On Thu, 2007-05-24 at 17:07 -0500, Mike McCarty wrote:
> Les wrote:
> > On Thu, 2007-05-24 at 11:38 +0100, Alan Cox wrote:
> >
> >>>The DEC KAL10 stored 5 characters in a 36 bit word with one bit
> >>>left over for anything you wanted to do with it! It used 7 bit ASCII,
> >>>of course.
> >>
> >>Or 6 BCD characters or 4 8 (as 9) bit ASCII values. Ditto the Honeywell
> >>L66. Character width as a compile option.
> >>
> >>Be very glad that today you don't have to fight that kind of thing. Don't
> >>however assume in portable code that int is 32bits, long is 32bits,
> >>pointers are 32bits and so on. C99 has proper types when you need to be
> >>specific.
> >>
> >>Also be aware that some Linux ports have char as unsigned by default and
> >>others as signed.
> >
> > Hi, Alan,
> > I haven't seen a signed char since the 80's. I did run into it once
> > on I think a ColecoVision package (don't ask!).
>
> $ cat char.c
> #include <stdio.h>
>
> int main(void) {
> char Chr;
>
> Chr = -1;
> printf("%d\n",Chr);
> return 0;
> }
>
> $ gcc -o char char.c
> $ ./char
> -1
> $ uname -a
> Linux Presario-1 2.6.10-1.771_FC2 #1 Mon Mar 28 00:50:14 EST 2005 i686
> i686 i386 GNU/Linux
>
> So, on FC2, char is signed by default. I suspect that to be true
> for all the Fedora Core releases.
>
> > That would really mess up any algorithm parsing into memory, and
> > would make dealing with solving some kinds of memory issues very
> > difficult, I would think, although
>
> I suspect you are conflating "signed char" with "a character in the
> execution environment which, when its code is stored into a signed char,
> results in a negative value."
>
> Mike
> --
> p="p=%c%s%c;main(){printf(p,34,p,34);}";main(){printf(p,34,p,34);}
> Oppose globalization and One World Governments like the UN.
> This message made from 100% recycled bits.
> You have found the bank of Larn.
> I can explain it for you, but I can't understand it for you.
> I speak only for myself, and I am unanimous in that!
>
Hi, Mike,
when you passed the char to the printf using %d, it was promoted to
an int which is signed. That promotion will automatically extend the
MSB, which in this case was a one (-1 decimal as a char is 0xff, and
extended is 0xffffffff expressed as an int, which prints as -1.
To find the true state of the char, you have to get very creative.
For example:
unsigned int x;
char y=-1;
char *z;
main()
{
z=&x+3;
*z=y;
printf ("%0x\n",x); /* if all went well, and if a signed
value was assigned to the char, you should see 0xff */
/* now to prove the situation you saw */
printf ("%0x\n", y); /* note that the msb propagated and you
now see 0xffffffff */
/* what happened is that regardless of Y, the sign is propagated
when the sign is extended */
/* now to show the effect another way */
{
unsigned char b=0x10; /* I have forced b to be unsigned */
printf ("%0x\n",b); /* if the compiler did not handle the
unsigned as truely unsigned, the bit was extended and you get 0xffffff80
*/
printf ("%0x\n",(unsigend int) b); /* now if everything
worked as it should you should see 0x00000080 */
printf ("%d\n", b); /* what do you see here? */
printf ("%d\n", (unsigned int) b); /* and now you should
see the number 128 */
/* isn't printf fun, also bits/bytes/ints and signs */
}
Note that I did not try these. I don't have my environment set up
yet.
Regards,
Les H
