2'complement for 8bit variables

[Ron Eggler]

Hi,

I want to write a a function that takes an 8 bit number and calculates the
2's complement of it and returns a signed character.
What I got so far:
The problem i'm having is, that the multiplication *-1 gives me the wrong
value, assuming i got 20 in num and then make a * -1, I get 236 and I can't
figure out why... anyone?

Thank you!
Ron
Code:

char bintodec(unsigned char bin) {
char num;
num = bin;
printf("0x%x -> 0x%x\n",bin,num);
if (num & 0x80) {
printf("bit 15 set 0x%x!\n",num);
num = ~num;
printf("0x%x\n",num);
num++;
printf("0x%x, %d\n",num,num);
num = num*-1;
printf("%d\n",num);
return num;
}
printf("%d is >=0\n",num);
return 0;
}

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[Jens Thoms Toerring]

Ron Eggler <ronDOT...@tscheemail.com> wrote:
> I want to write a a function that takes an 8 bit number and calculates the
> 2's complement of it and returns a signed character.
> What I got so far:
> The problem i'm having is, that the multiplication *-1 gives me the wrong
> value, assuming i got 20 in num and then make a * -1, I get 236 and I can't
> figure out why... anyone?

236 (0xEC) is the 2's complement of 20 (0x14). Using the '~' ope-
rator and then adding 1 to the result should work everywhere,
for it to work by multiplying by -1 requires that your machine
uses 2's complement internally (but most do so nowadays, I guess).

> Code:

> char bintodec(unsigned char bin) {
> char num;
> num = bin;
> printf("0x%x -> 0x%x\n",bin,num);
> if (num & 0x80) {
> printf("bit 15 set 0x%x!\n",num);
> num = ~num;
> printf("0x%x\n",num);
> num++;
> printf("0x%x, %d\n",num,num);
> num = num*-1;
> printf("%d\n",num);
> return num;
> }
> printf("%d is >=0\n",num);
> return 0;
> }

From your code it's impossible to say what is going on - for 20 (and
eveything below 127) the function always returns 0. And, if you start
doing bit-diffling you should take care to clearly distinguish between
signed and unsigned variables, e.g. whhy is 'num' a char (which could
be a signed or unsigned, that depends on your architecture) when 'bin'
is an unsigned char? For bit-fiddling it's usually better to stick
to unsigned quantities,
Regards, Jens
--
\ Jens Thoms Toerring ___ j...@toerring.de
\__________________________ http://toerring.de

[Ron Eggler]

Jens Thoms Toerring wrote:

> Ron Eggler <ronDOT...@tscheemail.com> wrote:
>> I want to write a a function that takes an 8 bit number and calculates
>> the 2's complement of it and returns a signed character.
>> What I got so far:
>> The problem i'm having is, that the multiplication *-1 gives me the wrong
>> value, assuming i got 20 in num and then make a * -1, I get 236 and I
>> can't figure out why... anyone?
>
> 236 (0xEC) is the 2's complement of 20 (0x14). Using the '~' ope-
> rator and then adding 1 to the result should work everywhere,
> for it to work by multiplying by -1 requires that your machine
> uses 2's complement internally (but most do so nowadays, I guess).

So 0xEC would be -20, right? I'm confdused...
how do I get -20 out of it?? :o
Thanks!
Ron

>
>> Code:
>
>> char bintodec(unsigned char bin) {
>> char num;
>> num = bin;
>> printf("0x%x -> 0x%x\n",bin,num);
>> if (num & 0x80) {
>> printf("bit 15 set 0x%x!\n",num);
>> num = ~num;
>> printf("0x%x\n",num);
>> num++;
>> printf("0x%x, %d\n",num,num);
>> num = num*-1;
>> printf("%d\n",num);
>> return num;
>> }
>> printf("%d is >=0\n",num);
>> return 0;
>> }
>
> From your code it's impossible to say what is going on - for 20 (and
> eveything below 127) the function always returns 0. And, if you start
> doing bit-diffling you should take care to clearly distinguish between
> signed and unsigned variables, e.g. whhy is 'num' a char (which could
> be a signed or unsigned, that depends on your architecture) when 'bin'
> is an unsigned char? For bit-fiddling it's usually better to stick
> to unsigned quantities,
> Regards, Jens

[Jens Thoms Toerring]

Ron Eggler <ronDOT...@tscheemail.com> wrote:
> Jens Thoms Toerring wrote:

> > Ron Eggler <ronDOT...@tscheemail.com> wrote:
> >> I want to write a a function that takes an 8 bit number and calculates
> >> the 2's complement of it and returns a signed character.
> >> What I got so far:
> >> The problem i'm having is, that the multiplication *-1 gives me the wrong
> >> value, assuming i got 20 in num and then make a * -1, I get 236 and I
> >> can't figure out why... anyone?
> >
> > 236 (0xEC) is the 2's complement of 20 (0x14). Using the '~' ope-
> > rator and then adding 1 to the result should work everywhere,
> > for it to work by multiplying by -1 requires that your machine
> > uses 2's complement internally (but most do so nowadays, I guess).

> So 0xEC would be -20, right? I'm confdused...
> how do I get -20 out of it?? :o

0xEC would be -20 if it is interpreted as a signed int.
Question is how you intent to print it out. Here's some
way it should work

#include <stdio.h>

char bintodec( unsigned char in ) {
return ~in + 1;
}

int main( void ) {
printf( "%d\n", ( signed char ) bintodec( 20 ) );
return 0;
}

The cast to signed char in the printf() call may be superfluous
on your system if a char is signed by default but on those were
char is unsigned it makes all the difference;-) And then, accor-
ding to the C rules for promotions the signed char is (silenty)
converted to an int and that is printed. Of course, on a machine
that doesn't use 2's complement the output wouldn't be -20 since
then using '~' and adding 1 doesn't "negate" the value, one a
1's complement machine you would have to drop adding 1 to get
the negative of the input value...

[jgharston]

Ron Eggler wrote:
> So 0xEC would be -20, right? I'm confdused...
> how do I get -20 out of it?? :o

EC -20
ED -19
EE -18
EF -17
F0 -16
F1 -15
F2 -14
F3 -13
F4 -12
F5 -11
F6 -10
F7 -9
F8 -8
F9 -7
FA -6
FB -5
FC -4
FD -3
FE -2
FF -1

JGH

[Wayne C. Morris]

In article <jpm4bt$1d0b$4...@adenine.netfront.net>,

Ron Eggler <ronDOT...@tscheemail.com> wrote:

> Jens Thoms Toerring wrote:
>
> > Ron Eggler <ronDOT...@tscheemail.com> wrote:
> >> I want to write a a function that takes an 8 bit number and calculates
> >> the 2's complement of it and returns a signed character.
> >> What I got so far:
> >> The problem i'm having is, that the multiplication *-1 gives me the wrong
> >> value, assuming i got 20 in num and then make a * -1, I get 236 and I
> >> can't figure out why... anyone?
> >
> > 236 (0xEC) is the 2's complement of 20 (0x14). Using the '~' ope-
> > rator and then adding 1 to the result should work everywhere,
> > for it to work by multiplying by -1 requires that your machine
> > uses 2's complement internally (but most do so nowadays, I guess).
>
> So 0xEC would be -20, right? I'm confdused...
> how do I get -20 out of it?? :o
> Thanks!
> Ron

char can be either signed or unsigned; it's implementation-defined. Your
compiler treats char as unsigned, so ((char) 0xEC) is 236. If you want signed
characters, you have to use signed char instead of char.

[Ike Naar]

On 2012-05-24, Ron Eggler <ronDOT...@tscheemail.com> wrote:
> char bintodec(unsigned char bin) {
> char num;

If you want signed char, use signed char, not char.
Plain char can be either signed or unsigned, depending on
the implementation you're using.

> num = bin;
> printf("0x%x -> 0x%x\n",bin,num);
> if (num & 0x80) {
> printf("bit 15 set 0x%x!\n",num);

Bit 15 or bit 7 ?

> num = ~num;
> printf("0x%x\n",num);
> num++;
> printf("0x%x, %d\n",num,num);
> num = num*-1;
> printf("%d\n",num);
> return num;
> }
> printf("%d is >=0\n",num);
> return 0;

Why 0 ?

> }