How to use pragma in a compound literal in GCC 9

[Jakub Jelinek]

On Tue, Feb 12, 2019 at 06:19:56PM +1100, Daurnimator wrote:
> A project I help maintain has had a report that it fails to compile with GCC 9
> https://github.com/wahern/cqueues/issues/212#issuecomment-461693111
> 
> I've attached a minimal reproduction of the issue.
> Trying to compile it results in:
> 
> <source>: In function 'main':
> <source>:46:15: error: lvalue required as unary '&' operand
>    46 |     void *x = &quietinit((struct bar){ 0, .a = 0 });
>       |               ^
> Compiler returned: 1

Guess you need to restructure the code somewhat, with the way these macros
are used it is not possible.  GCC doesn't accept _Pragma at arbitrary spots
within a single expression.

You could just add -Wno-override-init to the options and use the macro
without pragma.

Or better, don't do what the macro complains about, if I understand it
right, at least in the package I was looking at, there is just 0, 0, { 0, 0 }
style initialization in some macros and a macro that uses quietinit
sometimes ammends it with , .a = whatever, .b = whateverelse.
Those 0, 0, { 0, 0 } are useless, that is the default behavior, if the
compound literal is just (struct bar){}, it will be initialized to all
zeros, and (struct bar){ .a = 1 } will initialize just the designated
elements.  And you won't get any warnings for that (at least not those in
-Wall/-Wextra).

Or use those quietinit-like macros as toplevel statements only and pass
to the macro also a name of the pointer you assign the address to, so in the
end it will be:
  _Pragma("GCC diagnostic push")
  _Pragma("GCC diagnostic ignored \"-Woverride-init\"")
  void *x = &(struct bar){ 0, .a = 0 };
  _Pragma("GCC diagnostic pop")

Or use the no pragmas quietinit and add another set of macros,
quietinit_start/quietinit_end or whatever, define those macros to empty
on everything but GCC9+, on GCC9+ define those to the starting or ending
pragmas and put those in the source around the whole statements with that,
rather than just in subexpressions.

	Jakub