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

["jquinsey at entrenet dot com" <gcc-bugzilla at gcc dot gnu dot org>]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=78657

            Bug ID: 78657
           Summary: Using macro with _Pragma gives error: '#pragma' is not
                    allowed here
           Product: gcc
           Version: 5.4.0
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: c
          Assignee: unassigned at gcc dot gnu.org
          Reporter: jquinsey at entrenet dot com
  Target Milestone: ---

The following two-line source file test.c:

   #define FOO _Pragma("GCC diagnostic push") 42
   int foo = FOO;

gives:

gcc test.c -E
# 1 "test.c"
# 1 "<built-in>"
# 1 "<command-line>"
# 1 "test.c"

int foo =
# 2 "test.c"
#pragma GCC diagnostic push
# 2 "test.c"
 42;

gcc test.c   
test.c:2:1: error: expected expression before '#pragma'
 int foo = FOO; 
 ^

g++ -x c++ test.c
test.c:2:1: error: '#pragma' is not allowed here
 int foo = FOO; 
 ^

This appears to be consistent across many versions of gcc and g++. For example,
5.4.0 from Cygwin, 5.1 from this test sample at http://ideone.com/6unk5c, and
versions 4.4 through to 6.2 and 7 at https://godbolt.org/.

The icc and clang compilers (from 3.2 upwards) seem handle this correctly.

The text within the _Pragma doesn't seem relevant. Reversing the order of the
two parts of the macro gives a slightly different error message for C:

    error: expected ',' or ';' before '#pragma'