Java function overloading query

Question

A extends B

public class Test {

public static void print(A obj) {
    System.out.println("print A");
}

public static void print(B obj) {
    System.out.println("print B");
}

public static void main(String [] args ) {  
    A x = new B();

    print(x);
}

}

Why it'll print "print A"? Why function overloading doesn't look up the real type of x in runtime?

score 4 · Accepted Answer · answered May 15 '14 at 01:50

4

Because overloading is not overriding. The compiler binds methods based on the declared types of arguments.

answered May 15 '14 at 01:50

Sotirios Delimanolis

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score 1 · Answer 2 · answered May 15 '14 at 02:03

Why function overloading doesn't look up the real type of x in runtime?

Because function overloading doesn't look up the real type of x at runtime, by JLS #8.4.9: "the number of actual arguments (and any explicit type arguments) and the compile-time types of the arguments are used, at compile time, to determine the signature of the method that will be invoked".

Java function overloading query

2 Answers2