F＃讓執行順序

Question

我是F＃的新手，所以這段代碼對我來說很奇怪

let randomTest avgWait avgBusyTime numExp numClients labsRules =
    let clients, _ = mkClientsAndLabs numClients labsRules 
    doTest [for i in 0..numClients-1 -> randomTestClient clients i avgWait avgBusyTime numExp  ]

do let clients, _ = mkClientsAndLabs 5 [rulesA; rulesB] 
   doTest [scheduledClient clients 0 [(0, 500, A)];     // Request a lab at the very start, use for "A" for 0.5 seconds
           scheduledClient clients 1 [(200, 300, Mix (Mix (A,Mix (A,A)),B))] ;   // Request after 0.2s, release 0.3s later.

           scheduledClient clients 2 [(300, 200, Mix (A,Mix (A,A)))];   // These three will all be waiting for a lab.
           scheduledClient clients 3 [(400, 200, Mix (A,A))];           // Client 2 should include the others as guests.
           scheduledClient clients 4 [(400, 200, A)]
          ]

我不確定的是do let聲明-顯然是在randomTest之后聲明的，但randomTest仍然可以調用該函數。 該代碼執行的順序是什么？

Answer 1

它的編寫方式可能會令人困惑。 沒有do let聲明之類的東西。 事實上，它是一個整體的do {code}使用let里面綁定塊{code} 。 這意味着它不是函數聲明，do塊只是要執行的代碼，它沒有聲明函數或值。

這樣應該更容易閱讀：

do 
   let clients, _ = mkClientsAndLabs 5 [rulesA; rulesB] 
   doTest [scheduledClient clients 0 [(0, 500, A)];     // Request a lab at the very start, use for "A" for 0.5 seconds
           scheduledClient clients 1 [(200, 300, Mix (Mix (A,Mix (A,A)),B))] ;   // Request after 0.2s, release 0.3s later.

           scheduledClient clients 2 [(300, 200, Mix (A,Mix (A,A)))];   // These three will all be waiting for a lab.
           scheduledClient clients 3 [(400, 200, Mix (A,A))];           // Client 2 should include the others as guests.
           scheduledClient clients 4 [(400, 200, A)]
          ]

所以執行的順序是let randomTest ... ，然后是do塊。