c# - 为什么 f(double) 对于 f(long) 比 f(long, int=0) 更匹配？

Question

我的一位同事发现了这个奇怪的案例。我在这里发布一个简单的例子:

using System;

namespace Test
{
    public class F
    {
        public void f(double d) { Console.WriteLine("public void F.f(double d)"); }
        public virtual void f(long l, int q = 0) { Console.WriteLine("public virtual void F.f(long l, int q = 0)"); }
    }

    public class FDerived : F
    {
        public override void f(long l, int q) { Console.WriteLine("public override FDerived.f(long l, int q)"); }
        public void g() { f(2L); }
    }

    public class G : FDerived
    {
        public void h1() { F fd = new FDerived(); fd.f(2L); }
        public void h2() { FDerived fd = new FDerived(); fd.f(2L); }
    }

    class Program
    {
        static void Main(string[] args)
        {
            new FDerived().g();
            new G().h1();
            new G().h2();
            Console.ReadLine();
        }
    }
}

例子的输出是:

public void F.f(double d)
public override FDerived.f(long l, int q)
public void F.f(double d)

我不明白这有什么意义。

为什么对于 f(long)，f(double) 比 f(long, int=0) 更匹配？为什么它取决于“fd”的类型？!

Answer 1

我这里没有 C# 规范，但是可选参数值不会被覆盖的方法继承。

尝试更改 FDerived

public override void f(long l, int q)

到

public override void f(long l, int q = 0)

它会按预期工作。

请注意，这可以在一个简单的示例中显示:

public class F
{
    public virtual void f(long l, int q = 0) { Console.WriteLine("public virtual void F.f(long l, int q = 0)"); }
}

public class FDerived : F
{
    public override void f(long l, int q) { Console.WriteLine("public override FDerived.f(long l, int q)"); }
}

// Doesn't compile: No overload for method 'f' takes 1 arguments
new FDerived().f(5L);

显然这确实编译:

new F().f(5L);

甚至这个:

F obj = new FDerived();
obj.f(5L);

(这个会输出public override FDerived.f(long l, int q))