Implement -Wimplicit-fallthrough (version 7)

[Marek Polacek]

On Tue, Aug 30, 2016 at 01:08:22PM -0400, Jason Merrill wrote:
> On Tue, Aug 30, 2016 at 10:25 AM, Marek Polacek <polacek@redhat.com> wrote:
> >> > @@ -10585,15 +10610,26 @@ cp_parser_statement (cp_parser* parser, tree in_statement_expr,
> >> >         }
> >> >        /* Look for an expression-statement instead.  */
> >> >        statement = cp_parser_expression_statement (parser, in_statement_expr);
> >> > +
> >> > +      /* Handle [[fallthrough]];.  */
> >> > +      if (std_attrs != NULL_TREE
> >> > +         && is_attribute_p ("fallthrough", get_attribute_name (std_attrs))
> >> > +         && statement == NULL_TREE)
> >> > +       {
> >> > +         tree fn = build_call_expr_internal_loc (statement_location,
> >> > +                                                 IFN_FALLTHROUGH,
> >> > +                                                 void_type_node, 0);
> >>
> >> Let's use the 'statement' variable rather than a new variable fn so
> >> that we set the call's location below.  Let's also warn here about
> >> [[fallthrough]] on non-null expression statements.
> >
> > Done.  But I couldn't figure out a testcase where the warning would trigger, so
> > not sure how important that is.
> 
> What happens for
> 
> [[fallthrough]] 42;
> 
> ?  What ought to happen is a warning that the attribute only applies
> to null statements.

This:
q.C: In function ‘void f(int)’:
q.C:8:8: error: expected identifier before ‘[’ token
       [[fallthrough]] 42;
        ^
q.C: In lambda function:
q.C:8:23: error: expected ‘{’ before numeric constant
       [[fallthrough]] 42;
                       ^~

And the same with e.g. [[unused]].  G++ 6 behaves exactly the same,
so it isn't something my patch changed.  It seems we never even reach
the code in question, at least std_attrs is null.

It seems g++ is trying to parse a lambda expression.

	Marek