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

编辑:在Swift 1.2 (Xcode 6.3)中引入了一个新的toString函数。

你现在可以使用.self和使用.dynamicType打印任意类型的需求类型:

struct Box<T> {}

toString("foo".dynamicType)            // Swift.String
toString([1, 23, 456].dynamicType)     // Swift.Array<Swift.Int>
toString((7 as NSNumber).dynamicType)  // __NSCFNumber

toString((Bool?).self)                 // Swift.Optional<Swift.Bool>
toString(Box<SinkOf<Character>>.self)  // __lldb_expr_1.Box<Swift.SinkOf<Swift.Character>>
toString(NSStream.self)                // NSStream

试着调用你的类。self和youobject . dynamictype。

参考:https://devforums.apple.com/thread/227425。

其他回答

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。

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.

请看看下面的代码片段,让我知道你是否在寻找下面这样的东西。

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

var nowDataType = Mirror(reflecting: now)
print("Now is of type: \(nowDataType.subjectType)")

var soonDataType = Mirror(reflecting: soon)
print("Soon is of type: \(soonDataType.subjectType)")

Swift 3.0, Xcode 8

使用下面的代码,您可以向实例请求其类。你也可以比较两个实例,是否具有相同的类。

// CREATE pure SWIFT class
class MySwiftClass {
    var someString : String = "default"
    var someInt    : Int = 5
}

// CREATE instances
let firstInstance = MySwiftClass()
let secondInstance = MySwiftClass()
secondInstance.someString = "Donald"
secondInstance.someInt = 24

// INSPECT instances
if type(of: firstInstance) === MySwiftClass.self {
    print("SUCCESS with ===")
} else {
    print("PROBLEM with ===")
}

if type(of: firstInstance) == MySwiftClass.self {
    print("SUCCESS with ==")
} else {
    print("PROBLEM with ==")
}

// COMPARE CLASS OF TWO INSTANCES
if type(of: firstInstance) === type(of: secondInstance) {
    print("instances have equal class")
} else {
    print("instances have NOT equal class")
}

这就是如何获得对象或type的类型字符串,它是一致的,并考虑到对象定义属于哪个模块或嵌套在哪个模块中。适用于Swift 4.x。

@inline(__always) func typeString(for _type: Any.Type) -> String {
    return String(reflecting: type(of: _type))
}

@inline(__always) func typeString(for object: Any) -> String {
    return String(reflecting: type(of: type(of: object)))
}

struct Lol {
    struct Kek {}
}

// if you run this in playground the results will be something like
typeString(for: Lol.self)    //    __lldb_expr_74.Lol.Type
typeString(for: Lol())       //    __lldb_expr_74.Lol.Type
typeString(for: Lol.Kek.self)//    __lldb_expr_74.Lol.Kek.Type
typeString(for: Lol.Kek())   //    __lldb_expr_74.Lol.Kek.Type