Interface and c# using Add Program...pls Solve this problem

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1.00/5 (2 votes)

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//interface

C#

interface add
{
  int add(int a, int b);
}

C#

class adds : add
{
  public int c;
  public int add(int a, int b)
  {
    c = a + b;
    return c;
  }
}

//defult.cs

C#

adds c1 = new adds();
add val_add = new adds();
val_add.add(15, 15);
Response.Write(c1.c);

Posted 3-Aug-15 21:21pm

Member 10720577

Updated 3-Aug-15 21:31pm

Kornfeld Eliyahu Peter

v2

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Sergey Alexandrovich Kryukov 4-Aug-15 3:31am

What problem, for goodness sake?!
—SA

2 solutions

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DamithSL · Answer 1 · 2015-08-03T21:45:00

Solution 1

if you need to use add interface when you want to add. try as below

C#

add c1 = new adds();
c1.add(15, 15); // call interface method 
Console.WriteLine(((adds)c1).c); // get the value

Posted 3-Aug-15 21:45pm

DamithSL

Updated 3-Aug-15 21:46pm

v2

Comments

Member 10720577 5-Aug-15 1:41am

This is not working...

DamithSL 5-Aug-15 4:22am

DEMO[^]

OriginalGriff · Answer 2 · 2015-08-03T21:45:00

Solution 2

What problem?
You are creating two different instances of the adds class:

C#

adds c1 = new adds();
add val_add = new adds();

As a result, they will each have separate c fields, which will contain different values.
Since you only call the add method on one of them, only the val_add instance will have a non-zero value.
If you did this:

C#

adds c1 = new adds();
add val_add = c1;
val_add.add(15, 15);
Response.Write(c1.c);

Then you would get the value you expect.

Posted 3-Aug-15 21:45pm

OriginalGriff

Comments

Member 10720577 5-Aug-15 1:41am

Pls...interface using two values add method

OriginalGriff 5-Aug-15 2:19am

Please...remember that we can't see your screen, access your HDD, or read your mind.
So a tiny phrase doesn't tell me much about what problem you are still having...