macro definition containing #pragma - c++

Question

I am trying to define the following macro:
#if defined(_MSC_VER)
#define PRAGMA_PACK_PUSH(n) __pragma(pack(push, n))
#define PRAGMA_PACK_POP() __pragma(pack(pop))
#else
#define PRAGMA_PACK_PUSH(n) #pragma (pack(push, n))
#define PRAGMA_PACK_POP() #pragma (pack(pop))
#endif
But i get the following error on Linux -
error: '#' is not followed by a macro parameter
#define PRAGMA_PACK_PUSH(n) #pragma (pack(push, n))
and it points to the first ')' in the statment
How can i define a macro that contains a #?
Solution Update:
As stated in this thread Pragma in define macro the syntax that worked is:
#if defined(_MSC_VER)
#define PRAGMA_PACK_PUSH(n) __pragma(pack(push, n))
#define PRAGMA_PACK_POP() __pragma(pack(pop))
#else
#define PRAGMA_PACK_PUSH(n) _Pragma("pack(push, n)")
#define PRAGMA_PACK_POP() _Pragma("pack(pop)")
#endif

Answer 1

How can i define a macro that contains a #?
You can't (define a macro that contains a directive, that is. # can still be used in macros for stringization and as ## for token concatenation). That's why _Pragma was invented and standardized in C99. As for C++, it's definitely in the C++11 standard and presumably the later ones.
You can use it as follows:
#define PRAGMA(X) _Pragma(#X)
#define PRAGMA_PACK_PUSH(n) PRAGMA(pack(push,n))
#define PRAGMA_PACK_POP() PRAGMA(pack(pop))
With that,
PRAGMA_PACK_PUSH(1)
struct x{
int i;
double d;
};
PRAGMA_PACK_POP()
preprocesses to
# 10 "pack.c"
#pragma pack(push,1)
# 10 "pack.c"
struct x{
int i;
double d;
};
# 15 "pack.c"
#pragma pack(pop)
# 15 "pack.c"
As you can see, the _Pragmas are expanding to #pragma directives.
Since _Pragma is standard, you should be able to avoid the #ifdef here if Microsoft supports it.