为Swift中的Optional提供默认值? [英] Providing a default value for an Optional in Swift?
问题描述
在Swift中处理可选选项的习惯用法似乎过于冗长,如果您只想在nil为零的情况下提供默认值,就可以了:
The idiom for dealing with optionals in Swift seems excessively verbose, if all you want to do is provide a default value in the case where it's nil:
if let value = optionalValue {
// do something with 'value'
} else {
// do the same thing with your default value
}
涉及不必要的代码重复,或者
which involves needlessly duplicating code, or
var unwrappedValue
if let value = optionalValue {
unwrappedValue = value
} else {
unwrappedValue = defaultValue
}
要求unwrappedValue
不是常数.
Scala的Option monad(与Swift的Optional的想法基本相同)具有用于此目的的方法getOrElse
:
Scala's Option monad (which is basically the same idea as Swift's Optional) has the method getOrElse
for this purpose:
val myValue = optionalValue.getOrElse(defaultValue)
我错过了什么吗? Swift已经有一种紧凑的方式来做到这一点吗?否则,是否有可能在Optional扩展名中定义getOrElse
?
Am I missing something? Does Swift have a compact way of doing that already? Or, failing that, is it possible to define getOrElse
in an extension for Optional?
推荐答案
更新
Apple现在添加了一个合并运算符:
Apple has now added a coalescing operator:
var unwrappedValue = optionalValue ?? defaultValue
在这种情况下,三元运算符是您的朋友
The ternary operator is your friend in this case
var unwrappedValue = optionalValue ? optionalValue! : defaultValue
您还可以为Optional枚举提供自己的扩展名:
You could also provide your own extension for the Optional enum:
extension Optional {
func or(defaultValue: T) -> T {
switch(self) {
case .None:
return defaultValue
case .Some(let value):
return value
}
}
}
然后您可以做:
optionalValue.or(defaultValue)
但是,我建议您坚持使用三元运算符,因为其他开发人员将无需研究or
方法就可以更快地了解这一点
However, I recommend sticking to the ternary operator as other developers will understand that much more quickly without having to investigate the or
method
注意:我启动了模块,以便在
Note: I started a module to add common helpers like this or
on Optional
to swift.
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