Consider this code: class Foo { public void doIt(String… strs) { System.out.println(this is varargs);

Question

0

Asked: May 30, 20262026-05-30T05:10:46+00:00 2026-05-30T05:10:46+00:00

Consider this code: class Foo { public void doIt(String… strs) { System.out.println(this is varargs);

0

Consider this code:

class Foo {
    public void doIt(String... strs) {
        System.out.println("this is varargs");
    }

    private void doIt(String str) {
        System.out.println("this is single");
    }
}

class Bar {
    public static void main (String[] args) {
        new Foo().doIt("");
    }
}

With javac version 1.6.0_29, it fails to compile, stating:

VarArgsError.java:14: doIt(java.lang.String) has private access in Foo
  new Foo().doIt("");
           ^
1 error

Yes, this is silly code and there are at least two obvious workarounds, but I’m curious. Based on section 15.12.2 of the specification, this compilation error seems like a bug in javac, because the first step should remove the non-varargs doIt, as it is inaccessible according to section 6.6.1. Am I missing some other details in the lookup algorithm or is this as obviously wrong as I think it is?