[Bug c/78657] Using macro with _Pragma gives error: '#pragma' is not allowed here

2016-12-06 Thread jquinsey at entrenet dot com
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=78657 --- Comment #2 from Joseph Quinsey --- The code int foo = #pragma GCC diagnostic push 42; gives exactly the same error messages. So this report is not about _Pragma per se, but about #pragma. But I cannot see where or if the standard

[Bug c/78657] Using macro with _Pragma gives error: '#pragma' is not allowed here

2016-12-06 Thread rguenth at gcc dot gnu.org
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=78657 --- Comment #1 from Richard Biener --- This isn't how preprocessing works I think (thus INVALID)