4
In Java:
public class Teste extends Testando
{
private String nome;
private int numero;
public Teste(String teste, String nome, int numero)
{
super(teste);
this.nome = nome;
this.numero = numero;
}
}
What is the command corresponding to super() in C++?
8
You call explicitly by the class name, because in C++ there is multiple inheritance and can have more than one ancestor. Roughly it would be this:
class Teste : public Testando {
String nome;
int numero;
public:
Teste(String teste, String nome, int numero) : Testando(teste) {
this.nome = nome;
this.numero = numero;
}
}
2
There is no direct relationship in C++.
It is possible to do something similar if you create a super typedef for all classes that want to use this feature, as in the case below:
class Derivada: public Base
{
private:
typedef Base super; // não deve ser público porque cada classe pode
// ter apenas uma classe super
} ;
And then you could use super in place of the name of the base class, as in the example:
super::meumetodo();
However, it is not recommended to abuse this resource. In C++ there is the possibility of multiple inheritance and this would complicate the solution, and may cause confusion.
And if you have multiple inheritance?
It won’t work, of course. That’s why I commented that it would create confusion... even if you choose one of the classes to be the super and the other consider interface...
That’s right, in the rush I didn’t even see :)
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I was writing a reply that was the second example, but since you did +1
– Guilherme Nascimento
I think the second seems to me more "organized" and even more common, I do not know how is the standard in C++11 and 14, if there is any pattern, I personally divide the . hpp and the . cpp and make the "super" call on the . cpp, but of course this is how I do, to which I researched a little (unfortunately old links) did not find any definitive pattern, so both examples seem good (although I prefer the second).
– Guilherme Nascimento
I think at first it can also be without the name of the class, for example:
::Testando(teste). Not?– Luiz Vieira
@Guilhermenascimento I prefer the second, I put the first to look like what he did. I remember something that becomes problematic when you do the first one, but I don’t remember what. I used little C++, never studied thoroughly. I arranged as Luizvieira said.
– Maniero
I have the impression that the first example is wrong: it is calling the base class constructor and creating a temporary one, which is discarded...probably the compiler should warn. The second mode is the correct way to call the constructor of a base class.
– zentrunix