Failed to display Address class fields when using object creation method expressed in C#

Question

Visual Studio 2017 suggested something that I was surprised, the creation of object without using the new operator, I was surprised, because I’m coming from Java and it’s been a year since I stopped and decided to go back to study.

Well, when creating a new Company object, it does not appear the fields of the Address class that are declared within the Company class, but within it also has the Plan class and this yes, appear in the complete.

I cannot see wrong, because the two classes are almost equal, apart from the fact that Plano receives inheritance from the Persistant class.

Below follows the code to see from the classes:

Business Class

namespace Cobranca.pkgModel
{
    public class Empresa : Persistant
    {
        private string nome;
        private string razaoSocial;
        private string cnpj;
        private string status;
        private Plano plano;
        private Endereco endereco;

        public Empresa()
        {
            Endereco = new Endereco();
            Plano = new Plano();
        }

        //métodos getter e setter
    }
}

Flat Class

namespace Cobranca.pkgModel
{

    public class Plano : Persistant 
    {
        private string nome;
        private string tipoCobranca;
        private double valor;

        //métodos getter e setter
    }
}

Address Class

namespace Cobranca.pkgModel
{
    public class Endereco
    {
        private string logradouro;
        private string numero;
        private string bairro;
        private string complemento;
        private string cep;
        private string cidade;
        private string estado;

        //métodos getter e setter
    }
}

Below follows the print I took from the screen

Can anyone answer me why I’m going through this, since I created an equal method in class PlanoDAO and does not present this flaw.

And what name of this feature for me to study, that has to do with lambda?

Thank you

---

Answer:

Just for the record I edited my code as the confreres informed and it was like this:

            this.Model = new Empresa
            {
                Id = Convert.ToInt32(data["id"]),
                Nome = Convert.ToString(data["nome"]),
                RazaoSocial = Convert.ToString(data["razao_social"]),
                Cnpj = Convert.ToString(data["cnpj"]),
                Status = Convert.ToString(data["status"]),
            };

            this.Model.Plano = new Plano
            {
                Id = Convert.ToInt32(data["plano_pk"]),
                Nome = Convert.ToString(data["plano_nome"]),
                Valor = Convert.ToDouble(data["plano_valor"])
            };

            this.Model.Endereco = new Endereco
            {
                Logradouro = Convert.ToString(data["logradouro"]),
                Numero = Convert.ToString(data["numero"]),
                Bairro = Convert.ToString(data["bairro"]),
                Complemento = Convert.ToString(data["complemento"]),
                Cep = Convert.ToString(data["cep"]),
                Cidade = Convert.ToString(data["cidade"]),
                Estado = Convert.ToString(data["estado"])
            };

Answer 1

This code should not be compiling:

public Empresa()
{
    Endereco = new Endereco();
    Plano = new Plano();
}

Note that you have declared the properties endereco and plano with the names completely on lowercase and how private, but here you are using the ones with the uppercase initial.

About the tiny

The c# is case-sensitive, then Plano¹ is different from plano².

^{1 - Refers to type Plano (the class declared)
2 - Refers to type property Plano whose name on this occurrence is plano}

In the latest versions of Visual Studio, the highlight featured in the IDE reinforces the compiler’s interpretation of written code.

Observe :

empresa = new Empresa 
{
    Endereco. // AQUI
}

The word Addressee is in the color green (or blue, or greenish blue, or bluish green or whatever), in the same way as Empresa or MySqlCommand, for example. This suggests that at that time you are referencing the type, not the instance. Therefore, only static members of the class will be presented if they exist Endereco.

About the private

Even if you were using the name of the property correctly, you would not be able to assign the value of it, as it is in java, the modifier private limits access to the property to the scope where it is declared (the class itself).

Solving

Then to correct it would be:

1 - Change the access modifiers of the fields in their classes to public, as in the example below:

public class Endereco
{
    public string logradouro;
    public string numero;
    public string bairro;
    public string complemento;
    public string cep;
    public string cidade;
    public string estado;
}

2 - Use the property instead of guy in builders:

var endereco = new Endereco
    {
        logradouro = "algum nome",
        numero = "22A",
        // ... por aí vai
    };

(...)the creation of object without using the new operator(...)

I think that your statement was the result of the confusion between guys and instances of the type. So by destroying the dreams, to create the instance of the kind you really need the new.

I hope I’ve helped.

Answer 2

The right thing would be:

Endereco end = new Endereco()
{
    logradouro = "Rua teste",
    numero = "1200"
};

Without using the class name before the attribute (as you were doing).

However, there is a second problem, because all the properties of your classes are private.

When I add in VS Code, the following message appears:

'Endereco.logradouro' is Inaccessible due to its Protection level.

Switching to public would function and/or add the assignment of values to some method, such as the constructor. The choice depends on the purpose of the class and its properties. If it is just a container, switch to public. However, if it is a necessity for the data to be populated in the instance creation, use the constructor.

[...]but inside it also has the Plan class and this yes, appear in the complete.

Since it is not in the examples where it is being created, I cannot affirm the differences. If it is being declared within the same scope of its class, no errors will occur.