
b->refcnt++;
release(&bcache.bufmap_locks[key]);
acquiresleep(&b->lock);
return b;
}
}
// no hitting cache
release(&bcache.bufmap_locks[key]);
acquire(&bcache.eviction_lock);
for(b = bcache.bufmap[key].next;b;b = b->next){
if(b->dev == dev && b->blockno == blockno){
acquire(&bcache.bufmap_locks[key]);
b->refcnt++;
release(&bcache.bufmap_locks[key]);
release(&bcache.eviction_lock); // 释放大锁
acquiresleep(&b->lock); // 获取睡眠锁
return b;
}
}
// 依然没有命中缓存
struct buf *before_least = 0;
uint holding_bucket = -1;
for(int i = 0; i < NBUFMAP_BUCKET; i++){
int newfound = 0;
acquire(&bcache.bufmap_locks[i]);
for(b = &bcache.bufmap[i]; b->next; b = b->next){
if(b->next->refcnt == 0 && (!before_least || b->next->lastuse < before_least->next->lastuse)){
before_least = b;
newfound = 1;
}
}
if(!newfound){
release(&bcache.bufmap_locks[i]);
}else{
if(holding_bucket != -1) release(&bcache.bufmap_locks[holding_bucket]);
holding_bucket = i;
}
}
if(!before_least){
panic("bget: no buffers!");
}
b = before_least->next;
if(holding_bucket !=key){
before_least->next = b->next;
release(&bcache.bufmap_locks[holding_bucket]);
acquire(&bcache.bufmap_locks[key]);
b->next= bcache.bufmap[key].next;
bcache.bufmap[key].next = b;
}
// 设置参数
b->dev = dev;
b->blockno = blockno;
b->refcnt = 1;
b->valid = 0;
release(&bcache.bufmap_locks[key]);
release(&bcache.eviction_lock);
acquiresleep(&b->lock);
return b;
}
// 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;
}
// Write b's contents to disk. Must be locked.
void
bwrite(struct buf *b)
{
if(!holdingsleep(&b->lock))
panic("bwrite");
virtio_disk_rw(b, 1);
}
// Release a locked buffer.
// Move to the head of the most-recently-used list.
void
brelse(struct buf *b)
{
if(!holdingsleep(&b->lock))
panic("brelse");
releasesleep(&b->lock);
uint key = BUFMAP_HASH(b->dev,b->blockno);
acquire(&bcache.bufmap_locks[key]);
b->refcnt--;
if (b->refcnt == 0) { // 事实上,不需做任何事情,更新 lastuse 就好
b->lastuse = ticks;
}
release(&bcache.bufmap_locks[key]);
}
void
bpin(struct buf *b) {
uint key = BUFMAP_HASH(b->dev, b->blockno);
acquire(&bcache.bufmap_locks[key]);
b->refcnt++;
release(&bcache.bufmap_locks[key]);
}
void
bunpin(struct buf *b) {
uint key = BUFMAP_HASH(b->dev, b->blockno);
acquire(&bcache.bufmap_locks[key]);
b->refcnt--;
release(&bcache.bufmap_locks[key]);
}
当锁被应用时，它与之前的实验相似。持有大锁意味着线程不会被抢占。因此，即使多个线程同时请求相同的 blockno，并且所有线程碰巧通过了“块缓存是否存在”的初始检查并发现不存在，只有第一个线程进入由 eviction_lock 保护的“驱逐和重新分配区域”才会执行驱逐和重新分配。
只有第一个线程进入并执行了驱逐和重新分配后释放了 eviction_lock，blockno 的缓存才从不存在转变为存在。后续命中相同 blockno 的线程会被第二次“blockno 缓存是否存在”的检查拦截，并直接返回已分配的缓存 buf，而不会重新执行驱逐和重新分配。
这种方法的优势在于确保在查找过程中不会发生死锁，并避免了一个块生成多个缓存的极端情况。然而，它引入了一个全局的 eviction_lock，降低了原本并发的桶遍历过程的并行性。此外，每次缓存未命中都会增加跨桶的遍历成本。
然而，缓存未命中本身是一个相对罕见的事件。对于这些缓存未命中，因为后续操作需要从磁盘获取数据，磁盘读取时间会比跨桶查找的成本高出几个数量级。因此，我认为这种方法的额外开销是可以接受的。
至此，实验就完成了：
== Test running kalloctest ==
$ make qemu-gdb
(195.1s)
== Test kalloctest: test1 ==
kalloctest: test1: OK
== Test kalloctest: test2 ==
kalloctest: test2: OK
== Test kalloctest: test3 ==
kalloctest: test3: OK
== Test kalloctest: sbrkmuch ==
$ make qemu-gdb
kalloctest: sbrkmuch: OK (29.8s)
== Test running bcachetest ==
$ make qemu-gdb
(61.1s)
== Test bcachetest: test0 ==
bcachetest: test0: OK
== Test bcachetest: test1 ==
bcachetest: test1: OK
== Test usertests ==
$ make qemu-gdb
usertests: OK (161.1s)
== Test time ==
time: OK
Score: 80/80
三、总结
- don’t share if you don’t have to
- start with a few coarse-grained locks
- instrument your code – which locks are preventing parallelism?
- use fine-grained locks only as needed for parallel performance
- use an automated race detector
参考：锁优化、讨论。
全文完，感谢阅读。