你可以通过如下方式获取类名:

class Person {}
String(describing: Person.self)

如果你不喜欢这个乱七八糟的名字，你可以口述你自己的名字:

@objc(CalendarViewController) class CalendarViewController : UIViewController {
    // ...
}

然而，从长远来看，最好是学会解析这个混乱的名称。格式是标准的、有意义的，不会改变。

更新到swift 5

我们可以通过String初始化器使用实例变量获得类型名的详细描述，并创建某个类的新对象

例如，print(String(description: type(of: object)))。其中object可以是一个实例变量，如数组，字典，Int, NSDate等。

因为NSObject是大多数Objective-C类层次结构的根类，你可以尝试为NSObject做一个扩展，以获得NSObject的每个子类的类名。是这样的:

extension NSObject {
    var theClassName: String {
        return NSStringFromClass(type(of: self))
    }
}

或者您可以创建一个静态函数，其参数类型为Any(所有类型隐式遵循的协议)，并将类名返回为String。是这样的:

class Utility{
    class func classNameAsString(_ obj: Any) -> String {
        //prints more readable results for dictionaries, arrays, Int, etc
        return String(describing: type(of: obj))
    }
} 

现在你可以这样做:

class ClassOne : UIViewController{ /* some code here */ }
class ClassTwo : ClassOne{ /* some code here */ }

class ViewController: UIViewController {

    override func viewDidLoad() {
        super.viewDidLoad()

        // Get the class name as String
        let dictionary: [String: CGFloat] = [:]
        let array: [Int] = []
        let int = 9
        let numFloat: CGFloat = 3.0
        let numDouble: Double = 1.0
        let classOne = ClassOne()
        let classTwo: ClassTwo? = ClassTwo()
        let now = NSDate()
        let lbl = UILabel()

        print("dictionary: [String: CGFloat] = [:] -> \(Utility.classNameAsString(dictionary))")
        print("array: [Int] = [] -> \(Utility.classNameAsString(array))")
        print("int = 9 -> \(Utility.classNameAsString(int))")
        print("numFloat: CGFloat = 3.0 -> \(Utility.classNameAsString(numFloat))")
        print("numDouble: Double = 1.0 -> \(Utility.classNameAsString(numDouble))")
        print("classOne = ClassOne() -> \((ClassOne).self)") //we use the Extension
        if classTwo != nil {
            print("classTwo: ClassTwo? = ClassTwo() -> \(Utility.classNameAsString(classTwo!))") //now we can use a Forced-Value Expression and unwrap the value
        }
        print("now = Date() -> \(Utility.classNameAsString(now))")
        print("lbl = UILabel() -> \(String(describing: type(of: lbl)))") // we use the String initializer directly

    }
}

而且，一旦我们可以获得类名String，我们就可以实例化该类的新对象:

// Instantiate a class from a String
print("\nInstantiate a class from a String")
let aClassName = classOne.theClassName
let aClassType = NSClassFromString(aClassName) as! NSObject.Type
let instance = aClassType.init() // we create a new object
print(String(cString: class_getName(type(of: instance))))
print(instance.self is ClassOne)

也许这能帮助一些人!

在我的例子中，String(description: self)返回如下内容:

< My_project。ExampleViewController: 0x10b2bb2b0 >

但我想在Android上有类似getSimpleName的东西。

所以我创建了一个小扩展:

extension UIViewController {

    func getSimpleClassName() -> String {
        let describing = String(describing: self)
        if let dotIndex = describing.index(of: "."), let commaIndex = describing.index(of: ":") {
            let afterDotIndex = describing.index(after: dotIndex)
            if(afterDotIndex < commaIndex) {
                return String(describing[afterDotIndex ..< commaIndex])
            }
        }
        return describing
    }

}

现在它返回:

ExampleViewController

扩展NSObject而不是UIViewController也可以。上面的函数也是故障安全的:)

斯威夫特5

 NSStringFromClass(CustomClass.self)

斯威夫特5:

方式1:

print("Class: \(String(describing: self)), Function: \(#function), line: \(#line)")

输出:

Class: <Test.ViewController: 0x7ffaabc0a3d0>, Function: viewDidLoad(), line: 15

方式2:

print("Class: \(String(describing: type(of: self))), Function: \(#function), line: \(#line)")

输出:

Class: ViewController, Function: viewDidLoad(), line: 16