關鍵性數據結構
- hmap: map 的 header結構
- bmap: map 的 bucket結構
- mapextra: map 的 拓展結構 不是每一個map都包含
golang map 是用 hash map實現的，首先，我們先看 hash map是怎麼實現的；然後我們再看 golang map 是怎麼基於 hash map 封裝的 map 類型。
Bucket
// 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 elems.
// NOTE: packing all the keys together and then all the elems together makes the
// code a bit more complicated than alternating key/elem/key/elem/... but it allows
// us to eliminate padding which would be needed for, e.g., map[int64]int8.
// Followed by an overflow pointer.
}
這裏面的 bucketCnt 是一個常量：
const (
// Maximum number of key/elem pairs a bucket can hold.
bucketCntBits = 3
bucketCnt = 1 << bucketCntBits
... ...
)
這個topHash 就是指hash值的前8位, 每一個桶是大小爲 8*8 大小。此外topHash 的元素還可能有以下狀態：
const(
emptyRest = 0 // 此位置未被佔用，且後面的位置也沒被佔用
emptyOne = 1 // 此位置未被佔用，在delete後會設置此標誌
evacuatedX = 2 // 此位置已經被佔用，對應數據已經被遷移到新的buckets的first半區
evacuatedY = 3 // 此位置已經被佔用，對應數據已經被遷移到新的buckets的second半區
evacuatedEmpty = 4 // 此位置未被佔用，但bucket已經被遷移
minTopHash = 5 // topHash的最小值，爲了與前面4個值進行區分
)
Hmap
// 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 // overflow buckets 的計數器;詳細信息看 incrnoverflow
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), 已遷移的bucket的計數器
extra *mapextra // optional fields，記錄next_overflow的地址,和已經分配的overflow
}
const (
// flags
iterator = 1 // there may be an iterator using buckets，有一個iter在使用bucket
oldIterator = 2 // there may be an iterator using oldbuckets，有一個iter在使用old_bucket
hashWriting = 4 // a goroutine is writing to the map，有一個協程在寫入
sameSizeGrow = 8 // the current map growth is to a new map of the same size，新bucekt與old_bucket size相同.
)
// mapextra holds fields that are not present on all maps.
type mapextra struct {
// If both key and elem 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 elem 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
}
map解析
make map
在golang中可以通過 make(map[key]value, hint) 創建一個map實例，在runtime包中是通過如下函數實現的：
// 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 {
mem, overflow := math.MulUintptr(uintptr(hint), t.bucket.size)
if overflow || mem > maxAlloc {
hint = 0
}
// 1. 初始化hmap
if h == nil {
h = new(hmap)
}
h.hash0 = fastrand() // 取一個隨機值作爲hash seed
// Find the size parameter B which will hold the requested # of elements.
// For hint < 0 overLoadFactor returns false since hint < bucketCnt.
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.
// 分配空間
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 // 保存overflow區域的首地址
}
}
}
// 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) {
// 1. 計算要使用的空間大小
base := bucketShift(b) // base = 2^(b'), b' 32位系統時爲b的低5位,最大值爲31，64系統時爲b的低6位，最大值爲63
nbuckets := base
// For small b, overflow buckets are unlikely.
// Avoid the overhead of the calculation.
// 預分配內存
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) // nbuckets = base + 2^((b-4)')
sz := t.bucket.size * nbuckets
up := roundupsize(sz) // 內存頁對齊，golang 中pagesize 爲8kB
if up != sz {
nbuckets = up / t.bucket.size
}
}
// 2. 空間重複利用
if dirtyalloc == nil {
// 沒有髒空間，則新分配一個數組
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.ptrdata != 0 {
memclrHasPointers(buckets, size)
} else {
memclrNoHeapPointers(buckets, size)
}
}
// 3. 計算overflow 指針
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)) // 將overflow最後一個bucket的末尾位置存儲 buckets指針
}
return buckets, nextOverflow
}
// overLoadFactor reports whether count items placed in 1<<B buckets is over loadFactor.
// 檢查當前數量是不是超過負載係數
func overLoadFactor(count int, B uint8) bool {
return count > bucketCnt && uintptr(count) > loadFactorNum*(bucketShift(B)/loadFactorDen)
}
返回值類型是 hmap*, hmap.B 要符合以下規則
loadFactorNum = 13
loadFactorDen = 2
count = hint
uintptr(count) > loadFactorNum*(bucketShift(B)/loadFactorDen)
hmap.buckets 分配的內存: roudupsize(base=bucketsize*2^B + overflow=bucketsize*/2^B/4)，bucketsize = bitmap_size + 8*(key_size) + 8*(elem_size) + ptr_size，在32和64位系統中ptr_size不同，所以bucket_size會也不同。
從上面的代碼可以看出，在物理內存中是以一個數組來存儲hmap table,內存分佈大致如下：
# bucket
|bmap|key1~8|elem1~8|
#bucekts
|bucket1~N|overflow|
# overflow
|nextOverflow|...|last|
# last
|bmap|...|ptr_buckets|
bucket1~N 是base區域， overflow 是預留區域，降低內存重複分配的次數。
insert into map
// 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 {
... ...
// 1. 生成key的hash
alg := t.key.alg
hash := alg.hash(key, uintptr(h.hash0))
if h.buckets == nil {
h.buckets = newobject(t.bucket) // newarray(t.bucket, 1)
}
again:
// 2. 分桶
// 取出hash的低 B 位，作爲 bucket的編號
bucket := hash & bucketMask(h.B)
if h.growing() {
growWork(t, h, bucket)
}
// 找到桶對應的數組首地址
b := (*bmap)(unsafe.Pointer(uintptr(h.buckets) + bucket*uintptr(t.bucketsize)))
// hash 的高 8 位, 如果小於5 則 +5 返回
top := tophash(hash)
var inserti *uint8 // bmap中寫入 tophash 的地址
var insertk unsafe.Pointer // 寫入key的地址
var elem unsafe.Pointer // 寫入值的地址
bucketloop:
for {
for i := uintptr(0); i < bucketCnt; i++ { // 查找可使用的位置
if b.tophash[i] != top {
if isEmpty(b.tophash[i]) && inserti == nil {
inserti = &b.tophash[i]
insertk = add(unsafe.Pointer(b), dataOffset+i*uintptr(t.keysize))
elem = add(unsafe.Pointer(b), dataOffset+bucketCnt*uintptr(t.keysize)+i*uintptr(t.elemsize))
}
if b.tophash[i] == emptyRest { // 找到可使用的位置，跳出循環
break bucketloop
}
continue
}
// 找到相同topHash,比較key
k := add(unsafe.Pointer(b), dataOffset+i*uintptr(t.keysize))
if t.indirectkey() {
k = *((*unsafe.Pointer)(k))
}
if !alg.equal(key, k) { // key不相同繼續查找
continue
}
// already have a mapping for key. Update it.
// key相同，找到elem位置，跳轉到done
if t.needkeyupdate() {
typedmemmove(t.key, k, key)
}
elem = add(unsafe.Pointer(b), dataOffset+bucketCnt*uintptr(t.keysize)+i*uintptr(t.elemsize))
goto done
}
// 在N bucket 中沒有找到，查找overflow
ovf := b.overflow(t)
if ovf == nil {
break // overflow 爲空跳出循環
}
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.
// 是否觸發內存增長
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 { // 沒有找到可以插入的位置，新分配一個overflow
// all current buckets are full, allocate a new one.
newb := h.newoverflow(t, b)
inserti = &newb.tophash[0]
insertk = add(unsafe.Pointer(newb), dataOffset)
elem = add(insertk, bucketCnt*uintptr(t.keysize))
}
// store new key/elem at insert position
if t.indirectkey() {
kmem := newobject(t.key)
*(*unsafe.Pointer)(insertk) = kmem
insertk = kmem
}
if t.indirectelem() {
vmem := newobject(t.elem)
*(*unsafe.Pointer)(elem) = 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.indirectelem() {
elem = *((*unsafe.Pointer)(elem))
}
return elem
}
從上面代碼解析我們能清楚一個寫入的過程：
- 對key做hash，取hash的低B位，確定bucket的編號 N;
- 遍歷 N bucket中每一個 位置 ，找到沒有寫入的 位置，寫入topHash即hash的高8位；
- 如果 N bucket 中有相同的 topHash，則需要去出對應的key做比較，如果相同則修改elem，如果不同則繼續向後遍歷，尋找空閒的 位置 寫入，以此來解決衝突的問題。
那如果出現N bucket 滿了怎麼辦？雖然這種概率很低但是也難免會遇到，畢竟一個bmap中只能裝的下8個key，這裏就要用到我們剛纔說的預分配內存 overflow，分配overflow的邏輯如下：
func (h *hmap) newoverflow(t *maptype, b *bmap) *bmap {
var ovf *bmap
if h.extra != nil && h.extra.nextOverflow != nil { // 存在overflow區域
// We have preallocated overflow buckets available.
// See makeBucketArray for more details.
ovf = h.extra.nextOverflow
if ovf.overflow(t) == nil { // ovf不是最後一塊，將nextOverflow 指針向後移動
// We're not at the end of the preallocated overflow buckets. Bump the pointer.
h.extra.nextOverflow = (*bmap)(add(unsafe.Pointer(ovf), uintptr(t.bucketsize)))
} else { // 最後一塊 overflow，將nextOverflow 置空
// This is the last preallocated overflow bucket.
// Reset the overflow pointer on this bucket,
// which was set to a non-nil sentinel value.
ovf.setoverflow(t, nil) // 抹平末尾指針
h.extra.nextOverflow = nil
}
} else {// overflow 用光，新建一塊
ovf = (*bmap)(newobject(t.bucket))
}
h.incrnoverflow() // 增加overflow 使用計數
if t.bucket.ptrdata == 0 {
h.createOverflow()
*h.extra.overflow = append(*h.extra.overflow, ovf)
}
b.setoverflow(t, ovf) // 將overflow的地址加到末尾
return ovf
}
通過上面的函數我們可以明白，當N bucket被用光後如何擴充，即從預留區域查找一塊未使用的區域，將該區域的指針放在 N bucket 的末尾，作爲 N bucket的擴充區域：
# N bucket 擴充
|bmap|key1~8|elem1~8|ptr_extra_bucket|......|extra_bucket|
^---------------^
然後我們就可以愉快的將新的key-value放到extra_bucket 中，我們在結合上面一節的內容可以更加清晰的明白hashmap table在內存裏的構造，即連續數組+跳轉指針，這也是爲什麼訪問能如此快速的原因，基本上都是在連續內存上指針位移操作。
看到這裏我們對map的實現應該有一個粗淺的認識，但是這只是一部分。我們可以注意到hmap中還有一個oldbuckets,還有其他的成員變量的含義沒有被解開。
在上面的段落中我們給出了bucket溢出的解決方案，但如果bucket溢出過多怎麼辦（即單桶數據過多）？？我們可以假設一個極端情況 一個map中除了 bucket1，剩餘的都是它的extra_bucket，當我在 bucket1中取查詢的時候要遍歷很長，最差的情況要遍歷整個map。怎樣去解決這個問題？
- 對key進行排序，使key變成有序的。可以使用插入排序，實現簡單但是內存位移的操作較多。使用二分查找時間複雜度可以優化到log（n）。當然也可以使用其它排序算法，來減少內存位移發生的機率，但是因爲底層存儲是使用的數組，內存位移難以避免。
- 重建，通過重新劃分桶，來解決單桶數據過多問題。
上面代碼中不難展示出在golang中使用的是第二種方案，那麼什麼情況下會觸發重建？如何重建？
map Grow
tooManyOverflowBuckets 和 overLoadFactor 這兩個函數會去判斷是否需要執行 Grow 流程：
func mapassign(t *maptype, h *hmap, key unsafe.Pointer) unsafe.Pointer {
... ...
// 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
}
... ...
}
const(
loadFactorNum = 13
loadFactorDen = 2
)
// 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)
}
首先是overLoadFactor，當 key_count + 1 > 8 and key_count+1 > 13*2^(B-1) 時會觸發 Grow 流程。其次是 tooManyOverflowBuckets，當noverflow >= 2^(B&15)，也就是說overflow bucket的數量大於2^(B&15)就會觸發Grow，這裏分兩種情況：
- B < 16，此時當
overflow bucket和base bucket相等的時候就會觸發增長流程。 - B >= 16 時，此時當
overflow bucket數量 > 2^15時就會觸發增長流程。
這裏需要注意的是，noverflow 並不是一個準確的計數，當數量過大的時候它只能顯示一個近似的數量：
// incrnoverflow increments h.noverflow.
// noverflow counts the number of overflow buckets.
// This is used to trigger same-size map growth.
// See also tooManyOverflowBuckets.
// To keep hmap small, noverflow is a uint16.
// When there are few buckets, noverflow is an exact count.
// When there are many buckets, noverflow is an approximate count.
func (h *hmap) incrnoverflow() {
// We trigger same-size map growth if there are
// as many overflow buckets as buckets.
// We need to be able to count to 1<<h.B.
if h.B < 16 {
h.noverflow++
return
}
// Increment with probability 1/(1<<(h.B-15)).
// When we reach 1<<15 - 1, we will have approximately
// as many overflow buckets as buckets.
mask := uint32(1)<<(h.B-15) - 1
// Example: if h.B == 18, then mask == 7,
// and fastrand & 7 == 0 with probability 1/8.
if fastrand()&mask == 0 {
h.noverflow++
}
}
可以很明顯的看出，當h.B>=16時候並不是每次都會累加，此時base bucket的數量至少爲2^16，可以存儲2^19=524288條數據，外加預分配的overflow 。桶的數量越多，分佈的越離散，出現overflow的概率更低，即使出現overflow單桶過長的概率也會降低。
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)
// 判斷是否超出loadFactor
if !overLoadFactor(h.count+1, h.B) { // 沒有超過則維持以前的size
bigger = 0
h.flags |= sameSizeGrow
}
oldbuckets := h.buckets
newbuckets, nextOverflow := makeBucketArray(t, h.B+bigger, nil)
// 更新flags
flags := h.flags &^ (iterator | oldIterator)
if h.flags&iterator != 0 {
flags |= oldIterator
}
// commit the grow (atomic wrt gc)
h.B += bigger
h.flags = flags
// 替換buckets
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().
}
根據上面的hashGrow 函數，我們可以看出map內存增長的規則，在overLoadFactor的情況下，h.B = h.B + 1，即base bucket數量翻倍, 否則維持原size不變，創建一個新的 buckets數組。
func mapassign(t *maptype, h *hmap, key unsafe.Pointer) unsafe.Pointer{
... ...
bucket := hash & bucketMask(h.B)
if h.growing() {
growWork(t, h, bucket)
}
... ...
}
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)
}
}
// 遷移數據
func evacuate(t *maptype, h *hmap, oldbucket uintptr) {
b := (*bmap)(add(h.oldbuckets, oldbucket*uintptr(t.bucketsize)))
newbit := h.noldbuckets() //oldbuckets 的 size
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.
// 1. 確定兩個潛在目的遷移地址，X/Y
// X 地址爲 new buckets中編號爲 oldbucket 的bucket
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.e = add(x.k, bucketCnt*uintptr(t.keysize))
if !h.sameSizeGrow() { // size 不變的情況下
// Only calculate y pointers if we're growing bigger.
// Otherwise GC can see bad pointers.
// Y 地址爲 new buckets中編號爲 oldbucket + noldbuckets, 即second 半區
y := &xy[1]
y.b = (*bmap)(add(h.buckets, (oldbucket+newbit)*uintptr(t.bucketsize)))
y.k = add(unsafe.Pointer(y.b), dataOffset)
y.e = add(y.k, bucketCnt*uintptr(t.keysize))
}
// 2. 開始遷移數據，包括bucket overflow部分的數據
for ; b != nil; b = b.overflow(t) {
k := add(unsafe.Pointer(b), dataOffset)
e := add(k, bucketCnt*uintptr(t.keysize))
for i := 0; i < bucketCnt; i, k, e = i+1, add(k, uintptr(t.keysize)), add(e, uintptr(t.elemsize)) {
top := b.tophash[i]
if isEmpty(top) {
b.tophash[i] = evacuatedEmpty
continue
}
if top < minTopHash {
throw("bad map state")
}
k2 := k
if t.indirectkey() {
k2 = *((*unsafe.Pointer)(k2))
}
var useY uint8
// 計算遷移到first 半區還是 second半區
if !h.sameSizeGrow() {
// Compute hash to make our evacuation decision (whether we need
// to send this key/elem 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 || evacuatedX^1 != evacuatedY {
throw("bad evacuatedN")
}
// 修改topHash 爲 evacuatedX 或 evacuatedY, 表示已經被遷移
b.tophash[i] = evacuatedX + useY // evacuatedX + 1 == evacuatedY
dst := &xy[useY] // evacuation destination
// 遷移數據
if dst.i == bucketCnt {
dst.b = h.newoverflow(t, dst.b)
dst.i = 0
dst.k = add(unsafe.Pointer(dst.b), dataOffset)
dst.e = 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 elem
}
if t.indirectelem() {
*(*unsafe.Pointer)(dst.e) = *(*unsafe.Pointer)(e)
} else {
typedmemmove(t.elem, dst.e, e)
}
dst.i++
// These updates might push these pointers past the end of the
// key or elem 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.e = add(dst.e, uintptr(t.elemsize))
}
}
// Unlink the overflow buckets & clear key/elem to help GC.
// 清理數據
if h.flags&oldIterator == 0 && t.bucket.ptrdata != 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 {
// 統計是否完全遷移，如果完全遷移後，oldbuckets 會被釋放掉（設置爲nil）
advanceEvacuationMark(h, t, newbit)
}
}
上面我們提到了在Grow的流程中，新申請的buckets 可能會大小不變即same_size，也可能會變成oldbuckets的兩倍即double_size，當double_size的情況下，會劃分爲兩個半區first 和 second：
# oldbuckets
|bucket1~N|
# newbuckets
|bucket1~N|bucketN+1~2N|
first second
我們以dobule_size爲例，當插入一個新的key觸發Grow操作的時候，整體的執行流程如下:
- 取hash(key)的低 B 位作爲bucket編號 N, bucket_i = bucket_N;判斷當前是否處於增長狀態，如果不處於
Grow_State，執行下一步；否則跳轉到第 6 步； - 如果bucket_i中可以找到插入的位置，則插入結束流程；否則執行下一步；
- 查找bucket_i 是否分配了overflow bucket，如果有分配overflow bucket，則 bucketi = overflwo_bucket，跳轉到第2步；否則執行下一步；
- 判斷是否需要執行
Grow流程，如果需要則執行下一步；否則執行第 7 步； - 創建新的
buckets並擴充到dobule_size，設置爲Grow_State，跳轉到第 1 步； - 將
oldbucket從oldbuckets遷移到newbuckets，oldbucket = N^(2^B-1)，odlbucket中的數據會分佈到first和second兩個半區中（下面會詳細說明）,跳轉到第2步執行； - 爲
bucket_i分配的oveflow_bucketr，插入key，結束流程;
此處對第6步進行一下詳細描述，我們假設 N = 12, B = 3，觸發增長後B = 4, X = N^(2^B - 1) = 4, Y = X + 2^(B-1) = 12，新的key會被插到second半區；如果N = 4則會被插入到first半區，但無論如何都是在oldbucket遷移過來的數據桶中，以此來保證hash的一致性，這就是重構hash map的過程。
在same_size的情況下執行過程是一樣的，因爲bucket總數不變，所以oldbucket對應遷移到new_buckets中相同編號的bucket中即可。
同時有一個細節值得我們注意，在bmap中被設置爲emptyone的表示是已經被刪除的數據，在遷移的過程中跳過即可，這樣遷移後的數據會變的更加緊湊。
# 遷移前
# oldbucket_4
# emptyreset = 0, emptyone = 1
|xx|yy|1|ww|zz|0|0|0|key1~8|elem1~8|
# 遷移後
# newbucket
# newbucket_4
|xx|ww|0|0|0|0|0|0|key_xx|key_ww|...|elem_xx|elme_ww|...|
# newbucket_12
|yy|zz|0|0|0|0|0|0|key_yy|key_zz|...|elem_yy|elme_zz|...|
# oldbucket_4
# evacuatedX = 2, evacuatedY = 3, evacuatedEmpty = 4
|2|3|1|2|3|4|4|4|key1~8|elem1~8|
map access
golang map中訪問一個map中數據有三種方式，我們以map[int]int爲例：
sets := map[int]int{1:2,3:4,5:6}
value := map[1] // 返回值
value, isExist := map[i] // 返回值和是否存在
for key, value := range sets{ // 遍歷
}
前兩種訪問方式相同，都是通過key來訪問，只是返回的值有所不同而已。
func mapaccess2(t *maptype, h *hmap, key unsafe.Pointer) (unsafe.Pointer, bool) {
if raceenabled && h != nil {
callerpc := getcallerpc()
pc := funcPC(mapaccess2)
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 {
if t.hashMightPanic() {
t.key.alg.hash(key, 0) // see issue 23734
}
return unsafe.Pointer(&zeroVal[0]), false
}
if h.flags&hashWriting != 0 {
throw("concurrent map read and map write")
}
// 1. 計算對應的桶編號，獲取桶地址
alg := t.key.alg
hash := alg.hash(key, uintptr(h.hash0))
m := bucketMask(h.B)
b := (*bmap)(unsafe.Pointer(uintptr(h.buckets) + (hash&m)*uintptr(t.bucketsize))) // 取hash值的低B位爲桶編號
if c := h.oldbuckets; c != nil { // 當存在舊桶的時候，數據可能尚未遷移
if !h.sameSizeGrow() { // 當擴充桶爲double_size時, 桶編號要取 hash 值的低 B-1 位
// There used to be half as many buckets; mask down one more power of two.
m >>= 1
}
oldb := (*bmap)(unsafe.Pointer(uintptr(c) + (hash&m)*uintptr(t.bucketsize)))
if !evacuated(oldb) { // 如果沒有遷移則去老的桶取
b = oldb
}
}
top := tophash(hash)
bucketloop:
// 2. 遍歷尋找對應的key
for ; b != nil; b = b.overflow(t) {
for i := uintptr(0); i < bucketCnt; i++ {
if b.tophash[i] != top {
if b.tophash[i] == emptyRest {
break bucketloop
}
continue
}
k := add(unsafe.Pointer(b), dataOffset+i*uintptr(t.keysize))
if t.indirectkey() {
k = *((*unsafe.Pointer)(k))
}
if alg.equal(key, k) {
e := add(unsafe.Pointer(b), dataOffset+bucketCnt*uintptr(t.keysize)+i*uintptr(t.elemsize))
if t.indirectelem() {
e = *((*unsafe.Pointer)(e))
}
return e, true
}
}
}
return unsafe.Pointer(&zeroVal[0]), false
}
整個查詢過程比較簡單，如果你是順序閱讀到這裏的話應該很好理解：
- 取低key_hash的低B位計算桶的編號N，
bucket_i = bucket_N，如果此時有舊桶存在，執行第2步，否則執行第 4 步； - 尋找
oldbucket，判斷Grow 類型，如果是same_size則編號編號仍然爲N，如果是dobule_size則取key_hash低B-1位作爲舊桶編號；執行下一步； - 判斷
odlbucket是否被遷移，如果沒被遷移則bucket_i = old_bucket;執行下一步； - 在
bucket_i和overflow_bucket中尋找與key_topHash相等的tophash,找到後取對應的key做equal比對，如果相等則返回 對應的elem，否則繼續遍歷，到結束爲止。
相比較於按照key訪問，遍歷訪問要更復雜一些：
// mapiterinit initializes the hiter struct used for ranging over maps.
// The hiter struct pointed to by 'it' is allocated on the stack
// by the compilers order pass or on the heap by reflect_mapiterinit.
// Both need to have zeroed hiter since the struct contains pointers.
func mapiterinit(t *maptype, h *hmap, it *hiter) {
if raceenabled && h != nil {
callerpc := getcallerpc()
racereadpc(unsafe.Pointer(h), callerpc, funcPC(mapiterinit))
}
if h == nil || h.count == 0 {
return
}
if unsafe.Sizeof(hiter{})/sys.PtrSize != 12 {
throw("hash_iter size incorrect") // see cmd/compile/internal/gc/reflect.go
}
it.t = t
it.h = h
// grab snapshot of bucket state
// 1. 創建當前map狀態快照
it.B = h.B
it.buckets = h.buckets
if t.bucket.ptrdata == 0 {
// Allocate the current slice and remember pointers to both current and old.
// This preserves all relevant overflow buckets alive even if
// the table grows and/or overflow buckets are added to the table
// while we are iterating.
h.createOverflow()
it.overflow = h.extra.overflow
it.oldoverflow = h.extra.oldoverflow
}
// decide where to start
// 2. 決定起始位置
// 取一個隨機值
r := uintptr(fastrand())
if h.B > 31-bucketCntBits {
r += uintptr(fastrand()) << 31
}
// 選取一個隨機的起始bucket
it.startBucket = r & bucketMask(h.B)
// 選取一個隨機的偏移量
it.offset = uint8(r >> h.B & (bucketCnt - 1))
// iterator state
it.bucket = it.startBucket
// Remember we have an iterator.
// Can run concurrently with another mapiterinit().
// 寫入標誌位
if old := h.flags; old&(iterator|oldIterator) != iterator|oldIterator {
atomic.Or8(&h.flags, iterator|oldIterator)
}
mapiternext(it)
}
func mapiternext(it *hiter) {
h := it.h
if raceenabled {
callerpc := getcallerpc()
racereadpc(unsafe.Pointer(h), callerpc, funcPC(mapiternext))
}
if h.flags&hashWriting != 0 {
throw("concurrent map iteration and map write")
}
t := it.t
bucket := it.bucket
b := it.bptr
i := it.i
checkBucket := it.checkBucket
alg := t.key.alg
next:
if b == nil {
// 判斷是否已經循環遍歷所有的Bucket
if bucket == it.startBucket && it.wrapped {
// end of iteration
it.key = nil
it.elem = nil
return
}
if h.growing() && it.B == h.B {
//map 在grow state 且 iterInit在grow state 或者是 `same_size`
// Iterator was started in the middle of a grow, and the grow isn't done yet.
// If the bucket we're looking at hasn't been filled in yet (i.e. the old
// bucket hasn't been evacuated) then we need to iterate through the old
// bucket and only return the ones that will be migrated to this bucket.
oldbucket := bucket & it.h.oldbucketmask()
b = (*bmap)(add(h.oldbuckets, oldbucket*uintptr(t.bucketsize)))
if !evacuated(b) { // 判斷是否遷移
checkBucket = bucket
} else {
b = (*bmap)(add(it.buckets, bucket*uintptr(t.bucketsize)))
checkBucket = noCheck
}
} else {
// map未處於`grow state`，或`grow state`爲`doubel_size`.
b = (*bmap)(add(it.buckets, bucket*uintptr(t.bucketsize)))
checkBucket = noCheck
}
bucket++
if bucket == bucketShift(it.B) {
bucket = 0
it.wrapped = true
}
i = 0
}
for ; i < bucketCnt; i++ {
offi := (i + it.offset) & (bucketCnt - 1)
// 跳過空閒位置
if isEmpty(b.tophash[offi]) || b.tophash[offi] == evacuatedEmpty {
// TODO: emptyRest is hard to use here, as we start iterating
// in the middle of a bucket. It's feasible, just tricky.
continue
}
k := add(unsafe.Pointer(b), dataOffset+uintptr(offi)*uintptr(t.keysize))
if t.indirectkey() {
k = *((*unsafe.Pointer)(k))
}
e := add(unsafe.Pointer(b), dataOffset+bucketCnt*uintptr(t.keysize)+uintptr(offi)*uintptr(t.elemsize))
if checkBucket != noCheck && !h.sameSizeGrow() {// 過濾不會被遷移過來的數據
// Special case: iterator was started during a grow to a larger size
// and the grow is not done yet. We're working on a bucket whose
// oldbucket has not been evacuated yet. Or at least, it wasn't
// evacuated when we started the bucket. So we're iterating
// through the oldbucket, skipping any keys that will go
// to the other new bucket (each oldbucket expands to two
// buckets during a grow).
if t.reflexivekey() || alg.equal(k, k) {
// If the item in the oldbucket is not destined for
// the current new bucket in the iteration, skip it.
hash := alg.hash(k, uintptr(h.hash0))
if hash&bucketMask(it.B) != checkBucket {
continue
}
} else {
// Hash isn't repeatable if k != k (NaNs). We need a
// repeatable and randomish choice of which direction
// to send NaNs during evacuation. We'll use the low
// bit of tophash to decide which way NaNs go.
// NOTE: this case is why we need two evacuate tophash
// values, evacuatedX and evacuatedY, that differ in
// their low bit.
if checkBucket>>(it.B-1) != uintptr(b.tophash[offi]&1) {
continue
}
}
}
if (b.tophash[offi] != evacuatedX && b.tophash[offi] != evacuatedY) ||
!(t.reflexivekey() || alg.equal(k, k)) {// 數據沒有遷移，直接訪問
// This is the golden data, we can return it.
// OR
// key!=key, so the entry can't be deleted or updated, so we can just return it.
// That's lucky for us because when key!=key we can't look it up successfully.
it.key = k
if t.indirectelem() {
e = *((*unsafe.Pointer)(e))
}
it.elem = e
} else { // 數據已經遷移，通過key定位訪問
// The hash table has grown since the iterator was started.
// The golden data for this key is now somewhere else.
// Check the current hash table for the data.
// This code handles the case where the key
// has been deleted, updated, or deleted and reinserted.
// NOTE: we need to regrab the key as it has potentially been
// updated to an equal() but not identical key (e.g. +0.0 vs -0.0).
rk, re := mapaccessK(t, h, k)
if rk == nil {
continue // key has been deleted
}
it.key = rk
it.elem = re
}
it.bucket = bucket
if it.bptr != b { // avoid unnecessary write barrier; see issue 14921
it.bptr = b
}
it.i = i + 1
it.checkBucket = checkBucket
return
}
b = b.overflow(t) // 繼續遍歷overflow bucket
i = 0
goto next
}
當我們去遍歷一個map的時候可能有三種情況：
- 在
iterInit與iterNext期間未發生Grow: 只要順序遍歷it.buckets數據即可; iterInit與iterNext都發生在Grow state: 上述代碼可以清晰看出iterInit時候是對h.buckets、h.B做了 snapshot，此時iterNext是以newBucket作爲基礎去遍歷的，那麼一個bucket_N可能有兩個狀態：已經遷移和尚未遷移。已經遷移的直接遍歷桶即可，未遷移的則需要去oldbucket中遍歷，不過需要注意的一點是，要過濾掉那些不可能被遷移到bucket_N的數據（在double_size情況下分上下半區）;iterInit在Grow state之前，iterNext在Grow State：此種情況更加複雜一些，因爲iterInit是在Grow之前，iterNext的時候it.buckets實際對應的是h.oldbucket，也就是說是基於oldbuckets去遍歷，此時bucket_N也有兩種情況：已經遷移和沒有遷移，沒有遷移的直接取數據返回，已經遷移的則直接通過key訪問，因爲此時可能在新bucekt中已經被更新或者刪除了。
map delete
上面我們也提到過，map的移除是通過修改topHash爲emptyOne完成，刪除邏輯要比插入邏輯簡單很多：
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 {
if t.hashMightPanic() {
t.key.alg.hash(key, 0) // see issue 23734
}
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() { // 判斷是否在增長，如果在增長，則對對應的oldbucket 進行遷移
growWork(t, h, bucket)
}
b := (*bmap)(add(h.buckets, bucket*uintptr(t.bucketsize)))
bOrig := b // bucket在base區域的位置，但key實際所在位置可能是overflow
top := tophash(hash)
search:
for ; b != nil; b = b.overflow(t) {
for i := uintptr(0); i < bucketCnt; i++ {
if b.tophash[i] != top {
if b.tophash[i] == emptyRest { // 搜索到emptyRest，停止搜索
break search
}
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.
// 清理內存
if t.indirectkey() {
*(*unsafe.Pointer)(k) = nil
} else if t.key.ptrdata != 0 {
memclrHasPointers(k, t.key.size)
}
e := add(unsafe.Pointer(b), dataOffset+bucketCnt*uintptr(t.keysize)+i*uintptr(t.elemsize))
if t.indirectelem() {
*(*unsafe.Pointer)(e) = nil
} else if t.elem.ptrdata != 0 {
memclrHasPointers(e, t.elem.size)
} else {
memclrNoHeapPointers(e, t.elem.size)
}
b.tophash[i] = emptyOne
// If the bucket now ends in a bunch of emptyOne states,
// change those to emptyRest states.
// It would be nice to make this a separate function, but
// for loops are not currently inlineable.
// 如果emptyOne 後面緊跟的 emptyRest，則把emptyOne設置爲emptyRest
if i == bucketCnt-1 {
if b.overflow(t) != nil && b.overflow(t).tophash[0] != emptyRest {
goto notLast
}
} else {
if b.tophash[i+1] != emptyRest {
goto notLast
}
}
for {
b.tophash[i] = emptyRest
if i == 0 {
if b == bOrig {
break // beginning of initial bucket, we're done.
}
// Find previous bucket, continue at its last entry.
c := b
for b = bOrig; b.overflow(t) != c; b = b.overflow(t) { //查找b前面的overflow
}
i = bucketCnt - 1
} else {
i--
}
if b.tophash[i] != emptyOne {
break
}
}
notLast:
h.count--
break search
}
}
if h.flags&hashWriting == 0 {
throw("concurrent map writes")
}
h.flags &^= hashWriting
}
執行流程如下：
- 查找key所在bucket，將對應位置的
topHash設置爲emptyOne，清除對應的key、elem數據; - 如果下一個位置爲
emptyRest(包括後面緊跟的overflow_bueckt)，則將emptyOne修改爲emptyRest，向下執行；否則結束流程； - 向前查找前一個位置，如果到了
base_bucekt的起始位置，則結束流程；否則跳轉到第2步。
總結
- golang map的底層實現是通過hash table實現的，每個bucket可以存貯8個
key-elem，通過key_hash的低B(總共劃分2^B個桶)bit劃分桶，桶內通過topHash(key_hash前8bit)做區分; - hash table在內存中使用
連續數組+跳轉指針存儲，跳轉指針指向overflow_bucket，根據key查找的時候都是在連續內存上操作，以此來保證O(1)時間複雜度; - 桶的
Grow可以剔除被刪除數據佔用的空間，使得數據更加緊湊，同時overflow_bucket的排序會發生改變，優先遷移的bucket對應的overflow_bucket地址靠前。有兩種形式：same_size（Grow前後桶數不變）,數據分桶格局不變;double_size（擴張後桶數*2）, 根據key_hash的B-1bit決定是劃分到first半區還是second半區，完成桶的重新劃分。 - 刪除數據的時候會見對應位置的
topHash設置爲emptyOne，如果一個bucekt（這裏的bucekt是指邏輯上的桶，包括base_bucekt和overflow_bucket）中最後的位置爲emptyOne則修改爲emptyreset。