java - Calling overloaded inherited methods using super class reference -

i not understand java behavior. have 2 classes:

class c1 {     public void m1(double num) {         system.out.println("inside c1.m1(): " + num);     } }  class c2 extends c1 {     public void m1(int num) {         system.out.println("inside c2.m1(): " + num);     } } 

and main:

public class main {      public static void main(string[] args) {         c1 c = new c2();         c.m1(10);     } } 

and result was:

inside c1.m1(): 10.0 

when expected:

inside c2.m1(): 10 

also when try complete code syntax, found this:

where other m1 of c2 class?

i check bytecode of main.class , saw this:

compiled "main.java" public class com.company.main {   public com.company.main();     code:        0: aload_0        1: invokespecial #1                  // method java/lang/object."<init>":()v        4: return    public static void main(java.lang.string[]);     code:        0: new           #2                  // class com/company/c2        3: dup        4: invokespecial #3                  // method com/company/c2."<init>":()v        7: astore_1        8: aload_1        9: ldc2_w        #4                  // double 10.0d       12: invokevirtual #6                  // method com/company/c1.m1:(d)v       15: return } 

the bytecode tell me invoke c1.m1 (d)v (line 12).

why method of c1? trying understand behavior.

your 2 methods named m1 not have same signature; 1 in superclass takes double, , 1 in subclass takes int. means compiler select method signature call based on compile-time type of variable, c1, , call m1(double). since @ runtime class c2 doesn't have overriding version of m1(double), version c1 invoked.

the rule method signatures computed @ compile time based on compile-time types; method calls dispatched @ runtime based on matching signatures.