Paul Salerno

Case A
class ExSuper { public void func(String p, String s) { System.out.println(p) } } public class Example extends ExSuper { public void func(String p, String s) { System.out.println(p + ":" + s) } static public void main(String arg[]) { ExSuper e1 = new ExSuper() e1.func("hello", "hi") // hello ExSuper e2 = new Example() e2.func("hello2", "hi2") // hello2, hi2 } }
Case B
class Super { public int getNumber(int a) { return 2; // prints 2 } } public class Subclass extends Super { public int getNumber(int a, char ch) { return 4; } public static void main(String[] args) { System.out.println(new Subclass().getNumber(4) } }
My question is why in case B we are accessing the method in the superclass vs the overridden method in Subclass and is this always the case.

round (double) < -- returns long
I was under the impression that methods of Math return either int or double.
Where am I wrong?
TIA

Thanks Rob!

Could someone explain how this is different from my code?
class TestClass { int i = getInt(); // not inside a method int k = 20; public int getInt() { return k + i // < -- A } public static void main(String[] args) { TestClass t = new TestClass() System.out.println(t.i + "" + t.k) } }

Does anyone have a clue about the last piece of code that I posted.
4 string objects are being created and I wanted to be sure that they WILL be GC'd.
TIA

Is this the same situation?
Here 4 string literal objects are created within a for loop and GC'd since temp is a local var.
public void process(int count) ---> pass 5 { for(int i = 1; i < count; i++) { Object temp = "Hello" + i; } }

you guys are awesome, thanks

word
Thank you Velentin

woops a typo, heres the code that should be:
class TestClass { int i = getInt(); int k = 20; public int getInt() { return k + i // < -- A } public static void main(String[] args) { TestClass t = new TestClass() System.out.println(t.i + "" + t.k) } }

chafule,

i is assigned to the returned value of getInt()
which returns k+1

I dont see any addition taking place here. What I do see is concatenation k + i where k is zero at that time. yes the return of the method is i but again where is this 1 at?

Thanks for your input Rick
I'm still unsure where this 1 is coming from
0 + 1
Yes I can see that K is 0, but how are we getting 1 for i ?
TIA

Duncan and Corey, yes that does make sense.
Im still not 100% on case 1 yet, if anyone knows.

class TestClass { int i = getInt() int k = 20; public int getInt() { return k + 1 // < -- A } public static void main(String[] args) { TestClass t = new TestClass() System.out.println(t.i + "" + t.k) // <-- B } }
A: initializing i to getInt() where k hasnt been initialized yet at this point, its zero. however, how is "1" being returned and printed?
B: prints ret value of method which is 1 and the value of k member var which is 20

I can see how this dg matrix holds two rg objects. Nevertheless since rg is set to null, the return value would be null, no?
Could you tell me if Case 1 is correct?

Corey this makes total sense. Thanks!