创建继承接口的对象

Question

我有两个模型接口ILoanModel和IAmortizationModel。

public interface ILoanModel
{
    string ID { get; set; }
    IEmployeeModel Borrower { get; set; }
    // ******* other propeties *********
    ICollection<Model.IAmortizationModel> AmortizationTable { get; set; }
}

public interface IAmortizationModel
{
    string ID { get; set; }
    ILoanModel Loan { get; set; }
    // ******* other propeties *********
}

我创建了另一个处理计算和其他函数的接口。

public interface IAmortization
{
    // Compute and generate amortization table
    Model.ILoanModel Generate(Model.ILoanModel loan);
    // **** Other function *****
}

目前，我有三种类型的贷款，它们都有相同的计算方法和生成摊销表。

public abstract class Amortization : Controller.Computation.Interface.IAmortization
{
    public virtual Model.ILoanModel Generate(Model.ILoanModel loan)
    {
        // **** other code *******
        //Create the amortization table
        decimal last_Amortization_RunningBalance = loan.Amount;
        loan.AmortizationTable = new List<Model.IAmortizationModel>();

        for (int x = 1; loan.NumberOfPayment >= x; x++)
        {
            // having problem here because Model.IAmortizationModel is a interface
            var newAmortizationRow = new Model.IAmortizationModel
            {
                ID = Guid.NewGuid().ToString(),
                AmortizationAmount = loan.AmortizationPaymentAmount,
                AmortizationInterest = loan.AmortizationPaymentAmount* loan.Nominal,
                AmortizationPrepayment = 
                    loan.AmortizationPaymentAmount 
                    - (loan.AmortizationPaymentAmount * loan.Nominal),
                AmortizationOutstandingBalance = 
                    last_Amortization_RunningBalance 
                    - (loan.AmortizationPaymentAmount 
                        - (loan.AmortizationPaymentAmount * loan.Nominal)),
                Sequence = x
            };

            last_Amortization_RunningBalance -= 
                newAmortizationRow.AmortizationOutstandingBalance;
            loan.AmortizationTable.Add(newAmortizationRow);
        }
    }
}

我现在的问题是生成新的IAmortizationModel，并将其添加到loan.AmortizationTable中。如何获得loan.AmortizationTable的类型并创建？

Answer 1

一种选择是将其变为泛型，并使用new()添加一个泛型约束。

public virtual Model.ILoanModel Generate<T>(Model.ILoanModel loan)
   where T:IAmortizationModel, new()

然后调用new T()

另一种选择是传递一个生成器函数：

public virtual Model.ILoanModel Generate(Model.ILoanModel loan,
  Func<IAmortizationModel> generator)

打电话给generator()