C#,使用运算符对通过属性获取的值(get)

// m_age is a private int in the class Employee
public int Age
{
    get {return m_age;}
    set {m_age = value;}
}

现在,当我这样做

static void Main()
{
    Employee e = new Employee(age: 28); // Create new Employee
    System.Console.WriteLine("Age: {0}",e.Age); // Prints 28

    // Now increase age by 1
    ++e.Age;

    System.Console.WriteLine("Age: {0}",e.Age); // Prints 29
}

为什么呢

++e.Age;

工作？我做了一些搜索,发现了Properties – by value or by reference?

这篇文章有一个答案：

Technically it’s always by value,but you have to understand what is being passed. Since it’s a reference type,you are passing a reference back (but by value).

Hope that makes sense. You always pass the result back by value,but if the type is a reference you are passing the reference back by value,which means you can change the object,but not which object it refers to.

(我对值类型和引用类型有很好的理解,因此我的困惑).

现在,如果确实如此

e.Age

返回m_age的副本(int是值类型),我们不会将增量应用于副本吗？

或者……以下是真的吗？

++e.Age;

完全相同/被翻译成

e.Age = e.Age + 1

只有这样

++e.Age;

返回一个值(e.Age增加后的值),而

e.Age = e.Age + 1

是一个赋值,并不返回一个值(例如C会做).