问题
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 ]
}
}
回答1:
When we produce
A ab = bb;
what exactlyab
refers to?
It references bb
, but as an A
, i.e. one can only call methods and attributes defined in A
. There is no new object constructed. You can see this by checking System.out.println(bb == ab);
, which will evaluate as true
. This concept is known as attribute- or field-hiding.
This also is the reason why ab.a
returns 1
, because the attribute a
(of type Integer
) within A
is accessed. If one would, on the other hand, access bb.a
, one would get the attribute a
(ob type int
) within B
, which is initialzed to be 5
. Keep in mind that, if you construct a B
, there is always an explicit or implicit call to a superclass constructor as defined in JLS, §12.5.
来源:https://stackoverflow.com/questions/59591349/why-upcasted-ab-a-ab-b-produce-result-1-and-1-0