在c#中跨越继承边界重载？

概述看完这篇 article& article – 我很困惑. 它说 ： If there are two methods at different levels of the hierarchy, the “deeper” one will be chosen first, even if it isn’t a “better function member” for the call. 还 – I 看完这篇 article& article – 我很困惑.

它说 ：

If there are two methods at different levels of the hIErarchy,the
“deeper” one will be chosen first,even if it isn’t a “better function
member” for the call.

还 –

It turns out that if you overrIDe a base class method in a child
class,that doesn’t count as declaring it.

现在让我们回到我的问题：

情况1

public class Base     {           public virtual voID  Foo(int x)  { "1".Dump();}     }    public class Child : Base     {          public voID Foo(object x) { "3".Dump();}            public overrIDe voID  Foo(int x)  { "2".Dump();}     }voID Main(){    Child c = new Child();    c.Foo(10); //emits 3}

好的.根据文章

“deeper” one will be chosen first,even if it isn’t a “better function. and it doesn’t count the overrIDe…

所以它是正确的,程序发出“3”. (执行Foo(对象x))

让我们改变1行的顺序：

案例2

public class Base         {                 public virtual voID  Foo(int x)  { "1".Dump();}                 public voID Foo(object x) { "3".Dump();} //<line being moved here         }        public class Child : Base         {              public overrIDe voID  Foo(int x)  { "2".Dump();}         }    voID Main()    {        Child c = new Child();        c.Foo(10); //emits 2 !!!!    }

现在它发出“2”.

现在让我们将所有int更改为object,将所有对象更改为int：

案例3

public class Base    {      public virtual voID  Foo(object x)  { "1".Dump();}      public voID Foo(int x) { "3".Dump();}     }    public class Child : Base    {         public overrIDe voID  Foo(object x)  { "2".Dump();}    }voID Main(){    Child c = new Child();    c.Foo(1); //emits "3"}

问题：

问题#1：在案例2中,Child从父亲继承了Foo(对象x)并且他也覆盖了一个方法.

但我们只是这样说：

It turns out that if you overrIDe a base class method in a child
class,that doesn’t count as declaring it

???

事实上,我们还没有宣布继承的功能……所以这种情况下的规则是什么？

问题#2：在案例3中,Child从父亲那里继承了Foo(int x)并且他也覆盖了一个方法.

但现在,他选择了父亲的功能……

似乎覆盖只有在完全匹配时才会获胜.

再次,这种情况下的规则是什么？

解决方法 请参阅类型T中名称为N的 member lookup process(在您的案例中,类型为Child的成员Foo)：

首先,构造在T中声明为N的所有可访问(第3.5节)成员的集合,并构造T的基本类型(第7.3.1节)：

virtual voID Foo(int x) // BasevoID Foo(object x) // BaseoverrIDe voID Foo(int x) // Child

包含覆盖修饰符的声明将从集合中排除.

virtual voID Foo(int x) // BasevoID Foo(object x) // Base

参数有整数类型.所以,这里最好的选择是(参数类型匹配参数类型)

virtual voID Foo(int x) // Base

而这种方法叫做.但它是虚方法.由于virtual method invocation机制而调用它：

For every virtual method declared in or inherited by a class,there
exists a most derived implementation of the method with respect to
that class. The most derived implementation of a virtual method M with
respect to a class R is determined as follows:

If R contains the introducing virtual declaration of M,then this is the most derived implementation of M. Otherwise,if R contains an overrIDe of M,the most derived implementation of M with respect to R is the same as the most derived implementation of M with respect to
the direct base class of R.

什么是关于类Child的虚拟voID Foo(int x)方法的最多派生实现？是的

overrIDe voID Foo(int x) // Child

这是被调用的.第三个样本中应用了相同的规则.但是当重写方法删除后剩下两个选项时,最佳选择(由于参数类型)是非虚方法.

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