从Xcode 6.0.1开始(至少，不确定他们什么时候添加的)，你原来的例子现在可以工作了:

class MyClass {
    var count = 0
}

let mc = MyClass()
mc.dynamicType === MyClass.self // returns `true`

更新:

要回答最初的问题，您实际上可以成功地将Objective-C运行时与普通Swift对象一起使用。

试试下面的方法:

import Foundation
class MyClass { }
class SubClass: MyClass { }

let mc = MyClass()
let m2 = SubClass()

// Both of these return .Some("__lldb_expr_35.SubClass"), which is the fully mangled class name from the playground
String.fromCString(class_getName(m2.dynamicType))
String.fromCString(object_getClassName(m2))
// Returns .Some("__lldb_expr_42.MyClass")
String.fromCString(object_getClassName(mc))

评论:我不明白@JérémyLapointe如何回答这个问题。使用type(of:)只能通过检查编译时信息来工作，即使实际类型是一个更特定的子类。现在在Swift 5.1中有一个更简单的方法来动态查询类型，而不需要像@Dash建议的那样诉诸dynamicType。有关我从哪里得到这些信息的更多详细信息，请参阅SE-0068:将Swift Self扩展到类成员和值类型。

Code

斯威夫特5.1

// Within an instance method context
Self.self

// Within a static method context
self

这允许使用Self作为引用包含类型(在结构、枚举和final类的情况下)或动态类型(在非final类的情况下)的简写。

解释

提案很好地解释了为什么这种方法改进了dynamicType:

Introducing Self addresses the following issues: dynamicType remains an exception to Swift's lowercased keywords rule. This change eliminates a special case that's out of step with Swift's new standards. Self is shorter and clearer in its intent. It mirrors self, which refers to the current instance. It provides an easier way to access static members. As type names grow large, readability suffers. MyExtremelyLargeTypeName.staticMember is unwieldy to type and read. Code using hardwired type names is less portable than code that automatically knows its type. Renaming a type means updating any TypeName references in code. Using self.dynamicType fights against Swift's goals of concision and clarity in that it is both noisy and esoteric. Note that self.dynamicType.classMember and TypeName.classMember may not be synonyms in class types with non-final members.

dynamicType。printClassName代码来自Swift书籍中的一个示例。我不知道如何直接获取自定义类名，但可以使用is关键字检查实例类型，如下所示。这个例子还展示了如何实现一个自定义的className函数，如果你真的想要类名作为字符串。

class Shape {
    class func className() -> String {
        return "Shape"
    }
}

class Square: Shape {
    override class func className() -> String {
        return "Square"
    }
}

class Circle: Shape {
    override class func className() -> String {
        return "Circle"
    }
}

func getShape() -> Shape {
    return Square() // hardcoded for example
}

let newShape: Shape = getShape()
newShape is Square // true
newShape is Circle // false
newShape.dynamicType.className() // "Square"
newShape.dynamicType.className() == Square.className() // true

注意:NSObject的子类已经实现了他们自己的className函数。如果您正在使用Cocoa，您可以使用这个属性。

class MyObj: NSObject {
    init() {
        super.init()
        println("My class is \(self.className)")
    }
}
MyObj()

从Xcode 6.0.1开始(至少，不确定他们什么时候添加的)，你原来的例子现在可以工作了:

class MyClass {
    var count = 0
}

let mc = MyClass()
mc.dynamicType === MyClass.self // returns `true`

更新:

要回答最初的问题，您实际上可以成功地将Objective-C运行时与普通Swift对象一起使用。

试试下面的方法:

import Foundation
class MyClass { }
class SubClass: MyClass { }

let mc = MyClass()
let m2 = SubClass()

// Both of these return .Some("__lldb_expr_35.SubClass"), which is the fully mangled class name from the playground
String.fromCString(class_getName(m2.dynamicType))
String.fromCString(object_getClassName(m2))
// Returns .Some("__lldb_expr_42.MyClass")
String.fromCString(object_getClassName(mc))

//: Playground - noun: a place where people can play

import UIKit

class A {
    class func a() {
        print("yeah")
    }

    func getInnerValue() {
        self.dynamicType.a()
    }
}

class B: A {
    override class func a() {
        print("yeah yeah")
    }
}

B.a() // yeah yeah
A.a() // yeah
B().getInnerValue() // yeah yeah
A().getInnerValue() // yeah

老问题，但这符合我的需要(Swift 5.x):

print(type(of: myObjectName))