Tue, 01 Sep 2015 18:28:11 +0530
8068901: Surprising behavior with more than one functional interface on a class
8068903: Can't invoke vararg @FunctionalInterface methods
Reviewed-by: attila, hannesw
sundar@606 | 1 | SyntaxError:32:for each can only be used with for..in |
sundar@406 | 2 | for each(var v=0;false;); |
sundar@406 | 3 | ^ |