Question

所以，我想做类似的事情：

 Class A
{
    public static A operator + (A left, A right)
    {
        //Calculations
        return new A();
    }
}

Class B : A
{
    public static B operator + (B left, B right)
    {
        //Calculations
        base.+(left,right);
        return new A();
    }
}

修改

编辑：

在没有类型转换为基类的情况下实现这一点会很好，因为它变得丑陋：

public static Sums operator -(Sums operand1, Sums operand2)
{
   Sums res = new Sums(operand1._parent);
   res = (Sums) (operand1 - (srp_AccountPension.PensionSums)operand2);
   return res;

 }

我不喜欢向基类进行反向转换，也不喜欢派生类...

Answer 1

只需简单地调用+运算符即可获得两个参数。其中一个类型为A（左侧应始终为类型，即重载运算符的容器）

public class A
{
    public static A operator + (A left, A right)
    {
        //Calculations
        Console.WriteLine ("base A was called");
        return new A();
    }
}

public class B : A
{
    public static B operator + (B left, B right)
    {
        //Calculations
        A res = right + (A)left;
        return new B();
    }
}

现在正在呼叫

var B = new B() + new B();

将打印：

base A was called

Answer 2

重新定义这样的静态方法看起来并不好看。我的建议是使用模板方法设计模式，这使得+运算符成为模板方法，

class A
{
    public virtual A Plus(A right)
    {
        return new A();
    }

    public static A operator + (A left, A right)
    {
        //Calculations and handle left == null case.
        return left.Plus(right);
    }
}

class B : A
{
    public override B Plus(B right)
    {
        //Calculations
        base.Plus(right);
        return new A();
    }
}

如何调用基类运算符？

2 个答案: