Question

如果首先打破如何？

let WithdrawalCash (account, amount) = seq {        

    if ( account.Balance.CurrentAmount - amount < 0.0m) then
        yield NotEnoughMoneyForWithdrawal(account, amount)   
        // How to break here?     

    let newBalance = account.Balance.CurrentAmount - amount
    yield CashWithdrawnEvent(account, newBalance)
}

Answer 1

不确定这会有所帮助，为什么不使用else子句？

let WithdrawalCash (account, amount) = seq {        

   if ( account.Balance.CurrentAmount - amount < 0.0m) then
       yield NotEnoughMoneyForWithdrawal(account, amount)   
       // How to break here?     
   else 
      let newBalance = account.Balance.CurrentAmount - amount
      yield CashWithdrawnEvent(account, newBalance) 
}

另请看：

hubsFs - Break in F#?

Imperative computation in F# (II.) - Writing break and continue

hubsFs - yield break: deprecated or as-yet-unimplemented?

Answer 2

发布的代码只会返回一个CashWithdrawlEvent，然后结束序列...你需要有一个循环来返回多个值。另外，您是否考虑过使用“匹配”来处理多个案例？

（未经测试工作......）

let WithdrawalCash (account, amount) = seq {         

    let bDone = ref false

    while not (!bDone) do
        match amount with
        | v when account.Balance.CurrentAmount - v < 0 ->
           yield NotEnoughMoneyForWithdrawal(account, amount)
           bDone := true    // break

        // more when clauses can go here

        | _ ->
          let newBalance = account.Balance.CurrentAmount - amount 
          yield CashWithdrawnEvent(account, newBalance) 
          // let the sequence continue

}

但是，即使这看起来不像你想要的那样，因为每次从序列中提取值时它总是会撤回相同的“数量”，因为在创建序列时帐户和金额是固定的。所以，我放弃'seq'并使其成为一个简单的函数，如：

let WithdrawalCash (account, amount) =

    match amount with
    | v when account.Balance.CurrentAmount - v < 0 ->
       NotEnoughMoneyForWithdrawal(account, amount)

    // more when clauses can go here

    | _ ->
      let newBalance = account.Balance.CurrentAmount - amount 
      CashWithdrawnEvent(account, newBalance)

对于枚举某些序列并在满足特定条件时停止的一般情况，请考虑“Seq.takeWhile”，如：

let MySeq = seq {
    while true do
        yield someValue
    }

let code () =
    MySeq
    |> Seq.takeWhile ( fun v -> ShouldIContinueWorkingTest(v) )
    |> Seq.iter ( fun v -> DoWork(v) )

类似于F＃中的收益率突破

2 个答案: