macro definition containing #pragma

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

VS can handle `#pragma` syntax just fine, why do you use `__pragma` one? Just use `#pragma` everywhere. — user7860670, Jul 16 '17 at 16:17
@aschepler There is no point in defining this macro if in both cases `#pragma` syntax woks. — user7860670, Jul 16 '17 at 16:45
@VTT the defines are here because it's a pain to write _pragma(pack(push, n)) or __pragma(pack(pop)) over and over again. It's used for shortcut purposes — darkThoughts, Jul 16 '17 at 16:55
@darkThoughts These macros are shorter only by 6 symbols, it doesn't seem to save that much, if any, considering that you will need to #include header where these macros are defined. — user7860670, Jul 16 '17 at 17:08

Petr Skocik · Answer 1 · 2017-07-16T16:32:46.297

7

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.

edited Jul 16 '17 at 16:32

answered Jul 16 '17 at 16:11

Petr Skocik

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VS throws the error "this declaration has no storage class or type specifier" when using the proposed syntax. I have found a solution and will update my post accordingly – darkThoughts Jul 16 '17 at 16:51
@darkThoughts With your solution, the `n` argument will be just the letter `n`. Its value won't get interpolated into the `pragma`. – Petr Skocik Jul 16 '17 at 16:56
So at least in Linux where it's supported i should use your solution? – darkThoughts Jul 16 '17 at 17:00
@darkThoughts My solution should work on every standard-compliant compiler. It definitely does work on gcc/linux and clang/linux. – Petr Skocik Jul 16 '17 at 17:03

macro definition containing #pragma

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