【Swift4.2】循环引用的例子讲解

class Example{

    private var name = "123";

    /// 正确用法1
    lazy private var rightClosure1:()->() = { [unowned self] in
        print("name is \(self.name)")
    }

    /// 正确用法2
    lazy private var rightClosure2:()->() = { [weak self] in
        print("name is \(self?.name ?? "")")
    }

    /// 函数用法不存在该问题
    private func commonFunction(){
        print("name is \(self.name)")
    }

    /// 错误用法
    lazy private var errorCycleClosure1:()->() = {
        print("name is \(self.name)")
    }
    
    /// 错误用法（函数赋值给errorCycleClosure2，errorCycleClosure2变成一个闭包对象被引用，函数是这个闭包的实现）
    lazy private var errorCycleClosure2 = commonFunction;

    func startRight1() {
        rightClosure1()
    }

    func startRight2() {
        rightClosure2()
    }

    func startRight3() {
        commonFunction()
    }
    
    func startError1() {
        errorCycleClosure1()
    }
    
    //需要特殊警惕的情况，一旦函数被赋值给一个被引用的对象后，也会出现循环引用的情况
    func startError2() {
        errorCycleClosure2()
    }

    deinit{
        print("Example deinit")
    }

}

var example:Example? = Example()
//调用正确例子1，example=nil后能正常销毁
//example?.startRight1();
//调用正确例子2，example=nil后能正常销毁
//example?.startRight2();
//调用正确例子3，example=nil后能正常销毁（函数不存在循环引用问题）
//example?.startRight3();
//调用错误例子1，example=nil后没有调用deinit，说明有残留成员变量
//example?.startError1();
//调用错误例子2，example=nil后没有调用deinit，说明有残留成员变量
example?.startError2();
example=nil;