Ok, if declared as final and static, variables behave just like constants and
can be used in constructors.
I don't know how it happened the first time that it didn't work, but after
another test it's confirmed, they work.
With an afterthough, it would have been a big mistake in the java languag
The java language specification is a little difficult to follow on final static
fields.
The best place to look is 13.4.8 - Binary Compatability. This says
"We call a field that is static, final, and initialized with a compile-time
constant expression a primitive constant."
and it is impl
At 11:05 AM 7/16/99 +0200, Kontorotsui wrote:
>
>On 14-Jul-99 Michael Sinz wrote:
>> The reason is that until the constructor is called, no instance data
>> can be used. Note that you did not make this static final but rather
>> final. This means there must be actual instance data to access this
Kontorotsui wrote:
>
> aClass extends aSuperClass
> {
> final int v = 1;
>
> aClass()
> {
> super(v);
> }
> }
First off, my understanding of the question is why doesn't v exist when
passed into super. "super" is not a method call like everything else.
It is a special keyword and a
The answer is: according the JLF, superclass constructor is called
*before* any initialization statement in subclass. super() does not
actually run in aClass() but java compiler generates method where
it is called before first initialization statement of aClass. The syntax
super(...) is allowed o
Have you tried using "static final int v = 1;" instead?
-Andy
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED]]On Behalf Of Kontorotsui
> Sent: Wednesday, July 14, 1999 12:41 PM
> To: [EMAIL PROTECTED]
> Subject: final variables
I wonder why if I write this:
aClass extends aSuperClass
{
aClass()
{
super(1);
}
}
where the superclass is:
aSuperClass
{
aSuperClass(int i)
{
...
}
}
anything works ok.
If instead I write
aClass extends aSuperClass
{
final int v = 1;
aClass()
{
super(v);
