Swift 4:以非包装的true或false或包装为nil的形式输出Bool可选件吗?

问题描述

基本上我想将以下值打印为"true"，"false"或"nil".如果我尝试仅使用包装的值，则会得到我想要的"nil"，但会得到不需要的"Optional(true)"或"Optional(false).如果我强制拆开该值，而该值为nil，则会出现致命错误.我尝试了下面的代码，因为我发现它适用于Strings，但是因为"nil"不是Bool类型，所以不被接受.对此有任何解决方案吗?

Basically I want to print the below value as either "true", "false" or "nil". If I try just using the wrapped value I get the "nil" I want but gain the unwanted "Optional(true)" or "Optional(false). If I force unwrap the value and the value is nil I get a fatal error. I tried the below code as I've seen it work for Strings but because "nil" is not of type Bool it's not accepted. Any solutions to this?

var isReal: Bool?
String("Value is \(isReal ?? "nil")")

我正在导出到一个csv文件，知道此值是true还是false或尚未检查该值很有用.

I'm exporting to a csv file and it's useful to know if this value is true or false or hasn't been checked yet.

推荐答案

如果您想要单线:

var isReal: Bool?
let s = "Value is \(isReal.map { String($0) } ?? "nil")"
print(s) // "Value is nil"

或带有自定义扩展名:

extension Optional {
    var descriptionOrNil: String {
        return self.map { "\($0)" } ?? "nil"
    }
}

你可以做

var isReal: Bool?
let s = "Value is \(isReal.descriptionOrNil)"
print(s) // "Value is nil"

，这适用于所有可选类型.

and this works with all optional types.

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