CODEWITHSUNDEEP
c#
grilletto
grilletto

Reputation: 19

How to solve this error with the interface?

Visual Studio gives me this error and cannot proceed. The compiler reports that IArea is less accessible as processingArea. What can I do?

How do I then enter the parameter in the method processingArea()?

I enter into an object that implements the IArea, but how do I create an object that implements the interface?

public class GeometricShapes
{
    double Area;
    double @base;
    double height;
    public GeometricShapes()
    {

    }

    public void processingArea(IArea poligono)
    {
        var area = poligono.CalculateArea();
        Console.WriteLine(area);
        Console.ReadLine();
    }
}

public class Quadrato : GeometricShapes, IArea
{
    double @base;
    double height;

    public double CalculateArea()
    {
        Random x = new Random();
        this.@base = x.Next(100);
        this.height = x.Next(100);

        var Area = @base * @base;
        return Area;
    }
}

public class Rettangolo : GeometricShapes, IArea
{
    double @base;
    double height;

    public double CalculateArea()
    {
        Random x = new Random();
        this.@base = x.Next(100);
        this.height = x.Next(100);

        var Area = @base * height;
        return Area;
    }
}

interface IArea
{            
    double CalculateArea();
}

I made other errors?

Upvotes: 0

Views: 107

Answers (5)

Greg
Greg

Reputation: 11480

In order for your interface to correctly work, you should ensure that it is marked public. As for the parameter issue, you should do the following:

public interface IArea
{
     double CalculateArea(double length, double width);
}

Now when you implicitly or explicitly implement your interface your declared as public and taking the parameters.

Now when you build your object:

public static class Rectangle : IArea
{  
     public double CalculateArea(double length, double width)
     {
          // Logic to calculate.
     }
}

That would be an implementation.

I modified the answer a bit, so in the Main you would do:

Rectangle rectangle = new Rectangle();
Console.WriteLine(rectangle.CalculateArea(10, 5));

That would build your object, then should return the area to be written to your console.

Upvotes: 0

jd6
jd6

Reputation: 429

To resolve your error, you should write : public interface IArea {...}.

Indeed, the public keyword is used for the classes, but not for the interface, which is internal by default. Or, the processingArea method is public, where as IArea is so internal. If you change thus your code, it will works !

Upvotes: 0

KeithS
KeithS

Reputation: 71563

By not specifying a visibility modifier for the interface IArea, C# uses the default of "internal", so it's only visible to other types in the same assembly. However, your shapes are public, so as written, it's possible to refer to the concrete class from a location in code that isn't allowed to know about the interface it inherits from. This isn't kosher.

To fix it, simply change IArea to be a public interface by adding the public keyword to the definition.

Upvotes: 1

BradleyDotNET
BradleyDotNET

Reputation: 61349

That error means you have a public method, that takes a parameter (in this case, an IArea) with a more restrictive access modifier. internal is the default, so your interface probably needs to be:

public interface IArea
{
  ...
}

When you think about it, the error makes sense. If you call a public method from another assembly, but can't see the parameter type (because its internal), you actually can't call the method! The compiler detects this for you, and throws the error.

Upvotes: 1

David Crowell
David Crowell

Reputation: 3813

Your classes that implement IArea are public. Therefore IArea should be public.

public interface IArea
{
    double CalculateArea();
}

Upvotes: 1

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