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Home/ Questions/Q 6028265
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T04:42:47+00:00 2026-05-23T04:42:47+00:00

If I have the following Interface structure; public interface IPaymentTypeBase { void PayNow(); double

  • 0

If I have the following Interface structure;

public interface IPaymentTypeBase
{
    void PayNow();
    double Amount { get; set; }
}

public interface IPaymentTypePayPal : IPaymentTypeBase
{
    string UserName { get; set; }
    string Password { get; set; }
}

public interface IPaymentMethod<T>
{
}

Then I have the following classes;

public class PaymentTypePayPal : IPaymentTypePayPal
{
    public string UserName { get; set; }
    public string Password { get; set; }

    public void PayNow()
    {
        throw new NotImplementedException();
    }
}

public class PaymentMethod<T> : IPaymentMethod<T> where T : IPaymentTypeBase
{
}

Then in my web application I have this;

IPaymentMethod<IPaymentTypePayPal> payer = (IPaymentMethod<IPaymentTypePayPal>) new PaymentMethod<PaymentTypePayPal>();

I’d now like to call payer.PayNow(); but I’m just getting lost in interfaces etc and can’t seem to make this work.

I believe this is a simple thing but am just missing the point entierly.

Can anyone help?

Edit

The intention here is to have a set of interface such as PayPal, Voucher, CreditCard all of which do their own payment gateway type of stuff.

So I’d like to instantiate a class that takes the Payment Type as an interface and call that interfaces PayNow method.

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  1. Editorial Team
    Editorial Team
    2026-05-23T04:42:48+00:00Added an answer on May 23, 2026 at 4:42 am

    payer is of type IPaymentMethod<IPaymentTypePayPal>.

    But IPaymentMethod<T> is defined as

    public interface IPaymentMethod<T>
    {
    }
    

    Therefore, it has no methods and you can’t call PayNow() on it.

    The same is true for PaymentMethod<T>, so you can’t call any method on an instance of PaymentMethod<PaymentTypePaypal> either.

    Maybe you can explain a little more what your intention is.

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