Paul-Peter Tournaris wrote: > No you don't need do. But if you want to call the base constructors you > just call super with the appropriate parameters.Also you can create new > constructors for your new class that extends the base one! For example!
This is basic Java. Check out http://docs.oracle.com/javase/tutorial/ > > public class base { > Please conform more or less to the Java coding conventions, particularly with regard to naming: http://www.oracle.com/technetwork/java/codeconv-138413.html You don't extend constructors in Java. http://docs.oracle.com/javase/specs/jls/se7/html/jls-8.html#jls-8.8 "Constructor declarations are not members. They are never inherited and therefore are not subject to hiding or overriding." So the original question is flawed. What's going on is that > public base(){ > int i = 0; > } > } > > > public class newBase extends base { > > public newBase(){ > super();//if you want to call the mother constructor > This is done for you automatically by the compiler. Adding 'super()' to a constructor implementation is a rookie move. You need to add a super-constructor call when you need to specify an overload of the parent constructor, i.e., one with parameters. For the no-arg constructor it's a waste of boilerplate. > int i=0; > How does this help the pedagogy? > } > > } > > Simon Giddings wrote: > >> This may seem a bit basic, but I come from a C++ background and it is >> confusing me a little. >> >> When I extend a base class, do I need to create a constructor for each >> base class constructor ? >> > You don't. You only need to create a constructor that is required by the child class itself. It is not uncommon to use a no-arg child-class constructor that invokes a superclass constructor with arguments. public class DynamicType<T> { final Class<? extends T> token; protected DynamicType(Class<? extends T> tok) { if (tok == null) { throw new IllegalArgumentException("Null token"); } token = tok; assert token != null : "No IllegalArgumentException"; } // more stuff ... } public class DynamicFoo extends DynamicType<Foo> { public DynamicFoo() { super(Foo.class); } // more stuff ... } > What I mean is this. Given : >> >> public class base >> { >> protected int m_iValue; >> > Please follow the naming conventions. > >> public base() >> { >> m_iValue = 0; >> } >> >> public base(int iVal) >> { >> m_iValue = iVal; >> } >> } >> >> If I want to create a new class based on this base class, will I need to >> declare the two constructors again for them to be visible ? >> > They won't be visible even if you do declare a bunch of constructors in the child class. Constructors are not inherited. > The following seems to hide the second constructor of the base class. >> > Doesn't. > >> public class Derived extends base >> { >> public Derived() >> { >> super(); >> } >> } > > You only declared one constructor. You didn't hide any. If you want a constructor that takes a parameter, you have to declare it. Check out the Java tutorials. > What is the correct way to do this ? > > Forget C++. Since constructors aren't inherited, you must explicitly call any no-arg super constructor that you wish used. You must write every constructor needed by the class itself. You don't need to invoke any super-constructors the child class doesn't need. BIG WARNING! Do not do too much in a constructor. Constructors' roles should be constrained to setting up INITIAL conditions only, and only those needed to construct the object itself, at that. It is a common mistake to cram too much logic into the constructor. The danger is that such logic perforce operates on an incomplete instance of 'this'. That is very bad. CONSTRUCT IN CONSTRUCTORS. PERFORM IN METHODS. -- Lew -- You received this message because you are subscribed to the Google Groups "Android Developers" group. To post to this group, send email to android-developers@googlegroups.com To unsubscribe from this group, send email to android-developers+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/android-developers?hl=en
