Swift 3 从 Int 到 String 的转换
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Swift 3 conversion from Int to String
提问by Don Giovanni
In Swift 3, the Stringstructure does not seem to have an initializer init(_: Int)that will allow the conversion from an Intto a String. My question is why does let i = String(3)work? What Stringmethod or initializer is it calling? Thanks.
在 Swift 3 中，该String结构似乎没有init(_: Int)允许从 an 转换Int为 a的初始化程序String。我的问题是为什么let i = String(3)有效？String它调用什么方法或初始化程序？谢谢。
采纳答案by Alexander - Reinstate Monica
It's calling init(_:)(or init(_:)for UnsignedInteger) arguments of the Stringclass.
它正在调用init(_:)（或init(_:)for UnsignedInteger）String类的参数。
Rather than defining separate initializers for Int, Int64, Int32, Int16, Int8, UInt, UInt64, UInt32, UInt16, and UInt8, Apple made two generic initializers: one for SignedIntegertypes, and one for UnsignedIntegertypes.
Apple 没有为Int, Int64, Int32, Int16, Int8, UInt, UInt64, UInt32, UInt16, and定义单独的初始值设定项，而是UInt8制作了两个通用初始值设定项：一个用于SignedInteger类型，另一个用于UnsignedInteger类型。
回答by Chuck Smith
For anyone just looking to convert an int to string in Swift 3:
对于任何只想在 Swift 3 中将 int 转换为字符串的人：
let text = "\(myInt)"
回答by Satheeshwaran
For people who want to convert optional Integers to Strings on Swift 3,
对于想要在 Swift 3 上将可选整数转换为字符串的人，
String(describing:YourInteger ?? 0)
回答by Mercedes
I saw this solution to somebody, thank you, to that person, I don't remember who.
我在某人那里看到了这个解决方案，谢谢你，对于那个人，我不记得是谁。
infix operator ???: NilCoalescingPrecedence
public func ???<T>(optional: T?, defaultValue: @autoclosure () -> String) -> String {
switch optional {
case let value?: return String(describing: value)
case nil: return defaultValue()
}
}
For example:
例如：
let text = "\(yourInteger ??? "0")"
