你可以试试这种方法:

self.classForCoder.description()

要在Swift 4中获得类型名为字符串(我没有检查早期版本)，只需使用字符串插值:

"\(type(of: myViewController))"

你可以在类型本身上使用.self，在实例上使用type(of:_)函数:

// Both constants will have "UIViewController" as their value
let stringFromType = "\(UIViewController.self)"
let stringFromInstance = "\(type(of: UIViewController()))"

斯威夫特5

 NSStringFromClass(CustomClass.self)

斯威夫特3.0

字符串(描述:MyViewController.self)

在Swift 4.1和现在的Swift 4.2中:

import Foundation

class SomeClass {
    class InnerClass {
        let foo: Int

        init(foo: Int) {
            self.foo = foo
        }
    }

    let foo: Int

    init(foo: Int) {
        self.foo = foo
    }
}

class AnotherClass : NSObject {
    let foo: Int

    init(foo: Int) {
        self.foo = foo
        super.init()
    }
}

struct SomeStruct {
    let bar: Int

    init(bar: Int) {
        self.bar = bar
    }
}

let c = SomeClass(foo: 42)
let s = SomeStruct(bar: 1337)
let i = SomeClass.InnerClass(foo: 2018)
let a = AnotherClass(foo: 1<<8)

如果你周围没有实例:

String(describing: SomeClass.self) // Result: SomeClass
String(describing: SomeStruct.self) // Result: SomeStruct
String(describing: SomeClass.InnerClass.self) // Result: InnerClass
String(describing: AnotherClass.self) // Result: AnotherClass

如果你有一个实例:

String(describing: type(of: c)) // Result: SomeClass
String(describing: type(of: s)) // Result: SomeStruct
String(describing: type(of: i)) // Result: InnerClass
String(describing: type(of: a)) // Result: AnotherClass

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:)