Python“或"等价于Swift?

问题描述

由于Swift很酷的行为，我一直在寻找Swift中的or等效项.

Due to Swift cool behaviors I was looking for an or equivalent in Swift.

类似这样的东西:

variable = value or default

我编码了我的:

func |<T>(a:T?, b:T) -> T {
    if let a = a {
        return a
    }
    return b
}

但是我想知道Swift中是否已经存在任何默认实现?

But I was wondering if any default implementation of this already exists in Swift?

多亏了答案，我在Swift书中找到了参考文献:

Thanks to answers I found the reference in the Swift book:

nil合并运算符(a ?? b)如果包含一个可选值a，则将其拆开；如果a为nil，则返回默认值b.表达式a始终是可选类型.表达式b必须与存储在a内部的类型匹配.

The nil coalescing operator (a ?? b) unwraps an optional a if it contains a value, or returns a default value b if a is nil. The expression a is always of an optional type. The expression b must match the type that is stored inside a.

nil合并运算符是以下代码的简写:

The nil coalescing operator is shorthand for the code below:

    a != nil ? a! : b

上面的代码使用三元条件运算符并强制展开(a!)，以在a不为nil时访问包装在a中的值，否则返回b. nil合并运算符提供了一种更简洁的方法，以简洁易读的形式封装此条件检查和展开. (源)

The code above uses the ternary conditional operator and forced unwrapping (a!) to access the value wrapped inside a when a is not nil, and to return b otherwise. The nil coalescing operator provides a more elegant way to encapsulate this conditional checking and unwrapping in a concise and readable form. (source)

推荐答案

使用??:

var optional:String?
var defaultValue:String = "VALUE"

var myString:String = optional ?? defaultValue

print(myString)

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