Why is 'virtual' optional for overridden methods in derived classes?

Question

When a method is declared as virtual in a class, its overrides in derived classes are automatically considered virtual as well, and the C++ language ma

Answer 1

Since the language can't enforce "good" style, C++ generally doesn't even try. At least IMO, it's open to question whether including redundant specifiers like this is good style in any case (personally, I hate when they're there).

(At least parts of) Google's coding standards may make sense under some circumstances, but as far as C++ in general goes, are generally considered mediocre advice at best. To an extent, they even admit that -- some of them they openly state are only really there to fit with their old code. Other parts they don't admit so directly, and (to be entirely honest) that argument wouldn't support some of their standards anyway (i.e., some of it seems to lack real justification).