public class Derived extends base
{
public Derived()
{
super();
}
public Derived(int iVal)
{
super(iVal);
}
}
On Thursday, November 29, 2012 12:34:56 PM UTC-5, 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 ?
> What I mean is this. Given :
>
> public class base
> {
> protected int m_iValue;
>
> 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 ?
> The following seems to hide the second constructor of the base class.
>
> public class Derived extends base
> {
> public Derived()
> {
> super();
> }
> }
>
> What is the correct way to do this ?
>
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