是否有一种方法可以在swift中打印变量的运行时类型?例如:

var now = NSDate()
var soon = now.dateByAddingTimeInterval(5.0)

println("\(now.dynamicType)") 
// Prints "(Metatype)"

println("\(now.dynamicType.description()")
// Prints "__NSDate" since objective-c Class objects have a "description" selector

println("\(soon.dynamicType.description()")
// Compile-time error since ImplicitlyUnwrappedOptional<NSDate> has no "description" method

在上面的例子中,我正在寻找一种方法来显示变量“soon”的类型是ImplicitlyUnwrappedOptional<NSDate>,或至少NSDate!


当前回答

不完全是你想要的,但你也可以检查Swift类型的变量类型,像这样:

let object: AnyObject = 1

if object is Int {
}
else if object is String {
}

为例。

其他回答

Xcode 7.3.1, Swift 2.2:

字符串(instanceToPrint.self) .componentsSeparatedByString .last(“。”)

在lldb beta 5中,你可以通过以下命令查看对象的类:

fr v -d r shipDate

输出如下:

(DBSalesOrderShipDate_DBSalesOrderShipDate_ *) shipDate = 0x7f859940

展开的命令是这样的:

帧变量(打印帧变量)-d run_target(展开动态类型)

需要知道的一点是,使用“Frame Variable”来输出变量值可以确保不执行任何代码。

你可以使用reflect来获取关于对象的信息。 例如对象类的名称:

var classname = reflect(now).summary

在Xcode 8, Swift 3.0

步骤:

1. 获取类型:

选项1:

let type : Type = MyClass.self  //Determines Type from Class

选项2:

let type : Type = type(of:self) //Determines Type from self

2. 转换类型为字符串:

let string : String = "\(type)" //String

我目前的Xcode版本是6.0 (6A280e)。

import Foundation

class Person { var name: String; init(name: String) { self.name = name }}
class Patient: Person {}
class Doctor: Person {}

var variables:[Any] = [
    5,
    7.5,
    true,
    "maple",
    Person(name:"Sarah"),
    Patient(name:"Pat"),
    Doctor(name:"Sandy")
]

for variable in variables {
    let typeLongName = _stdlib_getDemangledTypeName(variable)
    let tokens = split(typeLongName, { $0 == "." })
    if let typeName = tokens.last {
        println("Variable \(variable) is of Type \(typeName).")
    }
}

输出:

Variable 5 is of Type Int.
Variable 7.5 is of Type Double.
Variable true is of Type Bool.
Variable maple is of Type String.
Variable Swift001.Person is of Type Person.
Variable Swift001.Patient is of Type Patient.
Variable Swift001.Doctor is of Type Doctor.