英文:
Is there a way to have a computed property for optionals in Swift
问题
我在想是否有一种方法可以在可选值上实现一个计算属性，而不需要解包或默认值。为了举例说明，假设以下情况：
我们有一个UITextField的子类，我们想要添加一些验证。假设存在或不存在值（再次强调，这只是为了举例说明）。目前做到这一点的方法是：
var hasValue: Bool {
if let value = text {
return value.isEmpty
} else {
return false
}
}
我尝试在Swift文档中搜索，但未能找到这样的内容。是否有人有更优雅的解决方案？ 
英文:
I was wondering if there is a way to achieve a computed property on an optional, which won't require unwrapping or default value. For the sake of the example, imagine the following scenario:
We have a subclass of UITextField and we want to add some sort of a validation. Let's say if there is a value or not (again, this is for the sake of the example). Right now the way to do so is:
var hasValue: Bool {
if let value = text {
return value.isEmpty
} else {
return false
}
}
I've tried searching through the Swift documentation , but I was not able to find such thing. Does anyone have some more elegant solution to this?
答案1
得分: 0
当然可以：
extension Optional where Wrapped == String {
var hasValue: Bool {
return !(self ?? "").isEmpty
}
}
然而，一般情况下，我会避免添加这样的方法，因为它们隐藏了变量是可选的事实。我更愿意显式地使用 (text ?? "").isEmpty。
英文:
Of course you can:
extension Optional where Wrapped == String {
var hasValue: Bool {
return !(self ?? "").isEmpty
}
}
However, in general I would avoid adding such methods because they hide the fact that a variable is optional. I would rather use the (text ?? "").isEmpty explicitly.
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