ios – 迅速强制解包异常不传播

我已经遇到这种愚蠢的行为,在 swift中,强制解包可选不传播.

从文档：

Trying to use ! to access a non-existent optional value triggers a runtime error. Always make sure that an optional contains a non-nil value before using ! to force-unwrap its value.

复制：

func foo(bar:String?) throws{
    print(bar!);
}

和

try foo(nil);

这似乎不符合逻辑或一致,我找不到有关这个问题的任何文件.

这是设计吗？

解决方法

从 documentation：

Error Handling

Error handling is the process of responding to and recovering from
error conditions in your program. Swift provides first-class support
for throwing,catching,propagating,and manipulating recoverable
errors at runtime.

…

Representing and Throwing Errors

In Swift,errors are represented by values of types that conform to
the ErrorType protocol. This empty protocol indicates that a type can
be used for error handling.

(注意：在Swift 3中,ErrorType已重命名为Error)

所以使用try / catch可以处理抛出的Swift错误(符合ErrorType协议的类型的值).
这与运行时错误和运行时异常完全无关
(也与基金会图书馆的NSException无关).

请注意,关于错误处理的Swift文档甚至不使用
字“异常”,唯一的例外(！)在(强调我的)在：

NOTE

Error handling in Swift resembles exception handling in other
languages,with the use of the try,catch and throw keywords. Unlike
exception handling in many languages—including Objective-C—error
handling in Swift does not involve unwinding the call stack,a process
that can be computationally expensive. As such,the performance
characteristics of a throw statement are comparable to those of a
return statement.

不包括任何选项的解开不会抛出
Swift错误(可能会传播),无法处理
尝试.

你必须使用众所周知的技术可选绑定,可选链接,检查对等