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Dynamic Binding

 
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class B{
int x;
B(){ x = 10;}
void m1(){x = 20;}
void m1(int x){this.x = x;}
}
class A extends B{
int x;
A(){x = 20;}
void m1(int x){super.x = x;}
public static void main(String arf[]){
B x = new A();
System.out.println(x.x);
x.m1();
System.out.println(x.x);
x.m1(30);
System.out.println(x.x);
((A)x).m1(30);
System.out.println(x.x);
}
}
a. 10 20 20 20
b. 10 20 30 30
c. 20 20 20 30
d. 20 20 20 20
e. Compiler error or run time error.
f. None of the above, it will give another output.

Answer is given as b, can anyone explain pls.
-Arun
 
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The bottom line here is that field (member variable) access is done according to the declared type of the variable (on the left of the assignment) while methods are resolved according to the runtime type of the object (on the right of the assignment).
 
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Note that A extends B, so A inherits m1() and overrides m1(int) from A. The methods will be picked up based on signature and object type.
Let's keep a record of A.x and B.x values after each statement:
>>> A.x = 0, B.x = 0 //class initialization
B(){ x = 10;}; // This implicit super() call is invoked every time a new A() is created.
>>> A.x = 0, B.x = 10
B x = new A();
>>> A.x = 20, B.x = 10
System.out.println(x.x); >>> prints 10 (B.x) because x is a supertype type. Variables are picked up strictly based on reference type.
x.m1(); // m1 in B is invoked because it matched the signature.
>>> A.x = 20, B.x = 20
System.out.println(x.x); >>> prints 20 (B.x) because x is a supertype type. Variables are picked up strictly based on reference type.
x.m1(30); // m1 in A is invoked because it matched the signature, and the method is non-static and overriding. Dynamic method binding applies: Method is picked up based on object type.
>>> A.x = 20, B.x = 30
System.out.println(x.x); >>> prints 30 (B.x). Variables are picked up strictly based on reference type.
((A)x).m1(30); // The casting changes the reference type from supertype to subtype. m1 in A is invoked because of its object type, not because of the casting.
>>> A.x = 20, B.x = 30
System.out.println(x.x); >>> prints 30 (B.x) because x is a supertype type. Variables are picked up strictly based on reference type. The casting in the preceding statement is transient, does not hold through the following statement.
[ April 16, 2002: Message edited by: Doanh Nguyen ]
[ April 16, 2002: Message edited by: Doanh Nguyen ]
 
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