Go map原理剖析

在使用map的过程中,有两个问题是经常会遇到的:读写冲突和遍历无序性。为什么会这样呢,底层是怎么实现的呢?带着这两个问题,我简单的了解了一下map的增删改查及遍历的实现。

结构

hmap

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 // 有效数据的长度# live cells == size of map.  Must be first (used by len() builtin)
    flags     uint8 // 用于记录hashmap的状态
    B         uint8  // 2^B = buckets的数量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 // 随机的hash种子

    buckets    unsafe.Pointer // buckets数组array of 2^B Buckets. may be nil if count==0.
    oldbuckets unsafe.Pointer // 老的buctedts数据,map增长的时候会用到
    nevacuate  uintptr        // progress counter for evacuation (buckets less than this have been evacuated)

    extra *mapextra // 额外的bmap数组optional fields
}

mapextra

 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
}

bmap

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.
}

stringStruct

type stringStruct struct {
    str unsafe.Pointer
    len int
}

hiter

map遍历时用到的结构,startBucket+offset设定了开始遍历的地址,保证map遍历的无序性

type hiter struct {
  // key的指针
    key         unsafe.Pointer // Must be in first position.  Write nil to indicate iteration end (see cmd/internal/gc/range.go).
  // 当前value的指针
    value       unsafe.Pointer // Must be in second position (see cmd/internal/gc/range.go).
    t           *maptype
  // 指向map的指针
    h           *hmap
  // 指向buckets的指针
    buckets     unsafe.Pointer // bucket ptr at hash_iter initialization time
  // 指向当前遍历的bucket的指针
    bptr        *bmap          // current bucket
  // 指向map.extra.overflow
    overflow    *[]*bmap       // keeps overflow buckets of hmap.buckets alive
  // 指向map.extra.oldoverflow
    oldoverflow *[]*bmap       // keeps overflow buckets of hmap.oldbuckets alive
  // 开始遍历的bucket的索引
    startBucket uintptr        // bucket iteration started at
  // 开始遍历bucket上的偏移量
    offset      uint8          // intra-bucket offset to start from during iteration (should be big enough to hold bucketCnt-1)
    wrapped     bool           // already wrapped around from end of bucket array to beginning
    B           uint8
    i           uint8
    bucket      uintptr
    checkBucket uintptr
}

这里的keys和values、*overflow三个变量在结构体中并没有体现,但是在源码过程中,一直有为他们预留位置,所以这里的示意图中就展示出来了,keys和values其实8个长度的数组

demo

我们简单写个demo,通过go tool 来分析一下底层所对应的函数

func main() {
    m := make(map[interface{}]interface{}, 16)
    m["111"] = 1
    m["222"] = 2
    m["444"] = 4
    _ = m["444"]
    _, _ = m["444"]
    delete(m, "444")

    for range m {
    }
}
▶ go tool objdump -s "main.main" main | grep CALL
  main.go:4             0x455c74                e8f761fbff              CALL runtime.makemap(SB)                
  main.go:5             0x455ce1                e8da6dfbff              CALL runtime.mapassign(SB)              
  main.go:6             0x455d7b                e8406dfbff              CALL runtime.mapassign(SB)              
  main.go:7             0x455e15                e8a66cfbff              CALL runtime.mapassign(SB)              
  main.go:8             0x455e88                e89363fbff              CALL runtime.mapaccess1(SB)             
  main.go:9             0x455ec4                e84766fbff              CALL runtime.mapaccess2(SB)             
  main.go:10            0x455f00                e85b72fbff              CALL runtime.mapdelete(SB)              
  main.go:12            0x455f28                e804a7ffff              CALL 0x450631                           
  main.go:12            0x455f53                e8b875fbff              CALL runtime.mapiterinit(SB)            
  main.go:12            0x455f75                e88677fbff              CALL runtime.mapiternext(SB)            
  main.go:7             0x455f8f                e81c9cffff              CALL runtime.gcWriteBarrier(SB)         
  main.go:6             0x455f9c                e80f9cffff              CALL runtime.gcWriteBarrier(SB)         
  main.go:5             0x455fa9                e8029cffff              CALL runtime.gcWriteBarrier(SB)         
  main.go:3             0x455fb3                e8f87dffff              CALL runtime.morestack_noctxt(SB) 

初始化

makemap

makemap创建一个hmap结构体,并赋予这个变量一些初始的属性

func makemap(t *maptype, hint int, h *hmap) *hmap {
  // 首先判断map的大小是否合适
    if hint < 0 || hint > int(maxSliceCap(t.bucket.size)) {
        hint = 0
    }

    // initialize Hmap
  // 初始化hmap结构
    if h == nil {
        h = new(hmap)
    }
  // 生成一个随机的hash种子
    h.hash0 = fastrand()

    // find size parameter which will hold the requested # of elements
  // 根据hint,也就是map预设的长度,确定B的大小,以使map的装载系数在正常范围内,扩容那块再细讲
    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.
  // 如果B==0,则赋值的时候进行惰性分配,如果B!=0,则分配对应数量的buckets
    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
}

makeBucketArray

makeBucketArray初始化了map所需的buckets,最少分配2^b个buckets

func makeBucketArray(t *maptype, b uint8, dirtyalloc unsafe.Pointer) (buckets unsafe.Pointer, nextOverflow *bmap) {
    base := bucketShift(b)
    nbuckets := base
    // 如果b,也就是map比较大的情况,则多分配点数组,给nextOverflow使用
    if b >= 4 {
    // 计算应该多分配的buckets数量
        nbuckets += bucketShift(b - 4)
        sz := t.bucket.size * nbuckets
        up := roundupsize(sz)
        if up != sz {
            nbuckets = up / t.bucket.size
        }
    }
    // 如果不是 dirtyalloc,新分配map空间时,dirtyalloc为nil
    if dirtyalloc == nil {
    // 申请buckets数组
        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)
        }
    }
  // 判断是否多申请了buckets,多申请的buckets放在nextOverflow里面以备后用
    if base != nbuckets {
        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
}

初始化的过程到此就结束了,比较简单,就是根据初始化的大小,确定buckets的数量,并分配内存等

查找(mapaccess)

在上面的go tool 分析过程中可以发现

  • _ = m["444"] 对应 mapaccess1
  • , = m["444"] 对应 mapaccess2

两个函数的逻辑大致相同,我们以mapaccess1为例来分析

mapaccess

func mapaccess1(t *maptype, h *hmap, key unsafe.Pointer) unsafe.Pointer {
  // 如果h还没有实例化,或者还没有值,返回零值
    if h == nil || h.count == 0 {
        return unsafe.Pointer(&zeroVal[0])
    }
  // 判断当前map是否处于 写 的过程中,读写冲突
    if h.flags&hashWriting != 0 {
        throw("concurrent map read and map write")
    }
  // 根据初始化生产的hash随机种子hash0,计算key的hash值
    alg := t.key.alg
    hash := alg.hash(key, uintptr(h.hash0))
    m := bucketMask(h.B)
  // 根据key的hash值,计算出对应的bucket的位置,计算过程后面图示
    b := (*bmap)(add(h.buckets, (hash&m)*uintptr(t.bucketsize)))
  // 扩容的过程中,oldbuckets不为空,所以这时候,这时候需要判断,目标bucket是否已经迁移完成了,扩容的时候细讲
    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
        }
    // 如果目标bucket在扩容中还没有迁移,则到oldbuckets中找目标bucket
        oldb := (*bmap)(add(c, (hash&m)*uintptr(t.bucketsize)))
        if !evacuated(oldb) {
            b = oldb
        }
    }
  // 计算出key的tophash,用于比对
    top := tophash(hash)
    for ; b != nil; b = b.overflow(t) {
        for i := uintptr(0); i < bucketCnt; i++ {
      // 如果tophash不一致,key肯定不同,继续寻找下一个
            if b.tophash[i] != top {
                continue
            }
      // tophash一直,需要判断key是否一致
            k := add(unsafe.Pointer(b), dataOffset+i*uintptr(t.keysize))
            if t.indirectkey {
                k = *((*unsafe.Pointer)(k))
            }
      // key也是相同的,则返回对应的value
            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])
}

overflow

这个函数就是找bmap的overflow的地址,通过结构图中可以看出,找到bmap结构体的最后一个指针占用的内存单元就是overflow指向的下一个bmap的地址了

func (b *bmap) overflow(t *maptype) *bmap {
    return *(**bmap)(add(unsafe.Pointer(b), uintptr(t.bucketsize)-sys.PtrSize))
}

上面的逻辑比较简单,但是在这里有几个问题需要解决

  1. bucket(bmap结构体)是怎么确定的
  2. tophash是怎么确定的
  3. key和value的地址为什么是通过偏移来计算的

先放一下buckets和bmap的放大图

  1. bucket(bmap结构体)是怎么确定的

    bucket := hash & bucketMask(h.B)
    b := (*bmap)(unsafe.Pointer(uintptr(h.buckets) + bucket*uintptr(t.bucketsize)))

    加入B=5,则说明buckets的数量为2^5 = 32,则取hash的末5位,来计算出目标bucket的索引,图中计算出索引为6,所以,在buckets上偏移6个bucket大小的地址,即可找到对应的bucket

  2. tophash是怎么确定的

    func tophash(hash uintptr) uint8 {
        top := uint8(hash >> (sys.PtrSize*8 - 8))
        if top < minTopHash {
            top += minTopHash
        }
        return top
    }

    每个bucket的tophash数组的长度为8,所以,这里直接去hash值的前8位计算出来数值,既是tophash了

  3. key和value的地址为什么是通过偏移来计算的

    k := add(unsafe.Pointer(b), dataOffset+i*uintptr(t.keysize))
    val = add(unsafe.Pointer(b), dataOffset+bucketCnt*uintptr(t.keysize)+i*uintptr(t.valuesize))

    根据最开始的数据结构分析和上面的bmap图示,可以看出bmap中所有的key是放在一起的,所有的value是放在一起的,dataoffset是tophash[8]所占用的大小,所以,key所在的地址也就是 b的地址+dataOffset的偏移+对应的索引i*key的大小,同理value是排列在key的后面的

插入

mapassign

func mapassign(t *maptype, h *hmap, key unsafe.Pointer) unsafe.Pointer {
    if h == nil {
        panic(plainError("assignment to entry in nil map"))
    }
    // map并发读写的处理,直接抛异常
    if h.flags&hashWriting != 0 {
        throw("concurrent map writes")
    }
  // 根据map的hash种子 hash0,计算key的hash值
    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
  // 如果map没有buckets,就分配(make(map)不指定map长度的时候就会惰性分配buckets)
    if h.buckets == nil {
        h.buckets = newobject(t.bucket) // newarray(t.bucket, 1)
    }

again:
  // 根据计算出的hash值,来确定应该插入的bucket在buckets中的索引
    bucket := hash & bucketMask(h.B)
  // 判断是否在扩容map,growWork是来完成扩容操作的
    if h.growing() {
        growWork(t, h, bucket)
    }
  // 确认bucket的地址
    b := (*bmap)(unsafe.Pointer(uintptr(h.buckets) + bucket*uintptr(t.bucketsize)))
  // 根据计算出hash二进制前八位的值,作为tophash使用
    top := tophash(hash)

    var inserti *uint8
    var insertk unsafe.Pointer
    var val unsafe.Pointer
    for {
        for i := uintptr(0); i < bucketCnt; i++ {
      // 循环遍历tophash数组,如果数组的索引位置为空,先拿过来使用
            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
            }
      // 找到了tophash数组中找到了当前key的tophash一致的情况
            k := add(unsafe.Pointer(b), dataOffset+i*uintptr(t.keysize))
      // 如果key是指针,获取指针对应的数据
            if t.indirectkey {
                k = *((*unsafe.Pointer)(k))
            }
      // 判断这两个key是否相同,不同继续寻找
            if !alg.equal(key, k) {
                continue
            }
            // already have a mapping for key. Update it.
            if t.needkeyupdate {
                typedmemmove(t.key, k, key)
            }
      // 根据i找到value应该存放的位置,可以结合结构图中bmap的数据结构来理解
            val = add(unsafe.Pointer(b), dataOffset+bucketCnt*uintptr(t.keysize)+i*uintptr(t.valuesize))
            goto done
        }
    // buckets中没有找到空余的位置或者相同的key,则到overflow中查找
        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.
  // 判断是否需要扩容
    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
    }
    // inerti==nil,表示map的buckets都满了,则需要新加一个overflow挂载到map和对应的bmap下
    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
  // 存储key value到指定的位置
    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")
    }
  // 修改map的flags
    h.flags &^= hashWriting
    if t.indirectvalue {
        val = *((*unsafe.Pointer)(val))
    }
    return val
}

setoverflow

func (h *hmap) newoverflow(t *maptype, b *bmap) *bmap {
    var ovf *bmap
  // 先去找一下预先分配的有没有剩余的overflow
    if h.extra != nil && h.extra.nextOverflow != nil {
        // We have preallocated overflow buckets available.
        // See makeBucketArray for more details.
    // 预先分配的有,直接使用预先分配的,然后更新一下 下一个可以用overflow => nextOverflow
        ovf = h.extra.nextOverflow
        if ovf.overflow(t) == nil {
            // 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 {
            // 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)
            h.extra.nextOverflow = nil
        }
    } else {
        ovf = (*bmap)(newobject(t.bucket))
    }
  // 增加noverflow
    h.incrnoverflow()
    if t.bucket.kind&kindNoPointers != 0 {
        h.createOverflow()
        *h.extra.overflow = append(*h.extra.overflow, ovf)
    }
  // 把当前overflow,挂载到bmap的overflow链表后面
    b.setoverflow(t, ovf)
    return ovf
}

overflow指向的就是一个bmap结构,而bmap结构的最后一个地址,存储的是overflow的地址,通过bmap.overflow可以将bmap的所有overflow串联起来,hmap.extra.nextOverflow也是一样的逻辑

扩容

mapassign函数中可以看到,扩容发生的情况有两种

overLoadFactor(h.count+1, h.B) || tooManyOverflowBuckets(h.noverflow, h.B)
  1. 超过设定的负载值
  2. 有太多的overflow

先来看一下这两个函数

overLoadFactor

func overLoadFactor(count int, B uint8) bool {
  // loadFactorNum = 13; loadFactorDen = 2
    return count > bucketCnt && uintptr(count) > loadFactorNum*(bucketShift(B)/loadFactorDen)
}

uintptr(count) > loadFactorNum*(bucketShift(B)/loadFactorDen) 可以简化为 count / (2^B) > 6.5, 这个6.5便是代表loadFactor的负载系数

tooManyOverflowBuckets

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)
}

通过判断noverflow的数量来判断overflow是否太多

我们理解一下这两种情况扩容的原因

  1. 超过设定的负载值

    根据key查找的过程中,根据末B位确定bucket,高8位确定tophash,但是查找tophash的过程中,是需要遍历整个bucket的,所以,最优的情况是每个bucket只存储一个key,这样就达到了hash的O(1)的查找效率,但是空间却大大的浪费了;如果所有的key都存储到了一个bucket里面面,就退变成了链表,查找效率就变成了O(n),所以装载系数就是为了平衡查找效率和存储空间的,当装载系数过大,就需要增加bucket了,来提高查找效率,即增量扩容

  2. 有太多的overflow

    当bucket的空位全部填满的时候,装载系数就达到了8,为什么还会有tooManyOverflowBuckets的判断呢,map不仅有增加还有删除的操作,当某一个bucket的空位填满后,开始填充到overflow里面,这时候再删除bucket里面的数据,其实整个过程很有可能并没有触发 超过负载扩容机制的,(因为有较多的buckets),但是查找overflow的数据,就首先要遍历bucket的数据,这个就是无用功了,查找效率就低了,这时候需要不增加bucket数量的扩容,也就是等量扩容

扩容的工作是由hashGrow开始的,但是真正进行迁移工作的是evacuate, 由growWork进行d调用;在每一次的maassign和mapdelete的时候,会判断这个map是否正在进行扩容操作,如果是的,就迁移当前的bucket;所以,map的扩容并不是一蹴而就的,而是一个循序渐进的过程

hashGrow

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.
  // 判断是等量扩容还是增量扩容
    bigger := uint8(1)
    if !overLoadFactor(h.count+1, h.B) {
        bigger = 0
        h.flags |= sameSizeGrow
    }
  // 为map根据新的B(h.B+bigger为新的h.B)重新分配新的buckets和overflow
    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)
  // 更新hmap相关的属性
    h.B += bigger
    h.flags = flags
    h.oldbuckets = oldbuckets
    h.buckets = newbuckets
    h.nevacuate = 0
    h.noverflow = 0
    // 将老的map的extra和nextOverflow更新到新的map结构下面
    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().
}

hashGrow 这个前菜已经准备完成了,接下来就交给growWorkevacuate两个函数来完成的

growWork

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

讲hmap中的一个bucket搬移到新的buckets中,老的bucket里key与新的buckets中位置的对应,同样参考map的查找过程

这里如何判断这个bucket是否已经搬移过了呢,主要就是依据evacuated函数来判断

func evacuated(b *bmap) bool {
    h := b.tophash[0]
    return h > empty && h < minTopHash
}

看了源码就发现原理很简单,就是对tophash[0]值的判断,那么肯定是在搬移之后设置的这个值,我们通过evacuate函数l哎一探究竟吧

func evacuate(t *maptype, h *hmap, oldbucket uintptr) {
    b := (*bmap)(add(h.oldbuckets, oldbucket*uintptr(t.bucketsize)))
    newbit := h.noldbuckets()
  // 判断是否搬移过
    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.
    // 吧bucket原先对应的索引赋值给x
        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))
        // 如果是增量扩容,扩容后的bucket有变,假如以B=5为例,B+1= 6,这时候去倒数6位计算bucket的索引,但是倒数第6位只能是0或者1,也就是说索引只能是,x或y(x+newbit),这里计算出来y,以备后用
        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))
        }
        // 进行搬移
        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]
        // 空的跳过
                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).
          // 判断hash计算出来,是使用x还是y,等量扩容是使用x
                    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
                // 如果目标的bucket已经满了,则新建overflow,挂载到bucket上,并使用这个overflow
                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))
                }
        // 拷贝key value,设置tophash数组的对应索引的值
                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)
    }
}

扩容是逐步进行的,一次搬运一个bucket

我们以原先的B=5为例,现在增量扩容后B=6,但是hash的倒数第6位只能是0或1,也就是说,如果原先计算出来的bucket索引为6的话,即 00110,那么新的bucket对应的索引只能是 100110(6+2^5)或 000110(6),x对应的就是6,y对应的就是(6+2^5);如果是等量扩容,那么索引肯定就是不变的,这时候就不需要y了

找到对应的新的bucket之后,按顺序依次存放就ok了

删除

mapdelete

删除的逻辑比较简单,根据key查找,找到就清空key和value及tophash

func mapdelete(t *maptype, h *hmap, key unsafe.Pointer) {
    if h == nil || h.count == 0 {
        return
    }
  // 读写冲突
    if h.flags&hashWriting != 0 {
        throw("concurrent map writes")
    }
    // 下面一大片的计算hash,查找bucket,查到bucket里面的key,逻辑一样,就不重复了
    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.
      // 这里找到了key,如果key是指针,设为nil,否则清空key对应内存的数据
            if t.indirectkey {
                *(*unsafe.Pointer)(k) = nil
            } else if t.key.kind&kindNoPointers == 0 {
                memclrHasPointers(k, t.key.size)
            }
      // 同理删除v
            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)
            }
      // 把tophash设置为0,并更新count属性
            b.tophash[i] = empty
            h.count--
            break search
        }
    }

    if h.flags&hashWriting == 0 {
        throw("concurrent map writes")
    }
    h.flags &^= hashWriting
}

遍历

按一般的思维来考虑,遍历值需要遍历buckets数组里面的每个bucket以及bucket下挂的overflow链表即可,但是map存在扩容的情况,这样就会导致遍历的难度增大了,我们看一下go是怎么实现的

根据go tool 的分析,我们可以简单看一下遍历时的流程信息

mapiterinit

func mapiterinit(t *maptype, h *hmap, it *hiter) {
    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
    }
  // 设置iter的属性
    it.t = t
    it.h = h

    // grab snapshot of bucket state
    it.B = h.B
    it.buckets = h.buckets
    if t.bucket.kind&kindNoPointers != 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
  // 随机生成一个种子,并根据这个随机种子计算出startBucket和offset,保证遍历的随机性
    r := uintptr(fastrand())
    if h.B > 31-bucketCntBits {
        r += uintptr(fastrand()) << 31
    }
    it.startBucket = r & bucketMask(h.B)
    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().
    if old := h.flags; old&(iterator|oldIterator) != iterator|oldIterator {
        atomic.Or8(&h.flags, iterator|oldIterator)
    }
    // 开始遍历
    mapiternext(it)
}

mapiternext

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:
  // b==nil说明bucket.overflow链表已经遍历完成了,遍历下一个bucket
    if b == nil {
    // 遍历到了开始的bucket,而且startBucket被遍历过了,则说明整个map遍历完成了
        if bucket == it.startBucket && it.wrapped {
            // end of iteration
            it.key = nil
            it.value = nil
            return
        }
    // 如果hmap正在扩容,则判断当前遍历的bucket是否搬移完了,搬移完了,使用新得bucket,否则使用oldbucket
        if h.growing() && it.B == h.B {
            // 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) {
                checkBucket = bucket
            } else {
                b = (*bmap)(add(it.buckets, bucket*uintptr(t.bucketsize)))
                checkBucket = noCheck
            }
        } else {
            b = (*bmap)(add(it.buckets, bucket*uintptr(t.bucketsize)))
            checkBucket = noCheck
        }
        bucket++
    // 遍历到了数组末尾,从数组头继续遍历
        if bucket == bucketShift(it.B) {
            bucket = 0
            it.wrapped = true
        }
        i = 0
    }
  // 遍历当前bucket或者bucket.overflow里面的数据
    for ; i < bucketCnt; i++ {
    // 通过offset与i,确定正在遍历的bucket的tophash的索引
        offi := (i + it.offset) & (bucketCnt - 1)
        if b.tophash[offi] == empty || b.tophash[offi] == evacuatedEmpty {
            continue
        }
    // 根据偏移量i,确定key和value的地址
        k := add(unsafe.Pointer(b), dataOffset+uintptr(offi)*uintptr(t.keysize))
        if t.indirectkey {
            k = *((*unsafe.Pointer)(k))
        }
        v := add(unsafe.Pointer(b), dataOffset+bucketCnt*uintptr(t.keysize)+uintptr(offi)*uintptr(t.valuesize))
        if checkBucket != noCheck && !h.sameSizeGrow() {
      // 说明增量扩容中,需要进一步判断
            // 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) {
        // 数据还没有从oldbucket迁移到新的bucket里面,判断这个key重新计算后是否与oldbucket的索引一致,不一致则跳过
                // 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)) {
      // 这里的数据不是正在扩容中的数据,可以直接使用
            // 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.indirectvalue {
                v = *((*unsafe.Pointer)(v))
            }
            it.value = v
        } else {
            // 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).
      // 在遍历开始之后,这个map进行了扩容,数据可能不正确,重新查找获取一下
            rk, rv := mapaccessK(t, h, k)
            if rk == nil {
                continue // key has been deleted
            }
            it.key = rk
            it.value = rv
        }
        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
    }
  // 遍历bucket.overflow链表
    b = b.overflow(t)
    i = 0
    goto next
}

整体思路如下:

  1. 首先从buckets数组中,随机确定一个索引,作为startBucket,然后确定offset偏移量,作为起始key的地址
  2. 遍历当前bucket及bucket.overflow,判断当前bucket是否正在扩容中,如果是则跳转到3,否则跳转到4
  3. 加入原先的buckets为0,1,那么扩容后的新的buckets为0,1,2,3,此时我们遍历到了buckets[0], 发现这个bucket正在扩容,那么找到bucket[0]所对应的oldbuckets[0],遍历里面的key,这时候是遍历所有的吗?当然不是,而是仅仅遍历那些key经过hash,可以散列到bucket[0]里面的部分key;同理,当遍历到bucket[2]的时候,发现bucket正在扩容,找到oldbuckets[0],然后遍历里面可以散列到bucket[2]的那些key
  4. 遍历当前这个bucket即可
  5. 继续遍历bucket下面的overflow链表
  6. 如果遍历到了startBucket,说明遍历完了,结束遍历

参考文章


go学习
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