late binding

[Leandro Oliveira]

I' m not very good with late binding.... I can't know exactly what method will be chosen by the virtual machine!!!
does late binding applies only to instance methods??? how about instance variables, instance methods or class variables???

[Dan Culache]

Hi Leandro,
Late binding only applies to instance methods. Variables (static or not) are bind at compile time based on the type of the reference used (see last two lines of output).
Briefly, if class B extends class A and overrides method m of class A then at runtime
for "A a = new B(); a.m();" method m of class B will be called.
see the code below:

package c01;
class LtA {
static int is=0;
int i = 0;
void m() {
System.out.println("in LtA static int is = " + is);
System.out.println("in LtA int i = " + i);
}
}
public class LtB extends LtA {
static int is=5;
int i = 5;
void m() {
System.out.println("in LtB static int is = " + is);
System.out.println("in LtB int i = " + i);
}
public static void main(String[] args) {
LtA A = new LtA();
LtB B = new LtB();
LtA C = new LtB();
A.m();
System.out.println("in main A.is = " + A.is);
System.out.println("in main A.i  = " + A.i);
B.m();
System.out.println("in main B.is = " + B.is);
System.out.println("in main B.i  = " + B.i);
C.m();
System.out.println("in main C.is = " + C.is);
System.out.println("in main C.i  = " + C.i);
}
}
//-------------------
// prints
in LtA static int is = 0
in LtA int i = 0
in main A.is = 0
in main A.i  = 0
in LtB static int is = 5
in LtB int i = 5
in main B.is = 5
in main B.i  = 5
in LtB static int is = 5
in LtB int i = 5
in main C.is = 0
in main C.i  = 0

[ February 04, 2003: Message edited by: Dan Culache ]

[Jose Botella]

Call binding from Thinking in Java

[Pallavi Chakraborty]

Hello Dan,
I wanted to ask you about what the print statements regarding C do in the main method.
I get what C.m() does, calls the m method of B.
But what do the last two print statements do?
Thank you
Pallavi

[Dan Culache]

Hi Pallavi,
The last two print statements are printing the variables of the object refered by C (one static, one instance). I was trying to show that binding the variables is resolved at compile time as oposed to method binding which is done at runtime.
Therefore because C is a reference to a LtA object (although it points to a LtB object) the static int is of the LtA class and the int i of a LtA instance are printed.
When method m() is called, the visible m() method of the refered object is called and because LtB had overriden m() when extending LtA and because C, although a LtA reference points to a LtB object then the only visible m() method is m() of LtB.
See the code below. It prints the variable corresponding to a LtB instance, the instance C really refers to. That is if you comment line 2. Here although C points to a LtB object the compiler doesn't know who k is because C is a reference to a LtA object and LtA has no k.
A bit too long of an explanation but I hope I was clear.

package c01;
class LtA {
static int is=0;
int i = 0;
}
public class LtB extends LtA {
static int is=5;
int i = 5;
int k = 7;
public static void main(String[] args) {
LtA C = new LtB();
C.m();
System.out.println("in main C.is = " + C.is);// 1
System.out.println("in main C.i  = " + C.k);// 2
System.out.println("in main C.is = " + ((LtB)C).is);// 3
System.out.println("in main C.i  = " + ((LtB)C).i); // 4
}
}

[ February 04, 2003: Message edited by: Dan Culache ]