C#•3d ago

Silly OOP question about constructors

I am learning OOP and came across this. A way to reuse constructors:

public Dog() : this("NoName") {}

public Dog(string name) : this(name, 1) {}

public Dog(string name, int age) : this(name, age, 0.3) {}

public Dog(string name, int age, double length) : this(name, age, length, new Collar()) {}

public Dog(string name, int age, double length, Collar collar)
{
    this.name = name;
    this.age = age;
    this.length = length;
    this.collar = collar;
}

public Dog() : this("NoName") {}

public Dog(string name) : this(name, 1) {}

public Dog(string name, int age) : this(name, age, 0.3) {}

public Dog(string name, int age, double length) : this(name, age, length, new Collar()) {}

public Dog(string name, int age, double length, Collar collar)
{
    this.name = name;
    this.age = age;
    this.length = length;
    this.collar = collar;
}

I was wondering that can I just use default values for parameters like this, to achieve same result:

public Dog(string name="NoName", int age=1, double length=0.3, Collar collar=new Collar())
{
    this.name = name;
    this.age = age;
    this.length = length;
    this.collar = collar;
}

public Dog(string name="NoName", int age=1, double length=0.3, Collar collar=new Collar())
{
    this.name = name;
    this.age = age;
    this.length = length;
    this.collar = collar;
}

If yes then is their any use case when we would prefer the 1st way over 2nd?

6 Replies

Pobiega•3d ago

Optional arguments isn't exactly the same as method/constructor overloading, especially when reflected over

Unknown User•3d ago

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Pobiega•3d ago

But if its a type only used internally or similar and that isn't a problem for you, yeah this works

yourFriendOP•3d ago

Hmm... I will see And yes this book is ancient 😅

GoldenFapple•2d ago

Your not going to be able to do the last parameter, Collar collar = new Collar() because its not a compile time constant. buuuut, you can do something like this: public Dog() : this("NoName", 1, 0.3, new Collar()) { } This will conflict with this constructor however: public Dog() : this("NoName") { } As that already defines a constructor with no parameters.

yourFriendOP•2d ago

Thanks, have to play around a bit to get used to it.

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Silly OOP question about constructors

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