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@louy2 louy2/E.java
Created Apr 24, 2017

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Solving A Java Inheritance Puzzler from https://llogiq.github.io/2015/07/14/inheritance.html
/*
* Solving A Java Inheritance Puzzler from
* https://llogiq.github.io/2015/07/14/inheritance.html
*/
class A {
int x = 3;
int a() { x++; return x + c(); }
int b() { return a(); }
int c() { return x; }
}
class B extends A {
@Override int a() { return super.a() + c(); }
}
class C extends B {
@Override int b() { return a() * 3; }
@Override int c() { return x + 1; }
}
public class E {
public static void main(String[] args) {
int x = new C().b();
/*
* Because C overrides b(), x = a() * 3
* C inherits B, which overrides a(), so a() = A.a() + c()
* x = (A.a() + c()) * 3
* Which c() is it? A.c() or C.c()?
* Without super, it is this.c(), and this == C()
* x = (A.a() + C.c()) * 3
* A.a() is x + 1 + c(), which c() is that?
* By the same logic, C.c()
* x = (x + 1 + C.c() + C.c()) * 3
* x = (x + 1 + x + 1 + x + 1) * 3
* x = (3 + 1 + 3 + 1 + 3 + 1) * 3
* x = 12 * 3 = 36
*/
System.out.println(x); // Wrong! x = 42
/*
* Override analysis was correct.
* Mutable state analysis was wrong.
* Assumption that x++ translate to x+1 is wrong.
* (Been programming pure functional too long?)
* C.b() -> B.a() * 3 x = 3
* B.a() -> A.a() + C.c() x = 3
* A.a() -> x + C.c() x = 4
* C.c() -> 5 x = 4
* A.a() -> 9 x = 4
* B.a() -> 14 x = 4
* C.b() -> 42 x = 4
*/
}
}
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