Golang map并发 读写锁

本文内容纲要：

-原文链接 golang并发 一：只有写操作 var( countint l=sync.Mutex{} m=make(map[int]int) ) //全局变量并发写导致计数错误 funcvari(){ fori:=0;i<10000;i++{ gofunc(iint){ //deferl.Unlock() //l.Lock() count++ }(i) } fmt.Println(count) } //map并发写不加锁fatalerror:concurrentmapwrites funcmp(){ fori:=0;i<1000;i++{ gofunc(){ deferl.Unlock() l.Lock() m[0]=0 }() } } funcmain(){ //vari() mp() time.Sleep(3*time.Second) }

sync.Mutex互斥锁多个groutine在同一时间只能有一个获取到互斥锁

二：读写都有 //不加锁的话有可能是读的错误的值 funcread(){ deferrwL.RUnlock() rwL.RLock() fmt.Println("read",m[0]) } //如果不加锁会报错fatalerror:concurrentmapwrites funcwrite(){ deferrwL.Unlock() rwL.Lock() m[0]=m[0]+1 } funcrwLock(){ fori:=0;i<10000;i++{ goread() } fori:=0;i<10000;i++{ gowrite() } } funcmain(){ //vari() //mp() rwLock() time.Sleep(3*time.Second) }

同时只能有一个goroutine能够获得写锁定同时可以有任意多个gorouinte获得读锁定同时只能存在写锁定或读锁定（读和写互斥）。

一：Question Whenmorethanonethread*needstomutatethesamevalue,alockingmechanismisneededtosynchronizesaccess. Withoutittwoormorethreads*couldbewritingtothesamevalueatthesametime,resultingincorruptmemory thattypicallyresultsinacrash. Theatomicpackageprovidesafastandeasywaytosynchronizeaccesstoprimitivevalues.Foracounteritisthefastestsynchronizationmethod. Ithasmethodswithwelldefinedusecases,suchasincrementing,decrementing,swapping,etc. Thesyncpackageprovidesawaytosynchronizeaccesstomorecomplicatedvalues,suchasmaps,slices,arrays,orgroupsofvalues. Youusethisforusecasesthatarenotdefinedinatomic. Ineithercaselockingisonlyrequiredwhenwriting.Multiplethreads*cansafelyreadthesamevaluewithoutalockingmechanism. 1.Letstakealookatthecodeyouprovided. typeStatstruct{ countersmap[string]*int64 countersLocksync.RWMutex averagesmap[string]*int64 averagesLocksync.RWMutex } func(s*Stat)Count(namestring){ s.countersLock.RLock() counter:=s.counters[name] s.countersLock.RUnlock() ifcounter!=nil{ atomic.AddInt64(counter,int64(1)) return } } 2.Whatsmissinghereishowthemapsthemselvesareinitialized.Andsofarthemapsarenotbeingmutated.Ifthecounternamesarepredeterminedandcannotbeaddedtolater,youdontneedtheRWMutex.Thatcodemightlooksomethinglikethis: typeStatstruct{ countersmap[string]*int64 } funcInitStat(names...string)Stat{ counters:=make(map[string]*int64) for_,name:=rangenames{ counter:=int64(0) counters[name]=&counter } returnStat{counters} } func(s*Stat)Count(namestring)int64{ counter:=s.counters[name] ifcounter==nil{ return-1//(int64,error)instead? } returnatomic.AddInt64(counter,1) }

(Note:Iremovedaveragesbecauseitwasntbeingusedintheoriginalexample.)

Now,letssayyoudidntwantyourcounterstobepredetermined.Inthatcaseyouwouldneedamutextosynchronizeaccess.

3.LetstryitwithjustaMutex.Itssimplebecauseonlyonethread*canholdLockatatime.Ifasecondthread*triestoLockbeforethefirstreleasestheirswithUnlock,itwaits(orblocks)**untilthen. typeStatstruct{ countersmap[string]*int64 mutexsync.Mutex } funcInitStat()Stat{ returnStat{counters:make(map[string]*int64)} } func(s*Stat)Count(namestring)int64{ s.mutex.Lock() counter:=s.counters[name] ifcounter==nil{ value:=int64(0) counter=&value s.counters[name]=counter } s.mutex.Unlock() returnatomic.AddInt64(counter,1) } 二：Thecodeabovewillworkjustfine.Buttherearetwoproblems. 1.IfthereisapanicbetweenLock()andUnlock()themutexwillbelockedforever,evenifyouweretorecoverfromthepanic.Thiscodeprobablywontpanic,butingeneralitsbetterpracticetoassumeitmight.

Problem#1iseasytosolve.Usedefer:

``` func(s*Stat)Count(namestring)int64{ s.mutex.Lock() defers.mutex.Unlock() counter:=s.counters[name] ifcounter==nil{ value:=int64(0) counter=&value s.counters[name]=counter } returnatomic.AddInt64(counter,1) } ```

ThisensuresthatUnlock()isalwayscalled.Andifforsomereasonyouhavemorethenonereturn,youonlyneedtospecifyUnlock()onceattheheadofthefunction.

2.Anexclusivelockistakenwhilefetchingthecounter.Onlyonethread*canreadfromthecounteratonetime. Problem#2canbesolvedwithRWMutex.Howdoesitworkexactly,andwhyisituseful? RWMutexisanextensionofMutexandaddstwomethods:RLockandRUnlock.ThereareafewpointsthatareimportanttonoteaboutRWMutex: RLockisasharedreadlock.Whenalockistakenwithit,otherthreads*canalsotaketheirownlockwithRLock.Thismeansmultiplethreads*canreadatthesametime.Itssemi-exclusive. Ifthemutexisreadlocked,acalltoLockisblocked**.Ifoneormorereadersholdalock,youcannotwrite. Ifthemutexiswritelocked(withLock),RLockwillblock**. AgoodwaytothinkaboutitisRWMutexisaMutexwithareadercounter.RLockincrementsthecounterwhileRUnlockdecrementsit.AcalltoLockwillblockaslongasthatcounteris>0. Youmaybethinking:Ifmyapplicationisreadheavy,wouldthatmeanawritercouldbeblockedindefinitely?No.ThereisonemoreusefulpropertyofRWMutex: Ifthereadercounteris>0andLockiscalled,futurecallstoRLockwillalsoblockuntiltheexistingreadershavereleasedtheirlocks,thewriterhasobtainedhislockandlaterreleasesit. Thinkofitasthelightabovearegisteratthegrocerystorethatsaysacashierisopenornot.Thepeopleinlinegettostaythereandtheywillbehelped,butnewpeoplecannotgetinline.Assoonasthelastremainingcustomerishelpedthecashiergoesonbreak,andthatregistereitherremainscloseduntiltheycomebackortheyarereplacedwithadifferentcashier. 3.LetsmodifytheearlierexamplewithanRWMutex: typeStatstruct{ countersmap[string]*int64 mutexsync.RWMutex } funcInitStat()Stat{ returnStat{counters:make(map[string]*int64)} } func(s*Stat)Count(namestring)int64{ varcounter*int64 ifcounter=getCounter(name);counter==nil{ counter=initCounter(name); } returnatomic.AddInt64(counter,1) } func(s*Stat)getCounter(namestring)*int64{ s.mutex.RLock() defers.mutex.RUnlock() returns.counters[name] } func(s*Stat)initCounter(namestring)*int64{ s.mutex.Lock() defers.mutex.Unlock() counter:=s.counters[name] ifcounter==nil{ value:=int64(0) counter=&value s.counters[name]=counter } returncounter } 三：WiththecodeaboveIveseparatedthelogicoutintogetCounterandinitCounterfunctionsto:

Keepthecodesimpletounderstand.ItwouldbedifficulttoRLock()andLock()inthesamefunction.

Releasethelocksasearlyaspossiblewhileusingdefer.

Thecodeabove,unliketheMutexexample,allowsyoutoincrementdifferentcounterssimultaneously.

AnotherthingIwantedtopointoutiswithalltheexamplesabove,themapmap[string]*int64containspointerstothecounters,notthecountersthemselves.Ifyouweretostorethecountersinthemapmap[string]int64youwouldneedtouseMutexwithoutatomic.Thatcodewouldlooksomethinglikethis:

typeStatstruct{ countersmap[string]int64 mutexsync.Mutex } funcInitStat()Stat{ returnStat{counters:make(map[string]int64)} } func(s*Stat)Count(namestring)int64{ s.mutex.Lock() defers.mutex.Unlock() s.counters[name]++ returns.counters[name] }

Youmaywanttodothistoreducegarbagecollection-butthatwouldonlymatterifyouhadthousandsofcounters-andeventhenthecountersthemselvesdonttakeupawholelotofspace(comparedtosomethinglikeabytebuffer).

WhenIsaythreadImeango-routine.Athreadinotherlanguagesisamechanismforrunningoneormoresetsofcodesimultaneously.Athreadisexpensivetocreateandtear-down.Ago-routineisbuiltontopofthreads,butre-usesthem.Whenago-routinesleepstheunderlyingthreadcanbeusedbyanothergo-routine.Whenago-routinewakesup,itmightbeonadifferentthread.Gohandlesallthisbehindthescenes.--Butforallintentsandpurposesyouwouldtreatago-routinelikeathreadwhenitcomestomemoryaccess.However,youdonthavetobeasconservativewhenusinggo-routinesasyoudothreads.

**Whenago-routineisblockedbyLock,RLock,achannel,orSleep,theunderlyingthreadmightbere-used.Nocpuisusedbythatgo-routine-thinkofitaswaitinginline.Likeotherlanguagesaninfinitelooplikefor{}wouldblockwhilekeepingthecpuandgo-routinebusy-thinkofthatasrunningaroundinacircle-youllgetdizzy,throwup,andthepeoplearoundyouwontbeveryhappy.

原文链接

https://stackoverflow.com/questions/19148809/how-to-use-rwmutex-in-golang

本文内容总结：原文链接，

原文链接：https://www.cnblogs.com/alin-qu/p/10632555.html