内容简介：[TOC]本文参考的是golang 1.10源码实现。golang中map是一个kv对集合。
[TOC]
本文参考的是golang 1.10源码实现。
golang中map是一个kv对集合。 底层使用hash table，用链表来解决冲突，通过编译器配合runtime，所有的map对象都是共用一份代码。
对比其他语言
c++使用红黑树组织，性能稍低但是稳定性很好。使用模版在编译期生成代码，好处是效率高，但是缺点是代码膨胀、编译时间也会变长。
java使用的是hash table+链表/红黑树，当bucket内元素超过某个阈值时，该bucket的链表会转换成红黑树。java为了所有map共用一份代码，规定了只有Object的子类才能使用作为map的key，缺点是基础数据类型必须使用object包装一下才能使用map。
1. 函数选择
hash函数，有加密型和非加密型。加密型的一般用于加密数据、数字摘要等，典型代表就是md5、sha1、sha256、aes256这种；非加密型的一般就是查找。在map的应用场景中，用的是查找。选择hash函数主要考察的是两点：性能、碰撞概率。
具体hash函数的性能比较可以看： http://aras-p.info/blog/2016/08/09/More-Hash-Function-Tests/
golang使用的hash算法根据硬件选择，如果cpu支持aes，那么使用aes hash，否则使用memhash，memhash是参考xxhash、cityhash实现的，性能炸裂。
把hash值映射到buckte时，golang会把bucket的数量规整为2的次幂，而有m=2 b ，则n%m=n&(m-1)，用位运算规避mod的昂贵代价。
2. 结构组成
首先我们看下map的结构：
// A header for a Go map.
type hmap struct {
// Note: the format of the hmap is also encoded in cmd/compile/internal/gc/reflect.go.
// Make sure this stays in sync with the compiler's definition.
count int // # live cells == size of map. Must be first (used by len() builtin)
flags uint8
B uint8 // log_2 of # of buckets (can hold up to loadFactor * 2^B items)
noverflow uint16 // approximate number of overflow buckets; see incrnoverflow for details
hash0 uint32 // hash seed
buckets unsafe.Pointer // array of 2^B Buckets. may be nil if count==0.
oldbuckets unsafe.Pointer // previous bucket array of half the size, non-nil only when growing
nevacuate uintptr // progress counter for evacuation (buckets less than this have been evacuated)
extra *mapextra // optional fields
}
// mapextra holds fields that are not present on all maps.
type mapextra struct {
// If both key and value do not contain pointers and are inline, then we mark bucket
// type as containing no pointers. This avoids scanning such maps.
// However, bmap.overflow is a pointer. In order to keep overflow buckets
// alive, we store pointers to all overflow buckets in hmap.extra.overflow and hmap.extra.oldoverflow.
// overflow and oldoverflow are only used if key and value do not contain pointers.
// overflow contains overflow buckets for hmap.buckets.
// oldoverflow contains overflow buckets for hmap.oldbuckets.
// The indirection allows to store a pointer to the slice in hiter.
overflow *[]*bmap
oldoverflow *[]*bmap
// nextOverflow holds a pointer to a free overflow bucket.
nextOverflow *bmap
}
// A bucket for a Go map.
type bmap struct {
// tophash generally contains the top byte of the hash value
// for each key in this bucket. If tophash[0] < minTopHash,
// tophash[0] is a bucket evacuation state instead.
tophash [bucketCnt]uint8
// Followed by bucketCnt keys and then bucketCnt values.
// NOTE: packing all the keys together and then all the values together makes the
// code a bit more complicated than alternating key/value/key/value/... but it allows
// us to eliminate padding which would be needed for, e.g., map[int64]int8.
// Followed by an overflow pointer.
}
一个map主要是由三个结构构成:
- hmap --- map的最外层的数据结构，包括了map的各种基础信息、如大小、bucket。
- mapextra --- 记录map的额外信息，例如overflow bucket。
- bmap --- 代表bucket，每一个bucket最多放8个kv，最后由一个overflow字段指向下一个bmap，注意key、value、overflow字段都不显示定义，而是通过maptype计算偏移获取的。
hmap.001.png
其中hmap.extra.nextOverflow指向的是预分配的overflow bucket，预分配的用完了那么值就变成nil。
hmap.noverflow是overflow bucket的数量，当B小于16时是准确值，大于等于16时是大概的值。
hmap.count是当前map的元素个数，也就是len()返回的值。
2.1 设计原理
介绍完结构，我们就细说一下这么设计的原因。
2.1.1 bmap细节
在golang map中出现冲突时，不是每一个key都申请一个结构通过链表串起来， 而是以bmap为最小粒度挂载，一个bmap可以放8个kv。这样减少对象数量，减轻管理内存的负担，利于gc。
如果插入时，bmap中key超过8，那么就会申请一个新的bmap（overflow bucket）挂在这个bmap的后面形成链表， 优先用预分配的overflow bucket，如果预分配的用完了，那么就malloc一个挂上去。注意golang的map不会shrink，内存只会越用越多，overflow bucket中的key全删了也不会释放
hash值的高8位存储在bucket中的tophash字段。每个桶最多放8个kv对，所以tophash类型是数组[8]uint8。 把高八位存储起来，这样不用完整比较key就能过滤掉不符合的key，加快查询速度。实际上当hash值的高八位小于常量minTopHash时，会加上minTopHash，区间[0, minTophash)的值用于特殊标记。 查找key时，计算hash值，用hash值的高八位在tophash中查找，有tophash相等的，再去比较key值是否相同。
????? 这里我不太清楚，1.为啥小于minTopHash才加 2.为什么不是位运算而用加。 刚好top在[0,minHash)，或着加上minHash之后溢出到这个区间，岂不是可能误判？
// tophash calculates the tophash value for hash.
func tophash(hash uintptr) uint8 {
top := uint8(hash >> (sys.PtrSize*8 - 8))
if top < minTopHash {
top += minTopHash
}
return top
}
bmap中所有key存在一块，所有value存在一块，这样做方便内存对齐。
当key大于128字节时，bucket的key字段存储的会是指针，指向key的实际内容；value也是一样。
我们还知道golang中没有范型，为了支持map的范型，golang定义了一个maptype类型，定义了这类key用什么hash函数、bucket的大小、怎么比较之类的，通过这个变量来实现范型。
2.1.2 扩容设计
bcuket挂接的链表越来越长，性能会退化，那么就要进行扩容，扩大bucket的数量。
当元素个数/bucket个数大于等于6.5时，就会进行扩容，把bucket数量扩成原本的两倍，当hash表扩容之后，需要将那些老数据迁移到新table上(源代码中称之为evacuate)， 数据搬迁不是一次性完成，而是逐步的完成（在insert和remove时进行搬移），这样就分摊了扩容的耗时。同时为了避免有个bucket一直访问不到导致扩容无法完成，还会进行一个顺序扩容，每次因为写操作搬迁对应bucket后，还会按顺序搬迁未搬迁的bucket，所以最差情况下n次写操作，就保证搬迁完大小为n的map。
扩容会建立一个大小是原来2倍的新的表，将旧的bucket搬到新的表中之后，并不会将旧的bucket从oldbucket中删除，而是加上一个已删除的标记。
只有当所有的bucket都从旧表移到新表之后，才会将oldbucket释放掉。 如果扩容过程中，阈值又超了呢？如果正在扩容，那么不会再进行扩容。
总体思路描述完，就看源码创建、查询、赋值、删除的具体实现。
3. 源码实现
3.1 创建
// makemap implements Go map creation for make(map[k]v, hint).
// If the compiler has determined that the map or the first bucket
// can be created on the stack, h and/or bucket may be non-nil.
// If h != nil, the map can be created directly in h.
// If h.buckets != nil, bucket pointed to can be used as the first bucket.
func makemap(t *maptype, hint int, h *hmap) *hmap {
if hint < 0 || hint > int(maxSliceCap(t.bucket.size)) {
hint = 0
}
// initialize Hmap
if h == nil {
h = new(hmap)
}
h.hash0 = fastrand()
// find size parameter which will hold the requested # of elements
B := uint8(0)
for overLoadFactor(hint, B) {
B++
}
h.B = B
// allocate initial hash table
// if B == 0, the buckets field is allocated lazily later (in mapassign)
// If hint is large zeroing this memory could take a while.
if h.B != 0 {
var nextOverflow *bmap
h.buckets, nextOverflow = makeBucketArray(t, h.B, nil)
if nextOverflow != nil {
h.extra = new(mapextra)
h.extra.nextOverflow = nextOverflow
}
}
return h
}
hint是一个启发值，启发初建map时创建多少个bucket，如果hint是0那么就先不分配bucket，lazy分配。大概流程就是设置一下hash seed、bucket数量、实际申请bucket之类的，流程很简单。
然后我们在看下申请bucket实际干了啥：
// makeBucketArray initializes a backing array for map buckets.
// 1<<b is the minimum number of buckets to allocate.
// dirtyalloc should either be nil or a bucket array previously
// allocated by makeBucketArray with the same t and b parameters.
// If dirtyalloc is nil a new backing array will be alloced and
// otherwise dirtyalloc will be cleared and reused as backing array.
func makeBucketArray(t *maptype, b uint8, dirtyalloc unsafe.Pointer) (buckets unsafe.Pointer, nextOverflow *bmap) {
base := bucketShift(b)
nbuckets := base
// For small b, overflow buckets are unlikely.
// Avoid the overhead of the calculation.
if b >= 4 {
// Add on the estimated number of overflow buckets
// required to insert the median number of elements
// used with this value of b.
nbuckets += bucketShift(b - 4)
sz := t.bucket.size * nbuckets
up := roundupsize(sz)
if up != sz {
nbuckets = up / t.bucket.size
}
}
if dirtyalloc == nil {
buckets = newarray(t.bucket, int(nbuckets))
} else {
// dirtyalloc was previously generated by
// the above newarray(t.bucket, int(nbuckets))
// but may not be empty.
buckets = dirtyalloc
size := t.bucket.size * nbuckets
if t.bucket.kind&kindNoPointers == 0 {
memclrHasPointers(buckets, size)
} else {
memclrNoHeapPointers(buckets, size)
}
}
if base != nbuckets {
// We preallocated some overflow buckets.
// To keep the overhead of tracking these overflow buckets to a minimum,
// we use the convention that if a preallocated overflow bucket's overflow
// pointer is nil, then there are more available by bumping the pointer.
// We need a safe non-nil pointer for the last overflow bucket; just use buckets.
nextOverflow = (*bmap)(add(buckets, base*uintptr(t.bucketsize)))
last := (*bmap)(add(buckets, (nbuckets-1)*uintptr(t.bucketsize)))
last.setoverflow(t, (*bmap)(buckets))
}
return buckets, nextOverflow
}
默认创建2 b 个bucket，如果 b大于等于4，那么就预先额外创建一些overflow bucket。除了最后一个overflow bucket，其余overflow bucket的overflow指针都是nil，最后一个overflow bucket的overflow指针指向bucket数组第一个元素，作为哨兵，说明到了到结尾了.
创建简单流程
3.2 查询
// mapaccess1 returns a pointer to h[key]. Never returns nil, instead
// it will return a reference to the zero object for the value type if
// the key is not in the map.
// NOTE: The returned pointer may keep the whole map live, so don't
// hold onto it for very long.
func mapaccess1(t *maptype, h *hmap, key unsafe.Pointer) unsafe.Pointer {
if raceenabled && h != nil {
callerpc := getcallerpc()
pc := funcPC(mapaccess1)
racereadpc(unsafe.Pointer(h), callerpc, pc)
raceReadObjectPC(t.key, key, callerpc, pc)
}
if msanenabled && h != nil {
msanread(key, t.key.size)
}
if h == nil || h.count == 0 {
return unsafe.Pointer(&zeroVal[0])
}
if h.flags&hashWriting != 0 {
throw("concurrent map read and map write")
}
alg := t.key.alg
hash := alg.hash(key, uintptr(h.hash0))
m := bucketMask(h.B)
b := (*bmap)(add(h.buckets, (hash&m)*uintptr(t.bucketsize)))
if c := h.oldbuckets; c != nil {
if !h.sameSizeGrow() {
// There used to be half as many buckets; mask down one more power of two.
m >>= 1
}
oldb := (*bmap)(add(c, (hash&m)*uintptr(t.bucketsize)))
if !evacuated(oldb) {
b = oldb
}
}
top := tophash(hash)
for ; b != nil; b = b.overflow(t) {
for i := uintptr(0); i < bucketCnt; i++ {
if b.tophash[i] != top {
continue
}
k := add(unsafe.Pointer(b), dataOffset+i*uintptr(t.keysize))
if t.indirectkey {
k = *((*unsafe.Pointer)(k))
}
if alg.equal(key, k) {
v := add(unsafe.Pointer(b), dataOffset+bucketCnt*uintptr(t.keysize)+i*uintptr(t.valuesize))
if t.indirectvalue {
v = *((*unsafe.Pointer)(v))
}
return v
}
}
}
return unsafe.Pointer(&zeroVal[0])
}
-
先定位出bucket，如果正在扩容，并且这个bucket还没搬到新的hash表中，那么就从老的hash表中查找。
-
在bucket中进行顺序查找，使用高八位进行快速过滤，高八位相等，再比较key是否相等，找到就返回value。如果当前bucket找不到，就往下找overflow bucket，都没有就返回零值。
这里我们可以看到， 访问的时候，并不进行扩容的数据搬迁。并且并发有写操作时抛异常 。
这里要注意的是，t.bucketsize并不是bmap的size，而是bmap加上存储key、value、overflow指针，所以查找bucket的时候时候用的不是bmap的szie。
查询简单流程
3.3 赋值
// Like mapaccess, but allocates a slot for the key if it is not present in the map.
func mapassign(t *maptype, h *hmap, key unsafe.Pointer) unsafe.Pointer {
if h == nil {
panic(plainError("assignment to entry in nil map"))
}
if raceenabled {
callerpc := getcallerpc()
pc := funcPC(mapassign)
racewritepc(unsafe.Pointer(h), callerpc, pc)
raceReadObjectPC(t.key, key, callerpc, pc)
}
if msanenabled {
msanread(key, t.key.size)
}
if h.flags&hashWriting != 0 {
throw("concurrent map writes")
}
alg := t.key.alg
hash := alg.hash(key, uintptr(h.hash0))
// Set hashWriting after calling alg.hash, since alg.hash may panic,
// in which case we have not actually done a write.
h.flags |= hashWriting
if h.buckets == nil {
h.buckets = newobject(t.bucket) // newarray(t.bucket, 1)
}
again:
bucket := hash & bucketMask(h.B)
if h.growing() {
growWork(t, h, bucket)
}
b := (*bmap)(unsafe.Pointer(uintptr(h.buckets) + bucket*uintptr(t.bucketsize)))
top := tophash(hash)
var inserti *uint8
var insertk unsafe.Pointer
var val unsafe.Pointer
for {
for i := uintptr(0); i < bucketCnt; i++ {
if b.tophash[i] != top {
if b.tophash[i] == empty && inserti == nil {
inserti = &b.tophash[i]
insertk = add(unsafe.Pointer(b), dataOffset+i*uintptr(t.keysize))
val = add(unsafe.Pointer(b), dataOffset+bucketCnt*uintptr(t.keysize)+i*uintptr(t.valuesize))
}
continue
}
k := add(unsafe.Pointer(b), dataOffset+i*uintptr(t.keysize))
if t.indirectkey {
k = *((*unsafe.Pointer)(k))
}
if !alg.equal(key, k) {
continue
}
// already have a mapping for key. Update it.
if t.needkeyupdate {
typedmemmove(t.key, k, key)
}
val = add(unsafe.Pointer(b), dataOffset+bucketCnt*uintptr(t.keysize)+i*uintptr(t.valuesize))
goto done
}
ovf := b.overflow(t)
if ovf == nil {
break
}
b = ovf
}
// Did not find mapping for key. Allocate new cell & add entry.
// If we hit the max load factor or we have too many overflow buckets,
// and we're not already in the middle of growing, start growing.
if !h.growing() && (overLoadFactor(h.count+1, h.B) || tooManyOverflowBuckets(h.noverflow, h.B)) {
hashGrow(t, h)
goto again // Growing the table invalidates everything, so try again
}
if inserti == nil {
// all current buckets are full, allocate a new one.
newb := h.newoverflow(t, b)
inserti = &newb.tophash[0]
insertk = add(unsafe.Pointer(newb), dataOffset)
val = add(insertk, bucketCnt*uintptr(t.keysize))
}
// store new key/value at insert position
if t.indirectkey {
kmem := newobject(t.key)
*(*unsafe.Pointer)(insertk) = kmem
insertk = kmem
}
if t.indirectvalue {
vmem := newobject(t.elem)
*(*unsafe.Pointer)(val) = vmem
}
typedmemmove(t.key, insertk, key)
*inserti = top
h.count++
done:
if h.flags&hashWriting == 0 {
throw("concurrent map writes")
}
h.flags &^= hashWriting
if t.indirectvalue {
val = *((*unsafe.Pointer)(val))
}
return val
}
-
hash表如果正在扩容，并且这次要操作的bucket还没搬到新hash表中，那么先进行搬迁（扩容细节下面细说）。
-
在buck中寻找key，同时记录下第一个空位置，如果找不到，那么就在空位置中插入数据；如果找到了，那么就更新对应的value；
-
找不到key就看下需不需要扩容，需要扩容并且没有正在扩容，那么就进行扩容，然后回到第一步。
-
找不到key，不需要扩容，但是没有空slot，那么就分配一个overflow bucket挂在链表结尾，用新bucket的第一个slot放存放数据。
3.4 删除
func mapdelete(t *maptype, h *hmap, key unsafe.Pointer) {
if raceenabled && h != nil {
callerpc := getcallerpc()
pc := funcPC(mapdelete)
racewritepc(unsafe.Pointer(h), callerpc, pc)
raceReadObjectPC(t.key, key, callerpc, pc)
}
if msanenabled && h != nil {
msanread(key, t.key.size)
}
if h == nil || h.count == 0 {
return
}
if h.flags&hashWriting != 0 {
throw("concurrent map writes")
}
alg := t.key.alg
hash := alg.hash(key, uintptr(h.hash0))
// Set hashWriting after calling alg.hash, since alg.hash may panic,
// in which case we have not actually done a write (delete).
h.flags |= hashWriting
bucket := hash & bucketMask(h.B)
if h.growing() {
growWork(t, h, bucket)
}
b := (*bmap)(add(h.buckets, bucket*uintptr(t.bucketsize)))
top := tophash(hash)
search:
for ; b != nil; b = b.overflow(t) {
for i := uintptr(0); i < bucketCnt; i++ {
if b.tophash[i] != top {
continue
}
k := add(unsafe.Pointer(b), dataOffset+i*uintptr(t.keysize))
k2 := k
if t.indirectkey {
k2 = *((*unsafe.Pointer)(k2))
}
if !alg.equal(key, k2) {
continue
}
// Only clear key if there are pointers in it.
if t.indirectkey {
*(*unsafe.Pointer)(k) = nil
} else if t.key.kind&kindNoPointers == 0 {
memclrHasPointers(k, t.key.size)
}
v := add(unsafe.Pointer(b), dataOffset+bucketCnt*uintptr(t.keysize)+i*uintptr(t.valuesize))
if t.indirectvalue {
*(*unsafe.Pointer)(v) = nil
} else if t.elem.kind&kindNoPointers == 0 {
memclrHasPointers(v, t.elem.size)
} else {
memclrNoHeapPointers(v, t.elem.size)
}
b.tophash[i] = empty
h.count--
break search
}
}
if h.flags&hashWriting == 0 {
throw("concurrent map writes")
}
h.flags &^= hashWriting
}
-
如果正在扩容，并且操作的bucket还没搬迁完，那么搬迁bucket。
-
找出对应的key，如果key、value是包含指针的那么会清理指针指向的内存，否则不会回收内存。
3.5 扩容
首先通过赋值、删除流程，我们可以知道， 触发扩容的是赋值、删除操作 ，具体判断要不要扩容的代码片段如下：
// overLoadFactor reports whether count items placed in 1<<B buckets is over loadFactor.
func overLoadFactor(count int, B uint8) bool {
return count > bucketCnt && uintptr(count) > loadFactorNum*(bucketShift(B)/loadFactorDen)
}
// tooManyOverflowBuckets reports whether noverflow buckets is too many for a map with 1<<B buckets.
// Note that most of these overflow buckets must be in sparse use;
// if use was dense, then we'd have already triggered regular map growth.
func tooManyOverflowBuckets(noverflow uint16, B uint8) bool {
// If the threshold is too low, we do extraneous work.
// If the threshold is too high, maps that grow and shrink can hold on to lots of unused memory.
// "too many" means (approximately) as many overflow buckets as regular buckets.
// See incrnoverflow for more details.
if B > 15 {
B = 15
}
// The compiler doesn't see here that B < 16; mask B to generate shorter shift code.
return noverflow >= uint16(1)<<(B&15)
}
{
....
// If we hit the max load factor or we have too many overflow buckets,
// and we're not already in the middle of growing, start growing.
if !h.growing() && (overLoadFactor(h.count+1, h.B) || tooManyOverflowBuckets(h.noverflow, h.B)) {
hashGrow(t, h)
goto again // Growing the table invalidates everything, so try again
}
....
}
翻译一下代码，意思就是：
func overLoadFactor(countint, Buint8) bool {
// return count>bucketCnt&&uintptr(count) >loadFactorNum*(bucketShift(B)/loadFactorDen)
return 元素个数>8 && count>bucket数量*6.5
其中loadFactorNum是常量13，loadFactorDen是常量2,所以是6.5
bucket数量不算overflow bucket
}
func tooManyOverflowBuckets(noverflowuint16, Buint8) bool{
if B > 15 {
B=15
}
// The compiler doesn't see here that B < 16; mask B to generate shorter shift code.
return noverflow>=uint16(1)<<(B&15)
}
if (不是正在扩容 && (元素个数/bucket数超过某个值 || 太多overflow bucket)) {
进行扩容
}
判断完扩容后，如果需要扩容，那么第一步需要做的，就是对hash表进行扩容：
//仅对hash表进行扩容，这里不进行搬迁
func hashGrow(t *maptype, h *hmap) {
// If we've hit the load factor, get bigger.
// Otherwise, there are too many overflow buckets,
// so keep the same number of buckets and "grow" laterally.
bigger := uint8(1)
if !overLoadFactor(h.count+1, h.B) {
bigger = 0
h.flags |= sameSizeGrow
}
oldbuckets := h.buckets
newbuckets, nextOverflow := makeBucketArray(t, h.B+bigger, nil)
flags := h.flags &^ (iterator | oldIterator)
if h.flags&iterator != 0 {
flags |= oldIterator
}
// commit the grow (atomic wrt gc)
h.B += bigger
h.flags = flags
h.oldbuckets = oldbuckets
h.buckets = newbuckets
h.nevacuate = 0
h.noverflow = 0
if h.extra != nil && h.extra.overflow != nil {
// Promote current overflow buckets to the old generation.
if h.extra.oldoverflow != nil {
throw("oldoverflow is not nil")
}
h.extra.oldoverflow = h.extra.overflow
h.extra.overflow = nil
}
if nextOverflow != nil {
if h.extra == nil {
h.extra = new(mapextra)
}
h.extra.nextOverflow = nextOverflow
}
// the actual copying of the hash table data is done incrementally
// by growWork() and evacuate().
}
扩容的函数hashGrow其实仅仅是进行一些空间分配，字段的初始化，实际的搬迁操作是在growWork函数中
func growWork(t *maptype, h *hmap, bucket uintptr) {
// make sure we evacuate the oldbucket corresponding
// to the bucket we're about to use
evacuate(t, h, bucket&h.oldbucketmask())
// evacuate one more oldbucket to make progress on growing
if h.growing() {
evacuate(t, h, h.nevacuate)
}
}
evacuate是进行具体搬迁某个bucket的函数，可以看出 growWork会搬迁两个bucket，一个是入参bucket；另一个是h.nevacuate。这个nevacuate是一个顺序累加的值 。可以想想如果每次仅仅搬迁进行写操作（赋值/删除）的bucket，那么有可能某些bucket就是一直没有机会访问到，那么扩容就一直没法完成，总是在扩容中的状态，因此会额外进行一次顺序迁移，理论上，有N个old bucket，最多N次写操作，那么必定会搬迁完。
然后我们再看下evacuate具体的实现
func evacuate(t *maptype, h *hmap, oldbucket uintptr) {
b := (*bmap)(add(h.oldbuckets, oldbucket*uintptr(t.bucketsize)))
newbit := h.noldbuckets()
if !evacuated(b) {
// TODO: reuse overflow buckets instead of using new ones, if there
// is no iterator using the old buckets. (If !oldIterator.)
// xy contains the x and y (low and high) evacuation destinations.
var xy [2]evacDst
x := &xy[0]
x.b = (*bmap)(add(h.buckets, oldbucket*uintptr(t.bucketsize)))
x.k = add(unsafe.Pointer(x.b), dataOffset)
x.v = add(x.k, bucketCnt*uintptr(t.keysize))
if !h.sameSizeGrow() {
// Only calculate y pointers if we're growing bigger.
// Otherwise GC can see bad pointers.
y := &xy[1]
y.b = (*bmap)(add(h.buckets, (oldbucket+newbit)*uintptr(t.bucketsize)))
y.k = add(unsafe.Pointer(y.b), dataOffset)
y.v = add(y.k, bucketCnt*uintptr(t.keysize))
}
for ; b != nil; b = b.overflow(t) {
k := add(unsafe.Pointer(b), dataOffset)
v := add(k, bucketCnt*uintptr(t.keysize))
for i := 0; i < bucketCnt; i, k, v = i+1, add(k, uintptr(t.keysize)), add(v, uintptr(t.valuesize)) {
top := b.tophash[I]
if top == empty {
b.tophash[i] = evacuatedEmpty
continue
}
if top < minTopHash {
throw("bad map state")
}
k2 := k
if t.indirectkey {
k2 = *((*unsafe.Pointer)(k2))
}
var useY uint8
if !h.sameSizeGrow() {
// Compute hash to make our evacuation decision (whether we need
// to send this key/value to bucket x or bucket y).
hash := t.key.alg.hash(k2, uintptr(h.hash0))
if h.flags&iterator != 0 && !t.reflexivekey && !t.key.alg.equal(k2, k2) {
// If key != key (NaNs), then the hash could be (and probably
// will be) entirely different from the old hash. Moreover,
// it isn't reproducible. Reproducibility is required in the
// presence of iterators, as our evacuation decision must
// match whatever decision the iterator made.
// Fortunately, we have the freedom to send these keys either
// way. Also, tophash is meaningless for these kinds of keys.
// We let the low bit of tophash drive the evacuation decision.
// We recompute a new random tophash for the next level so
// these keys will get evenly distributed across all buckets
// after multiple grows.
useY = top & 1
top = tophash(hash)
} else {
if hash&newbit != 0 {
useY = 1
}
}
}
if evacuatedX+1 != evacuatedY {
throw("bad evacuatedN")
}
b.tophash[i] = evacuatedX + useY // evacuatedX + 1 == evacuatedY
dst := &xy[useY] // evacuation destination
if dst.i == bucketCnt {
dst.b = h.newoverflow(t, dst.b)
dst.i = 0
dst.k = add(unsafe.Pointer(dst.b), dataOffset)
dst.v = add(dst.k, bucketCnt*uintptr(t.keysize))
}
dst.b.tophash[dst.i&(bucketCnt-1)] = top // mask dst.i as an optimization, to avoid a bounds check
if t.indirectkey {
*(*unsafe.Pointer)(dst.k) = k2 // copy pointer
} else {
typedmemmove(t.key, dst.k, k) // copy value
}
if t.indirectvalue {
*(*unsafe.Pointer)(dst.v) = *(*unsafe.Pointer)(v)
} else {
typedmemmove(t.elem, dst.v, v)
}
dst.i++
// These updates might push these pointers past the end of the
// key or value arrays. That's ok, as we have the overflow pointer
// at the end of the bucket to protect against pointing past the
// end of the bucket.
dst.k = add(dst.k, uintptr(t.keysize))
dst.v = add(dst.v, uintptr(t.valuesize))
}
}
// Unlink the overflow buckets & clear key/value to help GC.
if h.flags&oldIterator == 0 && t.bucket.kind&kindNoPointers == 0 {
b := add(h.oldbuckets, oldbucket*uintptr(t.bucketsize))
// Preserve b.tophash because the evacuation
// state is maintained there.
ptr := add(b, dataOffset)
n := uintptr(t.bucketsize) - dataOffset
memclrHasPointers(ptr, n)
}
}
if oldbucket == h.nevacuate {
advanceEvacuationMark(h, t, newbit)
}
}
在advanceEvacuationMark中进行nevacuate的累加，遇到已经迁移的bucket会继续累加，一次最多加1024。
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