老问题，但这符合我的需要(Swift 5.x):

print(type(of: myObjectName))

//: 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

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.

以下是我推荐的两种方法:

if let thisShape = aShape as? Square 

Or:

aShape.isKindOfClass(Square)

下面是一个详细的例子:

class Shape { }
class Square: Shape { } 
class Circle: Shape { }

var aShape = Shape()
aShape = Square()

if let thisShape = aShape as? Square {
    println("Its a square")
} else {
    println("Its not a square")
}

if aShape.isKindOfClass(Square) {
    println("Its a square")
} else {
    println("Its not a square")
}

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))

如果您只是需要检查变量是否属于X类型，或者它是否符合某种协议，那么您可以使用is或as?具体如下:

var unknownTypeVariable = …

if unknownTypeVariable is <ClassName> {
    //the variable is of type <ClassName>
} else {
    //variable is not of type <ClassName>
}

这相当于Obj-C中的isKindOfClass。

这等价于conformsToProtocol或isMemberOfClass

var unknownTypeVariable = …

if let myClass = unknownTypeVariable as? <ClassName or ProtocolName> {
    //unknownTypeVarible is of type <ClassName or ProtocolName>
} else {
    //unknownTypeVariable is not of type <ClassName or ProtocolName>
}