Question

我是F＃的新手，所以这段代码对我来说很奇怪

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)]
          ]

我不确定的是do let声明 - 它显然是在randomTest之后声明的，但randomTest仍然可以调用该函数。这段代码的执行顺序是什么？

Answer 1

它的写作方式可能令人困惑。没有do let声明这样的东西。事实上，它是一个整个do {code}块，在{code}内有一个let绑定。这意味着它不是函数声明，do块只是要执行的代码，它不会声明函数或值。

应该更容易阅读：

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)]
          ]

因此执行顺序首先是let randomTest ...，然后是do阻止。

F＃确实允许执行顺序

1 个答案: