> On Sat, May 24, 2008 at 09:58:47PM +0100, Jeremy Fitzhardinge wrote:
quoted text
> >
> > Would
> >
> > #define KCONFIG(x) (CONFIG_##x - 0)
> >
> > if (KCONFIG(PREEMPT)) {
> > ...
> > }
> >
> > work?
> 
> $ cat test.c
> #define KCONFIG(x) (CONFIG_##x - 0)
> 
> int main()
> {
> if (KCONFIG(PREEMPT))
> ;
> 
> return 0;
> }
> $ gcc -O2 -Wall test.c
> test.c: In function âmainâ:
> test.c:5: error: âCONFIG_PREEMPTâ undeclared (first use in this function)
> test.c:5: error: (Each undeclared identifier is reported only once
> test.c:5: error: for each function it appears in.)
> $ gcc --version
> gcc (Debian 4.3.0-5) 4.3.1 20080523 (prerelease)
>
> > J
>
> cu
> Adrian

How about this?

#include <stdio.h>

#define CONFIG_FOO	1
/* #define CONFIG_BAR	1 */

#define STR2(x) #x
#define STR(x) STR2(x)

#define PASTE2(x, y) x##y
#define PASTE(x, y) PASTE2(x, y)

#define KCONFIG2(x, y) STR(PASTE(CONFIG_, x))[y]
#define KCONFIG(x) (!((KCONFIG2(x, 0) == 'C') && \
					(KCONFIG2(x, 1) == 'O') && \
					(KCONFIG2(x, 2) == 'N') && \
					(KCONFIG2(x, 3) == 'F') && \
					(KCONFIG2(x, 4) == 'I') && \
					(KCONFIG2(x, 5) == 'G') && \
					(KCONFIG2(x, 6) == '_')))

int main()
{
	if (KCONFIG(FOO)) printf("FOO\n");
	if (KCONFIG(BAR)) printf("BAR\n");
	return 0;
}


It of course assumes that the result of all defined macros you want to test 
for do not start with "CONFIG_", so this doesn't work generally.  However, 
nearly all of the time the arguement to KCONFIG() will be defined to be "1" 
an integer, or some other known string that doesn't begin with "CONFIG_", so 
this trick is fairly robust.

Steven

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