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From: Les Cargill <lcargill AT worldnet DOT att DOT net>
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Newsgroups: comp.os.msdos.djgpp
Subject: Re: DJGPP: #define problems
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Date: Thu, 27 Dec 2001 22:54:09 GMT
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Matt wrote:
> 
> Hi All,
> 
> I have am having a problem with the way #define's are being handled. I
> have some sample code to demonstrate the problem.
> 
> I have commented the source to explain the problem.
> 
> Basically when i do some math with the #define's the compiler looks
> like it is selecting the wrong data type.
> 
> I have upgraded from 2.954 binaries to v3.03 binaries but this didn't
> fix the problem.
> 
> If anyone has a suggestion i would appreciate it.
> 
> Regards,
> Matt
> (matt AT mattshouseofpain DOT com)
> 
> /**START**/
> #include <stdio.h>
> #define BASEVAL1    233
> #define BASEVAL2    BASEVAL1 + 29
> #define BASEVAL3    233
> #define BASEVAL4    BASEVAL1 + 57
> 
> int main(void)
> {
>    printf("baseval1 [233] = %d\n", BASEVAL1);
>    printf("baseval2 [262] = %d\n\n", BASEVAL2);
>    printf("baseval3 [233] = %d\n", BASEVAL3);
>    printf("baseval4 [290] = %d\n\n", BASEVAL4);
> 
>    /* up to this point the define's are ok but when the math is done
> on
>       the next line it seems to be treating the result as an unsigned
> char
>       data type. */
>    printf("baseval4 - baseval2 [28] = %d\n\n", BASEVAL4 - BASEVAL2);
> 
>    /* here is a sanity check */
>    printf("290 - 262 [28] = %d\n", 290 - 262);
> 
>    return(0);
> }
> /**END**/


If you'll parenthesize the right-hand side of the #defines for BASEVAL2 and
BASEVAL4,
you'll get the expected results:

#define BASEVAL2    (BASEVAL1 + 29)
...
#define BASEVAL4    (BASEVAL1 + 57)

The macro preprocessor simply substitutes the text "33 + 29" when it sees
BASEVAL2. It
doesn't group them together into a single quantity. 

So "printf("baseval4 - baseval2 [28] = %d\n\n", BASEVAL4 - BASEVAL2);" becomes
"printf("baseval4 - baseval2 [28] = %d\n\n", 33 + 57 - 233 + 29);

--
http://home.att.net/~lcargill