swift - 我如何判断哪个守卫声明失败了？

Question

如果我有一堆链式 guard let 语句，除了将我的 guard let 分解成多个语句外，我如何诊断哪个条件失败了？

给出这个例子:

guard let keypath = dictionary["field"] as? String,
    let rule = dictionary["rule"] as? String,
    let comparator = FormFieldDisplayRuleComparator(rawValue: rule),
    let value = dictionary["value"]
    else
    {
        return nil
    }

我如何判断 4 个 let 语句中的哪一个失败并调用了 else block ？

我能想到的最简单的事情是将语句分解为 4 个连续的 guard else 语句，但感觉不对。

 guard let keypath = dictionary["field"] as? String
    else
    {
        print("Keypath failed to load.")

        self.init()
        return nil
    }

    guard let rule = dictionary["rule"] as? String else
    {
        print("Rule failed to load.")

        self.init()
        return nil
    }

    guard let comparator = FormFieldDisplayRuleComparator(rawValue: rule) else
    {
        print("Comparator failed to load for rawValue: \(rule)")

        self.init()
        return nil
    }

    guard let value = dictionary["value"] else
    {
        print("Value failed to load.")

        self.init()
        return nil
    }

如果我想将它们全部放在一个 guard 语句中，我可以想到另一种选择。检查 guard 语句中的 nils 可能会起作用:

guard let keypath = dictionary["field"] as? String,
    let rule = dictionary["rule"] as? String,
    let comparator = FormFieldDisplayRuleComparator(rawValue: rule),
    let value = dictionary["value"]
    else
    {

        if let keypath = keypath {} else {
           print("Keypath failed to load.")
        }

        // ... Repeat for each let...
        return nil
    }

我什至不知道这是否会编译，但我还不如使用一堆 if let 语句或 guard 开始。

什么是惯用的 Swift 方式？

Answer 1

Erica Sadun 刚刚就这个主题写了一篇不错的博文。

她的解决方案是劫持 where 子句并使用它来跟踪哪些 guard 语句通过。使用 diagnose 方法的每个成功的保护条件都会将文件名和行号打印到控制台。最后一个 diagnose 打印语句之后的警戒条件是失败的条件。解决方案如下所示:

func diagnose(file: String = #file, line: Int = #line) -> Bool {
    print("Testing \(file):\(line)")
    return true
}

// ...

let dictionary: [String : AnyObject] = [
    "one" : "one"
    "two" : "two"
    "three" : 3
]

guard
    // This line will print the file and line number
    let one = dictionary["one"] as? String where diagnose(),
    // This line will print the file and line number
    let two = dictionary["two"] as? String where diagnose(),
    // This line will NOT be printed. So it is the one that failed.
    let three = dictionary["three"] as? String where diagnose()
    else {
        // ...
}

可以找到 Erica 关于此主题的文章 here