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C# 小测试(一):类成员初始化与构造函数执行的顺序

2008-07-12 20:25 Anders Cui 阅读(...) 评论(...) 编辑 收藏

看看下面这段代码,你觉得它会输出什么呢?

    class Foo
    {
        public Foo(string s)
        {
            Console.WriteLine("Foo constructor: {0}", s);
        }
 
        public void Bar(){}
    }
 
    class Base
    {
        readonly Foo baseFoo = new Foo("Base initializer");
        public Base()
        {
            Console.WriteLine("Base constructor");
        }
    }
 
    class Derived : Base
    {
        readonly Foo derivedFoo = new Foo("Derived initializer.");
        public Derived()
        {
            Console.WriteLine("Derived constructor");
        }
    }
 
    class Program
    {
        static void Main(string[] args)
        {
            new Derived();
        }
    }

先猜一下吧,似乎应该是“Base initializer, Base constructor, Derived initializer, Derived constructor”。

事实上,应当是先执行类成员的初始化,顺序是从derived到base,然后是两个构造函数,顺序是从base从derived。

这种方式是很有意义的,在类继承体系中层次较深的类(离System.Object较远)将依赖于较浅的类(离System.Object较近)。但是很多人会相信调用的顺序应当等价于下面的伪代码:

// 期望的顺序
BaseConstructor()
{
    ObjectConstructor();
    baseFoo = new Foo("Base initializer");
    Console.WriteLine("Base constructor");
}
DerivedConstructor()
{
    BaseConstructor();
    derivedFoo = new Foo("Derived initializer");
    Console.WriteLine("Derived constructor");
}

而实际情况则是:

 // 实际的顺序
BaseConstructor()
{
    baseFoo = new Foo("Base initializer");
    ObjectConstructor();
    Console.WriteLine("Base constructor");
}
DerivedConstructor()
{
    derivedFoo = new Foo("Derived initializer");
    BaseConstructor();
    Console.WriteLine("Derived constructor");
} 

那么,这样处理是为什么呢?

...

...

...

我们来看一下,如果代码按期望的顺序(第一段伪代码)执行,会产生什么问题:

class Base
{
    public Base()
    {
        Console.WriteLine("Base constructor");
        if (this is Derived) (this as Derived).DoIt();
        // 如果是在创建Derived类的实例,就会遭遇null。
        Blah();
        // 如果是在创建MoreDerived类的实例,就会遭遇null。
    }
 
    public virtual void Blah() { }
}
 
class Derived : Base
{
    readonly Foo derivedFoo = new Foo("Derived initializer");
    public DoIt()
    {
        derivedFoo.Bar();
    }
}
 
class MoreDerived : Derived
{
    public override void Blah() { DoIt(); }
} 

看Base类的构造函数,如果按期望的顺序执行,那么在Base方法执行时,Derived类的实例成员并没有得到初始化,此时就会有NullReference异常了。

而按照实际执行的顺序,所有的实例成员都能确保被完整地初始化:)

当然了,如果readonly字段是在构造函数中进行的,那么上面的确保机制就不复存在了。

参考:

Why Do Initializers Run In The Opposite Order As Constructors? Part One 

Why Do Initializers Run In The Opposite Order As Constructors? Part Two