A superclass method is called instead of the subclass method

Question

Let's take a look at this code:

public class ParentClass {
    public void foo(Object o) {
        System.out.println("Parent");
    }
}

public class SubClass extends ParentClass {
    public void foo(String s) {
        System.out.println("Child");
    }

    public static void main(String args[]) {
        ParentClass p = new SubClass();
        p.foo("hello");
    }
}

I expected this to print out "Child", but the result is "Parent". Why does java call the super class instead, and what do I do to make it call the method in the subclass?

Answer 1

SubClass#foo() does not override ParentClass#foo() because it doesn't have the same formal parameters. One takes Object and the other takes a String . Therefore polymorphism at runtime is not applied and does not cause the subclass method to execute. From the Java Language Specification :

An instance method m C declared in or inherited by class C, overrides from C another method m A declared in class A, iff all of the following are true:

• A is a superclass of C.

• C does not inherit m A .

• The signature of m C is a subsignature (§8.4.2) of the signature of m A .

...

And this section defines method signatures:

Two methods or constructors, M and N, have the same signature if they have the same name, the same type parameters (if any) (§8.4.4), and, after adapting the formal parameter types of N to the the type parameters of M, the same formal parameter types.

The signature of a method m 1 is a subsignature of the signature of a method m 2 if either:

• m 2 has the same signature as m 1 , or

• the signature of m1 is the same as the erasure (§4.6) of the signature of m 2 .

Answer 2

The parent class has no method with the signature public void foo(String s) . Therefore, at compile time, the compiler can only choose the public void foo(Object o) method for executing p.foo("hello") .

Since the child class doesn't override this method (it doesn't have the same parameter type), the parent class's foo method is the one executed at runtime.

Answer 3

字符串是java中的对象，因为它没有被覆盖，但是当你传递一个字符串时，它将运行第一个可用于参数的方法，在这种情况下它是超类的方法。