pragma pack in a preprocessing directive

I want to put #pragma pack(n) as a preprocessor define.
eg

Code:

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#define PACK_ON(n)                  _Pragma("pack(n)")
PACK_ON(2)

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I get the compile warning:

Code:

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warning: unknown action 'n' for '#pragma pack' - ignored

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I've also tried this

Code:

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DO_PRAGMA(x)                  _Pragma(#x)
#define PACK_ON(n)                  DO_PRAGMA(pack(n))
PACK_ON(2)

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but get error:

Code:

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_Pragma takes a parenthesized string literal

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How do I get around this ?

Maybe try with concatenation to form the parenthesized string literal:

Code:

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#define PACK_ON(n) _Pragma("pack(" #n ")")

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Or maybe it is necessary to be explicit about the concatenation:

Code:

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#define PACK_ON(n) _Pragma("pack(" ## #n ## ")")

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Thanks for your help:

Code:

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#define PACK_ON(n) _Pragma("pack(" #n ")")
gave error: _Pragma takes a parenthesized string literal

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and

Code:

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#define PACK_ON(n) _Pragma("pack(" ## #n ## ")")
gave error: pasting ""pack("" and ""2"" does not give a valid preprocessing token,

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So still no luck

Like this?

Code:

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#include <stdio.h>

#define str(s) #s
#define PACK(x) _Pragma(str(pack(x)))

PACK(2)
struct foo {
    char a;
    short b;
    int c;
};


int main()
{
    printf("%zd\n",sizeof(struct foo));
    return 0;
}

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great that worked ! many thanks

I couldn't solve this one, either. But once you see the solution, it's "obvious". Macros are weird.