ios - Guard let Cause ，从整个功能 block 返回

Question

以下代码从guard let details返回，但不执行其他打印语句。

func application(application: UIApplication, didFinishLaunchingWithOptions launchOptions: [NSObject: AnyObject]?) -> Bool {
    // Override point for customization after application launch.

    let testDictionary:[String:AnyObject] = ["a":NSMutableArray()]

    let result = testDictionary["C"]

    print("Before details Check")
    guard let details = result where details.isKindOfClass(NSArray) else {
        return false
    }
    print("Returned Result \(details)")
    print("After Details Check")
    return true
}

gaurd后面的语句没有被执行有什么原因吗？我做错了什么？

Answer 1

来自"Statements"在 Swift 文档中:

Return Statement

A return statement occurs in the body of a function or method definition and causes program execution to return to the calling function or method.

和

Guard Statement

A guard statement is used to transfer program control out of a scope if one or more conditions aren’t met.
...
The else clause of a guard statement is required, and must either call a function marked with the noreturn attribute or transfer program control outside the guard statement’s enclosing scope ...

如果您的 guard let details = 条件失败，则返回“立即”从当前函数返回。你不能简单地继续，因为 details 将是未定义的。正是如此guard 语句是为之而创建的。

您可以通过引入本地范围来解决您的问题:

print("Before details Check")
checkLabel: do {
    guard let details = result where details.isKindOfClass(NSArray) else {
        break checkLabel
    }
    print("Returned Result \(details)")
}
print("After details Check")

这里，break checkLabel 在 do-scope 之后继续执行 if条件失败。

但更简单的方法是使用 if-let 而不是 guard-let:

print("Before details Check")
if let details = result where details.isKindOfClass(NSArray) {
    print("Returned Result \(details)")
}
print("After details Check")