Install all class members that implement an interface from an object of type interface without a lot of write code in C #

Question

This may seem a little complicated, but I will try to explain it clearly.

Is this possible in C # if I have a long interface:

public interface IInterface
{
    bool interface_1;
    bool interface_2;
    bool interface_3;
    bool interface_4;
    bool interface_5;
    ...
}

And the class that implements the interface

public class MyClass : IInterface
{
    ...
}

Imagine I have an IInterface object myInterface, is there a way to create a MyClass object myClass and set all myClass fields to one of myInterface without setting all the fields one after the other. For instance:

IInterface myInterface;
MyClass myClass;
myClass.SetIInterface(myInterface);

Answer 1

Imagine I have an IIterter myInterface object, is there a way to create a MyClass myClass object and set all myClass fields with one of myInterface

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Answer 2

, , . typeof(IInterface).GetProperties() ( ), , , (, PropertyInfo.GetValue() PropertyInfo.SetValue()).

Answer 3

, . , IInterface MyClass:

public interface IInterface
{
    bool interface_1 { get; set; }
}

public class MyClass : IInterface
{

}

public class ReflectionHelper
{
    private MyClass CreateByCopy(IInterface instance)
    {
        MyClass obj = new MyClass();
        foreach (PropertyInfo property in typeof( IInterface ).GetProperties())
        {
            property.SetValue( property, property.GetValue( instance, null ), null );
        }

        return obj;
    }
}

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Answer 4

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Answer 5

, . SetIInterface() (/ ) . , MyClass IInterface.