concurrency - F# 解决死锁

Question

我是并发编程的新手，所以我有一个死锁的小问题需要解决。

所以对于下面的这段代码，它没有打印出任何我怀疑一定存在死锁的东西，尽管我不太确定它是如何发生的。

let sleepMaybe() = if (random 4) = 1 then Thread.Sleep 5

type account(name:string) =
    let balance = ref 1000
    member this.Balance = lock this <| fun () -> !balance
    member this.Name = name
    member this.Withdraw amount = balance := !balance - (sleepMaybe(); amount)
    member this.Deposit amount  = balance := !balance + (sleepMaybe(); amount)

    member this.Transfer (toAcc:account) amount = 
        lock this <| fun () ->  lock toAcc <| fun () -> toAcc.Deposit amount
                                this.Withdraw amount


let doTransfers (acc:account) (toAcc:account) () = 
    for i in 1..100 do acc.Transfer toAcc 100
    printfn "%s balance: %d  Other balance: %d" acc.Name acc.Balance toAcc.Balance

let q2main() = 
    let acc1=account("Account1") 
    let acc2=account("Account2")

    startThread (doTransfers acc1 acc2)
    startThread (doTransfers acc2 acc1)

q2main()         

Answer 1

您正在锁定实例本身并要求两个实例都被锁定以传输某些内容。这是导致死锁的秘诀。

• 线程 1 锁定 acc1 以开始传输
• 线程 2 锁定 acc2 以开始传输
• 线程 1 等待 acc2 上的锁被释放，以便完成传输
• 线程 2 等待 acc1 上的锁被释放，以便完成传输

他们会无限期地等待对方释放他们的锁。

如果您必须一次获取多个锁，总是以相同的顺序获取锁。也就是说，尽量不要通过更改对象职责同时需要多个锁。

例如，取款和存款是两个不相关的独立操作，但它们会修改余额。你正试图用锁来保护余额。一旦帐户的余额发生变化，就没有必要再保留该锁定。此外，我建议知道如何转移到其他帐户不是帐户的责任。

考虑到这一点，以下是消除死锁的更改。

type Account(name:string) =
    let mutable balance = 1000
    let accountSync = new Object()

    member x.Withdraw amount = lock accountSync 
                                  (fun () -> balance <- balance - amount)
    member x.Deposit amount =  lock accountSync 
                                  (fun () -> balance <- balance + amount)

let transfer amount (fromAccount:Account) (toAccount:Account) =
    fromAccount.Withdraw(amount)
    toAccount.Deposit(amount)