2016年9月更新

Swift 3.0:使用type(of:)，例如type(of: someThing)(因为dynamicType关键字已被删除)

2015年10月更新:

我更新了下面的例子到新的Swift 2.0语法(例如println替换为print, toString()现在是String())。

Xcode 6.3发布说明:

@nschum在评论中指出，Xcode 6.3发布说明显示了另一种方式:

使用时，类型值现在打印为完整的需求类型名 Println或字符串插值。

import Foundation

class PureSwiftClass { }

var myvar0 = NSString() // Objective-C class
var myvar1 = PureSwiftClass()
var myvar2 = 42
var myvar3 = "Hans"

print( "String(myvar0.dynamicType) -> \(myvar0.dynamicType)")
print( "String(myvar1.dynamicType) -> \(myvar1.dynamicType)")
print( "String(myvar2.dynamicType) -> \(myvar2.dynamicType)")
print( "String(myvar3.dynamicType) -> \(myvar3.dynamicType)")

print( "String(Int.self)           -> \(Int.self)")
print( "String((Int?).self         -> \((Int?).self)")
print( "String(NSString.self)      -> \(NSString.self)")
print( "String(Array<String>.self) -> \(Array<String>.self)")

输出:

String(myvar0.dynamicType) -> __NSCFConstantString
String(myvar1.dynamicType) -> PureSwiftClass
String(myvar2.dynamicType) -> Int
String(myvar3.dynamicType) -> String
String(Int.self)           -> Int
String((Int?).self         -> Optional<Int>
String(NSString.self)      -> NSString
String(Array<String>.self) -> Array<String>

Xcode 6.3更新:

你可以使用_stdlib_getDemangledTypeName():

print( "TypeName0 = \(_stdlib_getDemangledTypeName(myvar0))")
print( "TypeName1 = \(_stdlib_getDemangledTypeName(myvar1))")
print( "TypeName2 = \(_stdlib_getDemangledTypeName(myvar2))")
print( "TypeName3 = \(_stdlib_getDemangledTypeName(myvar3))")

并将其作为输出:

TypeName0 = NSString
TypeName1 = __lldb_expr_26.PureSwiftClass
TypeName2 = Swift.Int
TypeName3 = Swift.String

最初的回答:

在Xcode 6.3之前，_stdlib_getTypeName获取变量的类型名。伊万·斯维克(Ewan Swick)的博客有助于解读这些字符串:

例如:_TtSi代表Swift的内部Int类型。

Mike Ash有一篇很棒的博客文章涉及了同样的主题。

我找到了这个解决方案，希望对其他人也有用。 我创建了一个类方法来访问该值。请记住，这只适用于NSObject子类。但至少是一个干净整洁的解决方案。

class var className: String!{
    let classString : String = NSStringFromClass(self.classForCoder())
    return classString.componentsSeparatedByString(".").last;
}

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.

这里的许多答案都不支持最新的Swift(本文撰写时为Xcode 7.1.1)。

目前获取信息的方法是创建一个Mirror并对其进行询问。对于类名，它很简单:

let mirror = Mirror(reflecting: instanceToInspect)
let classname:String = mirror.description

关于对象的其他信息也可以从镜像中检索。详情见http://swiftdoc.org/v2.1/type/Mirror/。

斯威夫特4:

// "TypeName"
func stringType(of some: Any) -> String {
    let string = (some is Any.Type) ? String(describing: some) : String(describing: type(of: some))
    return string
}

// "ModuleName.TypeName"
func fullStringType(of some: Any) -> String {
    let string = (some is Any.Type) ? String(reflecting: some) : String(reflecting: type(of: some))
    return string
}

用法:

print(stringType(of: SomeClass()))     // "SomeClass"
print(stringType(of: SomeClass.self))  // "SomeClass"
print(stringType(of: String()))        // "String"
print(fullStringType(of: String()))    // "Swift.String"

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