具有相同方法签名的多个接口 [英] Multiple Interface with same method signature

问题描述

我们不在这里做作业！



我想知道，如果我的应用程序中有多个接口具有相同的方法签名，如下所示:

We do not do your homework here!

Hi,

I wanted to know, if I have more than 1 interface in a application having same method signature as below:

public interface ISum1
    {
        int Sum(int a, int b);
    }

public interface ISum2
    {
        int Sum(int a, int b);
    }

现在，我已经在类中实现了这些接口，如下所示:

Now I have implemented these interfaces in to class as below:

public class MainClass : ISum1, ISum2
    {

        public int Sum(int a, int b)
        {
            return a + b;
        }

    }

它与接口ISum1的实现配合良好，如果我想在类中的ISum2中编写方法的定义，现在允许我将相同的方法放入相同的签名中，我该如何进一步进行操作....请帮助.

VJ

It is working fine with implementation of Interface ISum1, if I wanted to write the definition for method in ISum2 in a class, it is now allowing me to put the same method with same signatures, how can I proceed further.... Please help.

VJ

推荐答案

您需要显式实现接口

看到这里..

http://msdn.microsoft.com/en-us/library/aa288461％28v = vs.71％29.aspx [ ^ ]

也请看一下这个博客，这正是您要尝试做的事情

http://sandblogaspnet.blogspot.co.uk/2008/05/Implementation-two-interface-having-same.html [ ^ ]

You need to implement the interfaces explicitly

See here..

http://msdn.microsoft.com/en-us/library/aa288461%28v=vs.71%29.aspx[^]

Have a look at this blog as well, which is exactly what you are trying to do

http://sandblogaspnet.blogspot.co.uk/2008/05/implementing-two-interface-having-same.html[^]

使用接口名称对方法进行显式限定会有所帮助

Explicity qualifying the methods with Interface names would help

public interface ISum1
{
    int Sum(int a, int b);
}

public interface ISum2
{
    int Sum(int a, int b);
}
public class MainClass : ISum1, ISum2
{
    int ISum1.Sum(int a, int b)
    {
        
        return a + b;
    }
    int ISum2.Sum(int a, int b)
    {
        return a * b;
    }
}
public class Test
{
    
    public void  TestMethod()
    {
        int result;
        ISum1 is1 = new MainClass();
        result = is1.Sum(2,3);
        ISum2 is2 = new MainClass();
        result = is2.Sum(2,3);

       
    }


}

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