java - method signature in inheritance -

in code below

class {     public void v(int... vals) {         system.out.println("super");     } }  class b extends {     @override     public void v(int[] vals) {         system.out.println("sub");     } } 

then can call new b().v(1, 2, 3);//print sub rather super ridiculous does work well. if change b

class b {     public void v(int[] vals) {         system.out.println("not extending a");     } } 

the call new b().v(1, 2, 3); invalid. have call new b().v(new int[]{1, 2, 3});, why?

under jdk 1.7, neither of examples compiles, , don't believe should.

it's easy see why second version doesn't - when b doesn't extend a, there's no indication of varargs @ all, there's no reason why compiler should possibly convert argument list of 3 int arguments single int[]. more interesting situation b does extend a.

the compiler finds signature of v(int[] vals) doesn't use varargs. there's nothing in spec should inheritance chain find whether 1 of other declarations (there multiple ones) uses varargs. fact appears in version of jdk 1.6 suggests compiler bug has since been fixed. (i've reproduced bug in jdk 1.6.0_39 well.)

basically, if want able invoke b.v() in varargs-style syntax, b.v() should declared using varargs. of course, if change compile-time type a, work:

a = new b(); a.v(1, 2, 3); 

or (ick):

((a) new b()).v(1, 2, 3); 

note if compile -xlint warning b anyway:

test.java:14: warning: v(int[]) in b overrides v(int...) in a; overriding method  missing '...'     public void v(int[] vals) {                 ^