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Why is super class constructor always called [duplicate]
(3 answers)
Closed 4 months ago.
I don't get how we managed to invoke constructor without parameters in class A at all.
How upcasting works in this particular example? When we produce A ab = bb; what exactly ab refers to?
public class A {
public Integer a;
public Float b;
public A() {
a = 1;
b = 1.0f;
}
public A(Integer x) {
a = 2;
b = 2.0f;
}
public int f(long x) {
return 3;
}
public int f(double x) {
return 4;
}
}
public class B extends A {
public int a;
public B(float x) {
a = 5;
}
public int f(int x) {
return 6;
}
public int f(float x) {
return 7;
}
public int f(double x) {
return 8;
}
}
public class M {
public static void main(String[] args) {
A aa = new A(1);
System.out.println(aa.a + " " + aa.b);// OUT: [ 2 ] [2.0]
int ret = aa.f(aa.b);
System.out.println(ret); // OUT: [ 4 ]
B bb = new B(6);
A ab = bb;
System.out.println(bb.a); // OUT: [ 5 ]
System.out.println(ab.a + " " + ab.b);// OUT: [ 1 ] [1.0]
ret = bb.f(1);
System.out.println(ret); // OUT: [ 6 ]
ret = ab.f(1.0f);
System.out.println(ret); // OUT: [ 8 ]
ret = ab.f(aa.a);
System.out.println(ret); // OUT: [ 3 ]
ret = bb.f(aa.b);
System.out.println(ret); // OUT: [ 7 ]
}
}
