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
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) // Base void Foo(object x) // Base override 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) // Base void 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
