为什么在继承中调用固定数据类型参数的对象class参数方法linseed

Question

案例一：

public class BaseClass
{
    public virtual void Print(int i)
    {
        Console.WriteLine("BaseClass Print(int)");
    }
}

public class DerivedClass : BaseClass
{
    public override void Print(int i)
    {
        Console.WriteLine("DerivedClass Print(int)");
    }

    public void Print(object obj)
    {
        Console.WriteLine("DerivedClass Print(object)");
    }
}
static void Main(string[] args)
{
    DerivedClass objDerivedClass = new DerivedClass();
    int i = 10;
    objDerivedClass.Print(i);
}

输出为 DerivedClass Print(object).

案例二：

public class SomeClass
{
    public void Print(int i)
    {
        Console.WriteLine("DerivedClass Print(int)");
    }

    public void Print(object obj)
    {
        Console.WriteLine("DerivedClass Print(object)");
    }
}

static void Main(string[] args)
{
    SomeClass objSomeClass = new SomeClass();
    int i = 10;
    objSomeClass.Print(i);
}

输出为 DerivedClass Print(int)。

调用objDerivedClass.Print(i);方法后，输出为DerivedClass Print(object)。我不明白为什么调用方法 Print(object obj) 而不是 Print(int i).

如果DerivedClass不继承BaseClass class则输出为DerivedClass Print(int).

请解释......

Answer 1

这是重载解析如何与继承一起工作：

包含 override 修饰符的函数已从候选集中排除。
既然可以使用带object参数的函数，那么将在base中声明的函数从候选集合中移除。

获胜者是带有 object 参数的函数。

根据Eric Lippert (from Microsoft)：

This is by design and for a good reason. This design helps prevent the Brittle Base Class problem. C# was designed to make it easier and safer to write "versioned" components, and this rule is a big part of that.

当不使用继承时，两个函数都是候选的，使用更具体的那个。获胜者是带有 int 参数的函数。

为什么在继承中调用固定数据类型参数的对象class参数方法linseed

Why call object class parameter method linseed of fixed data type parameter in inheritance

c#

inheritance

overload-resolution