Java virtual function calls

Question

From my understanding all function-calls in Java are virtual, and numeral literals have the type int. But why does the Output in the example below differ?

public class A {
    public int f(long d) {
        return 2;
    }
}
public class B extends A {
    public int f(int d) {
        return 1;
    }
}
public class M {
    public static void main(String[] args) {
        B b = new B();
        A ab = b;
        System.out.println(b.f(1));
        System.out.println(ab.f(1));
    }
}

`A` doesn't define any `f(int)` method, so obviously, `f(long)` is called in the second case. There is no method overriding in your code . — Arnaud, Feb 29 '16 at 16:30

Nikolas Charalambidis · Accepted Answer · 2016-02-29T16:51:23.647

You dont override anything.

The first calling System.out.println(b.f(1)); returns 1, because it works with class B, even the method is named same, but parameters are different (long is not the same as int).
In case when parameters are same (int d), the result would be 1, because it overrides (@Override) the method from the class A.
Now, you know why the second calling System.out.println(ab.f(1)); returns 2. Look from what class it's called from.

massimiliano coppola · Answer 2 · 2021-05-27T20:45:30.617

Actually the subclass B has inherited the method f(with a long) and has added (overloaded) another method f(with a int). When you write down a value such as 1 in such a way the compiler, even before assigning it to a typed reference does parse it as an int. More here: Java's L number (long) specification .

When you call the method f using the reference ab (which is A class) it (the reference) says I can only send you to a method f(with a long) and then implicitly cast the type int 1 into a long.

Let's try to change the type of the method in A class to f(short), then System.out.println(ab.f(1)); will give you this error: "The method f(short) in the type A is not applicable for the arguments (int)"

Java virtual function calls

2 Answers2