Date: Wed, 3 Feb 1999 23:48:23 +0100
From: anarko <anarko AT flashback DOT net>
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To: Eli Zaretskii <djgpp AT delorie DOT com>
Subject: Re[2]: what is wrong with this?
References: <Pine DOT SUN DOT 3 DOT 91 DOT 990203185620 DOT 21364D-100000 AT is>
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Reply-To: djgpp AT delorie DOT com

Hi Eli, onsdag, den 3 februari 1999, you wrote:


> You forget that C promotes all integral data types to int's in an
> expression.  In other words, ((tmp << 2) >> 2) is a no-op, since tmp is
> loaded into a 32-bit register, where the high two bits aren't shifted out
> by the innermost parentheses.  In the first example, the 2 high bits are
> chopped off when you store the result of (tmp << 2) into an 8-bit
> variable.

yeah, i solved the problem later on, guess i was to quick to ask for
help, thanks anyway :-)

> Btw, to reset bit 6, which is what you seem to want, you don't need to go
> to such great lengths.  All you need is this:

>         bittmp = tmp & 0xCF;

hm, this is not true,
0xCF = 11001111 in binary,
i don't see how this could reset bit 6,
in fact it doesnt,
for example
0xFF & 0xCF = 11001111
0x10 & 0xCF = 00000000

well, i got the function working at least, here it is:

#define RBIT(c,b) ((uchar) (c >> (1+b)) << (1+b)) | ((uchar) (c << (8-b)) >> (8-b))
#define SBIT(c,b,b2) (((uchar) (c << (8-b))) >> (8-b)) | (((uchar) (c >> (1+b))) << (1+b)) | b2

RBIT(c,b)    resets bit 'b' in char c
SBIT(c,b,b2) sets bit 'b' in char c, need to OR it with b2 also, wich
             is 0x1 for bit 0, 0x2 for bit 1, 0x4 for bit 2 etc..
             i'm sure it can be solved more easy, but i'm not that
             talent in math so this is what i managed to code :-)

/anarko <anarko AT flashback DOT net>

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